Experiments with summarise(); or, when does x=x[[1]]?

Here’s another innocent-eye exploration, this time about the Tidyverse’s summarise() function. I’d been combining data tables by nesting and joining them, which gave me a tibble with nested tibbles in. I wanted to check the sizes of these inner tibbles, by mapping nrow() over the columns containing them. The Tidyverse provides several ways to do this, one of which (I thought) would be summarise(). So I tried calling it with the argument s=nrow(tibbles), where tibbles was the column with the tibbles in. It crashed. Why? And how should I make it work?

The insight I got from these experiments is that summarise() passes its summarising functions a slice of a column, not an element. To illustrate the difference in meaning: a slice of a list is a list, possibly with fewer elements; a slice of a table is a table, possibly with fewer rows. But an element of a list is not a list, and an element of a table is not a table. I’d overlooked the distinction because I’m used to table columns being atomic vectors. In these, there’s no difference between an element of the vector and a slice. This is because R is strange, and regards all numbers and other primitive values as one-element vectors.

But when the columns are lists, there is a difference. The summarising functions get passed a list containing the element rather than the element itself, so they have to unwrap it. And, by the way, if they return a result that’s to go into a list column, they must wrap it. That’s important if I want them to return tibbles.

With that as my introduction, here is my code, with comments explaining what I was trying.

# try_summarise.R
#
# Some experiments with summarise(),
# to work out why it didn't seem
# to work when applied to columns
# that are lists of tibbles. 

library( tidyverse )

library( stringr )
# For string_c() .


t <-
  tribble( ~a, ~b   , ~c  ,   ~d   , ~e         , ~f
         , 3 , FALSE, "AA", c(1,11), list(x=1)  , tibble(x=c(1))
         , 1 , TRUE , "B" , c(2,22), list()     , tibble(y=c(1,2))
         , 2 , TRUE , "CC", c(3,33), list(1,2,3), tibble(x=1,y=2)
         )

summarise( t, s=min(a) )
#
#   A tibble: 1 x 1
#         s
#     <dbl>
#   1     1
# So a tibble with one element,
# 1.

summarise( t, s=max(a) )
#
# A tibble with one element,
# 3.  

summarise( t, s=mean(a) )
#
# A tibble with one element,
# 2.  

summarise( t, s=str_c(c) )
#
# Gives an error:
#   Error in summarise_impl(.data, dots) : 
#   Column `s` must be length 1 (a summary value), not 3
 
summarise( t, s=str_c(c,collapse='') )
#
#
# A tibble with one element,
# 'AABCC'. 

summarise( t, s=any(b) )
#
#
# A tibble with one element,
# TRUE.

summarise( t, s=all(b) )
#
#
# A tibble with one element,
# FALSE.

summarise( t, s=show(a) )
#   [1] 3 1 2
# Then gives error:
#   Error in summarise_impl(.data, dots) : 
#   Column `s` is of unsupported type NULL

# The above all show that the
# entire column (t$a or t$b or t$c)
# gets passed to the expression
# after =. If that expression can't
# reduce it to a single atomic
# value, we get an error.


# This was confirmed by the 
#   summarise( t, s=show(a) )
# and also by the two below.

summarise( t, s=identity(a) )
# Gives error:
#   Error in summarise_impl(.data, dots) : 
#   Column `s` must be length 1 (a summary value), not 3

summarise( t, s=a[[1]] )
#
# A tibble with one element,
# 3.

summarise( t, s=a[[3]] )
#
# A tibble with one element,
# 2.

# These are consistent with the
# entire column t$a , which is
#   c(3,1,2)
# being passed.


# What happens if I group by a?

t %>% group_by(a) %>% summarise( s=min(a) )
#
#   A tibble: 3 x 2
#         a     s
#     <dbl> <dbl>
#   1     1     1
#   2     2     2
#   3     3     3
# So now I get a tibble with
# as many rows as t has. 

t %>% group_by(a) %>% summarise( s=show(a) )
#
# Shows 1 and then gives an error.

# So now, t is sliced into three
# groups. There are three calls to
# the expression after =, and in
# each, the appropriate slice of t$a
# is substituted for 'a' in the expression.


# Let's confirm this.

t %>% group_by(a) %>% summarise( s=nchar(c) )
#
# Gives a tibble whose single column
# s is three elements, 1 2 2.
# These are the lengths of the elements
# of t$c .


# Does this work with list columns?
# That's where I had trouble, and is
# what inspired me to try these
# calls.

t %>% group_by(a) %>% summarise( s=length(e) )
#
#   A tibble: 3 x 2
#   a     s
#   <dbl> <int>
#   1     1     1
#   2     2     1
#   3     3     1
# So here I get lengths of 1. But
# I'd expect 0, 3, and 1.

t %>% group_by(a) %>% summarise( s=nrow(f) )
#
# And here, I get an error:
#   Error in summarise_impl(.data, dots) : 
#   Column `s` is of unsupported type NULL


# Why don't these work? The calls
#   summarise( s=length(e) )
#   summarise( s=nrow(f) )
# are surely analogous to
#   summarise( s=nchar(c) ) .

# Epiphany! The reason, I realise, is
# that in each of the expressions after =,
# the column variable is getting
# substituted for by a single-row
# slice of that column. (This is in
# my grouped examples. In the others,
# it gets substituted for by the entire
# column.)

# When the columns are atomic vectors,
# this single-row slice is an atomic
# vector too, with just one element.
# But in R, these get treated like
# single numbers, strings or Booleans.
# So the call to nchar() gets passed
# a vector containing a single string
# element of column c. It works, because
# such single elements are vectors anyway.

# But when the columns are lists,
# the single-row slice is also a list.
# So nrow(f), for example, gets passed
# not a tibble but a list containing
# a tibble. It then crashes. Similarly,
# length(e) gets passed a list containing
# whichever single slice of column e,
# and always returns 1.  

# I'll confirm this by doing these calls.

summarise( t%>%group_by(a), s=nrow(f[[1]]) )
#
# Works! Returns a tibble with the table
# lengths.

summarise( t%>%group_by(a), s=length(e[[1]]) )
#
# Also works. Returns a tibble with the
# list lengths.


# The key to all this is that if v
# is a length-1 atomic vector, v=v[[1]].
#   > v <- c(1)
#   > v
#   [1] 1
#   > v[[1]]
#   [1] 1

# But if l is a length-1 list, l!=l[[1]]:
#   > l <- list(tibble())
#   > l
#   [[1]]
#   A tibble: 0 x 0
# Whereas:
#   > l[[1]]
#   A tibble: 0 x 0
# The latter loses a level of subscripting.


# By the way, I note that summarise()
# works with more than one column name,
# and more than use of a column name,
# and substitutes them all appropriately.

summarise( t%>%group_by(a), s=a/a )
#
# Returns a tibble with three
# rows, all 1.

summarise( t%>%group_by(a), s=str_c(a,b,c,collapse='') )
#
# Returns a tibble where each
# row is the result of concatenating
# the first three elements in
# the corresponding row of t.


# Finally, I also note that if
# I want to _return_ a tibble
# from a summary function, I have
# to wrap it in a list.

summarise( t%>%group_by(a), s=tibble(x=a) )
#
# Runs, but doesn't do what I wanted.
# (Puts a double into each row.)

summarise( t%>%group_by(a), s=list(tibble(x=a)) )
#
# Runs, and does do what I wanted:
#   A tibble: 3 x 2
#   a                s
#   <dbl>           <list>
#   1     1 <tibble [1 x 1]>
#   2     2 <tibble [1 x 1]>
#   3     3 <tibble [1 x 1]>

# I suppose that this is because 
# the expression in aummarise() has
# to return something that is 
# a row of a column, not an element.
# If the column is an atomic vector,
# these are the same, but they aren't
# if it's a list.

My Testing Was Hurt By a List With No Names

I spied a strange thing with my innocent eye. I was testing spread_to_list, the function I wrote about in “How Best to Convert a Names-Values Tibble to a Named List?”. One of my tests passed it a zero-row tibble, expecting that the result would be a zero-element list:

test_that( "Test on zero-element tibble", {
  t <- tribble( ~a, ~b )
  l <- spread_to_list( t )
  expect_that( length( l ), equals( 0 ) )
  expect_that( l, is_identical_to( list() ) )
})
But it wasn’t. Or at least, it was, but with an unsuspected excrescence.

The package I used to run that test is Hadley Wickham’s testthat, explained in his 2011 article “testthat: Get Started with Testing”. Even if you don’t know it, you can predict from his well-chosen function names what two things I expected. The first thing, that the result be length zero, came true. The second thing, that it be an empty list, didn’t.

Output from the test gave me a hint:

Error: Test failed: 'Test on zero-element tibble'
* `l` not identical to list().
Attributes: < Modes: list, NULL >
Attributes: < Lengths: 1, 0 >
Attributes: < names for target but not for current >
Attributes: < current is not list-like >

Displaying l and a list() told me the rest:

> l
named list()
> list()
list()
R evidently thinks a named list is not identical to a list, even when the set of names is empty. I don’t know whether this is reasonable or not. An empty set is different from no set. On the other hand, a dictionary which just happens to have no definitions in it is still a dictionary. The R manual’s section on object attribute lists implies that a list’s names are an attribute. Let’s see how that behaves:
> attributes(l)
$names
character(0)

> attributes(list())
NULL
So a list can either have no names and no box to put them in, or the box with nothing inside.

Multinest: Factoring Out Hierarchical Names by Converting Tibbles to Nested Named Lists

Consider this tibble:

vat_rates <- tribble( ~l1,   ~l2         , ~l3         , ~val
                    , 'VAT', 'Cigarettes', '' 	       , 'standard'
                    , 'VAT', 'Tobacco'   , ''          , 'standard'
                    , 'VAT', 'Narcotics' , ''	       , 0
                    , 'VAT', 'Clothing'  , 'Adult'     , 'standard'
                    , 'VAT', 'Clothing'  , 'Children'  , 0
                    , 'VAT', 'Clothing'  , 'Protective', 0
                    )
It contains three name columns followed by a value. The name columns are hierarchical. The top category is VAT (Value Added Tax). The second column subdivides this into VAT on cigarettes, other tobacco, narcotics, and clothing. And the third column further subdivides clothing into adult, children’s, or protective. This is part of the VAT parameters worksheet that provides our economic model with its knowledge of the British tax system. Here’s the same fragment as a screen shot:
I want to convert this into a tree-structured named list in which I can access parameter values by multiple selection. That is, I want to be able to ask for params$VAT$Tobacco and get 'standard', or for params$VAT$Clothing$Children and get 0. This is how I did so.

I’m going to show my code below. I’ve annotated it in the same way I did in my previous posts, so all the explanation is in the code as comments. What I did was to write a function called multinest, correcting it a step at a time as I found cases it couldn’t handle. I built it up that way because I found it easier to explain. It seemed to be easier for the reader (or at least, for me acting as reader) to visualise than presenting a finished version of the function immediately and explaining how its recursion and base cases worked. I’m not sure why I found the latter harder, but it’s something to do with the fact that nest transforms an entire table in one go. If you then want to perform an inner transformation on the nested tables it inserts, you have to loop or recurse over them. That’s different from how a more traditional algorithm would work. Anyway, here’s my code. The final version of my function is at the end, multinest_4.

# try_multinest_4.R
#
# This is a version of multinest which
# I hope is easier to understand than
# that in try_multinest_3.R . I built
# it by successive enhancement, starting
# with nest() and adding one correction
# at a time.


library( tidyverse )


t <- tribble( ~l1, ~l2, ~val
            , 'A', 'a', 'Aa'
            , 'B', 'a', 'Ba'
            , 'A', 'b', 'Ab'
            , 'B', 'b', 'Bb'
            )
#
# I will try my functions on this
# tibble. I want to write a function
# that factors out the two name
# columns l1 and l2, converting it
# to the nested named list
#   list( A = list( a = "Aa", b = "Ab" )
#       , B = list( a = "Ba" ,b = "Bb" )
#       )
# So if this list is l, l$A$a must be
# "Aa", and l$B$b must be "Bb", etc.
     

# First, let's try nest() interactively.

t_nested <- t %>% group_by( l1 ) %>%
                 nest
#
# Gives a tibble where the second
# and third columns of t have been
# collapsed to inner, nested, tibbles:
#   # A tibble: 2 x 2
#        l1             data
#     <chr>           <list>
#   1     A <tibble [2 x 2]>
#   2     B <tibble [2 x 2]>

t_nested[[1,2]]
#
#   # A tibble: 2 x 2
#        l2   val
#     <chr> <chr>
#   1     a    Aa
#   2     b    Ab
# So the data column in t_nested
# is indeed tibbles.


# Now I want to "spread" t_nested, 
# flipping it into a list where the column-1
# elements "A" and "B" become the list's
# names, and the tibbles in column 2
# becomes the values associated with those
# names.

# In my blog post 
# http://www.j-paine.org/blog/2017/10/how-best-to-convert-a-names-values-tibble-to-a-named-list.html 
# I tried functions for doing this. I could
# use spread(). But it turns out I don't need
# to. setNames() can do the job much more 
# efficiently. Here's my function. See
# the post for why it's as it is.

spread_to_list <- function( t )
{
  setNames( as.list( t[[2]] ), t[[1]] ) 
}


# Let's try it.

t_nested_spread <- spread_to_list( t_nested )
#
#   $A
#   # A tibble: 2 x 2
#        l2   val
#     <chr> <chr>
#   1     a    Aa
#   2     b    Ab
#
#   $B
#   # A tibble: 2 x 2
#        l2   val
#     <chr> <chr>
#   1     a    Ba
#   2     b    Bb
# So that does what I wanted.
# Let's check by displaying one
# of the elements.

t_nested_spread$A
#
#   #  A tibble: 2 x 2
#        l2   val
#     <chr> <chr>
#   1     a    Aa
#   2     b    Ab
# Good.

# So we can convert the outer level of
# a table t to a named list whose keys are
# the names in t[,1].

# I should note that spread_to_list() 
# is safe because nest() will ensure 
# that spread_to_list()'s argument 
# never has a name more than
# once in column 1.


# Let's make this into a function.

multinest <- function( t )
{
  colname_1 <- as.name( names(t)[1] )

  t %>% group_by( !!colname_1 ) %>%
        nest %>%
        spread_to_list    
}

t_nested_spread_ <- multinest( t )
identical( t_nested_spread, t_nested_spread_ )
#
# Which gives the same result as above,
# so multinest() appears OK.


# Now I want to make multinest() recursive so
# that it factors out the inner tables
# in the same way. I'll assume all the name
# columns are at the front of the table. I
# do have to give an extra argument saying
# how many there are.

multinest_2 <- function( t, no_of_name_cols )
{
  colname_1 <- as.name( names(t)[1] )
  #
  # For as.name and !! , see e.g. Henrik's
  # answer to https://stackoverflow.com/questions/26724124/standard-evaluation-in-dplyr-summarise-on-variable-given-as-a-character-string ,
  # "[u]se as.name if you have a character 
  # string that gives a variable name".

  listed <- t %>% group_by( !!colname_1 ) %>%
                  nest %>%
                  spread_to_list    

  if ( no_of_name_cols > 1 ) 
    return( map( listed, multinest_2, no_of_name_cols-1 ) )
  else
    return( listed )
}

identical( multinest_2( t, 1 ), t_nested_spread )
#
# So this function called for just one column
# works as before.


# Let's see what happens with two columns.

t_nested_spread_2 <- multinest_2( t, 2 )

for ( X in c('A','B') ) 
  for( Y in c('a','b') ) 
    cat( t_nested_spread_2[[X]][[Y]] ) 
#
# But ...
#   Error in cat(listed_2[[X]][[Y]]) : 
#   argument 1 (type 'list') cannot be handled by 'cat'

t_nested_spread_2$A$a
#   # A tibble: 1 x 1
#       val
#     <chr>
#   1    Aa

# The innermost list elements are tibbles, not
# simple values. That would be OK if I wanted to
# carry more than one value cell with each
# innermost element, but it wastes space when
# I have only one. What's causing this is the
# inner call to nest(), which is converting
# a tibble such as 
#   # A tibble: 2 x 2
#        l2   val
#     <chr> <chr>
#   1     a    Aa
#   2     b    Ab
# into
#   # A tibble: 2 x 2
#       l2             data
#    <chr>           <list>
#  1     a <tibble [1 x 1]>
#  2     b <tibble [1 x 1]>


# Let's rewrite to avoid this. If there's
# only one column left, just convert to list
# without calling nest().

multinest_3 <- function( t, no_of_name_cols )
{
  colname_1 <- as.name( names(t)[1] )
 
  if ( no_of_name_cols > 1 )
    listed <- t %>% group_by( !!colname_1 ) %>%
                    nest %>%
                    spread_to_list %>% 
                    map( multinest_3, no_of_name_cols-1 )
  else if ( no_of_name_cols == 1 )
    listed <- t %>% spread_to_list

  listed
}

identical( multinest_3( t, 1 ), t_nested_listed )
#
# This actually returns FALSE. So asked to nest
# just one column, the function doesn't do the same
# as its predecessors. It returns the list
#   list(A = "a", B = "a", A = "b", B = "b")
# But that's reasonable, because it's treating
# the second column as the values. So I'll not
# amend it.


# What happens with two columns?

t_nested_spread_3 <- multinest_3( t, 2 )

for ( X in c('A','B') ) 
  for( Y in c('a','b') ) 
    cat( t_nested_spread_3[[X]][[Y]] ) 
#
# Displays AaAbBaBb.
# So that looks OK. 

dput(t_nested_spread_3)
#
# So does a dput():
#   structure(list(A = structure(list(a = "Aa", b = "Ab"), .Names = c("a", 
#   "b")), B = structure(list(a = "Ba", b = "Bb"), .Names = c("a", 
#   "b"))), .Names = c("A", "B"))


# Now let's try three columns. 
# I'll reassign t for this.

t <- tribble( ~l1, ~l2, ~l3, ~val
            , 'A', 'a', '1', 'Aa1'
            , 'A', 'a', '2', 'Aa2'
            , 'A', 'b', '1', 'Ab1' 
            , 'A', 'b', '2', 'Ab2'
            , 'B', 'a', '1', 'Ba1'
            , 'B', 'a', '2', 'Ba2'
            , 'B', 'b', '1', 'Bb1' 
            , 'B', 'b', '2', 'Bb2'
            )

t_nested_spread_3 <- multinest_3( t, 3 )

for ( X in c('A','B') ) 
  for( Y in c('a','b') )
    for( Z in c('1','2') ) 
      cat( t_nested_spread_3[[X]][[Y]][[Z]] )
#
# Displays Aa1Aa2Ab1Ab2Ba1Ba2Bb1Bb2. Good.


# Now let's try on some real data, a fragment 
# of our VAT parameters.

vat_rates <- tribble( ~l1,   ~l2         , ~l3         , ~val
                    , 'VAT', 'Cigarettes', '' 	       , 'standard'
                    , 'VAT', 'Tobacco'   , ''          , 'standard'
                    , 'VAT', 'Narcotics' , ''	       , 0
                    , 'VAT', 'Clothing'  , 'Adult'     , 'standard'
                    , 'VAT', 'Clothing'  , 'Children'  , 0
                    , 'VAT', 'Clothing'  , 'Protective', 0
                    )

vat_rates_nested_spread <- multinest_3( vat_rates, 3 )
#
# But this goes awry on the first three table rows,
# where it generates results such as 
#   $VAT$Narcotics
#   $VAT$Narcotics[[1]]
#   [1] "0"
# The value is wrapped in an unwanted list.

# This is because I didn't check for names
# where some of the final components are
# empty. R has come across the '' and done its
# own thing. 


# I need to fix this. I'll do so below. If 
# the table passed to multinest() has '' 
# as its first element, return the final 
# element, the corresponding value. I don't think such a table can 
# ever have more than one row. If it did, it 
# would have been part of a table such as
#   Pars VAT ''  20
#   Pars VAT Low 17
# which would have to translate into the list
#   Pars$VAT <- 20
#   Pars$VAT$Low <- 17
# This would mean that Pars$VAT is 
# simultaneously a terminal and a non-terminal,
# which makes no sense.

multinest_4 <- function( t, no_of_name_cols )
{
  if ( t[[1,1]] == '' )
    return( t[[ 1, ncol(t) ]] )
  else {
    colname_1 <- as.name( names(t)[1] )

    if ( no_of_name_cols > 1 )
      listed <- t %>% group_by( !!colname_1 ) %>%
                      nest %>%
                      spread_to_list %>% 
                      map( multinest_4, no_of_name_cols-1 )
    else if ( no_of_name_cols == 1 )
      listed <- t %>% spread_to_list

    return( listed )
  }
}

vat_rates_nested_spread <- multinest_4( vat_rates, 3 )
#
#   $VAT
#   $VAT$Cigarettes
#   [1] "standard"
#
#   $VAT$Tobacco
#   [1] "standard"
#
#   $VAT$Narcotics
#   [1] "0"
#
#   $VAT$Clothing
#   $VAT$Clothing$Adult
#   [1] "standard"
#
#   $VAT$Clothing$Children
#   [1] "0"
#
#   $VAT$Clothing$Protective
#   [1] "0"
#
# Good.

How Best to Convert a Names-Values Tibble to a Named List?

Here, in the spirit of my “Experiments with by_row()” post, are some experiments in writing and timing a function spread_to_list that converts a two-column tibble such as:

x 1
y 2 
z 3
t 4
to a named list:
list( x=1, y=2, z=3, t=4 )
I need this for processing the parameter sheets shown in that by_row post, and I’ll explain why later. In this post, I’m just interested in how best to define spread_to_list. The best implementation looks like either spread_to_list_3 or spread_to_list_4 below.

# try_spread_to_list.R
#
# Consider a tibble t with two columns,
# where each cell in the second column
# represents the value associated with
# the string (assumed to be its name) 
# in the first column.
#
# I want to define a function spread_to_list() 
# which converts t into a named list whose names
# are the names in the first column, and whose
# values are the values in the second column.
#
# For example, if t is:
#        names no_of_days
#    DaysInMay         31
#   DaysInJune         30
# then
#   spread_to_list(t) 
# would be the list
#   list( DaysInMay = 31, DaysInJune = 30 )
#
# The code here tries various ways
# of implementing spread_to_list,
# and benchmarks them. Two are variants
# of one another, using spread() and converting
# its result. Another two take the values
# column as a list, and call setNames() to 
# convert to a named list.


library( tidyverse )
library( microbenchmark )
library( stringr )


t <- tribble( ~a , ~b 
            , 'x',  1
            , 'y',  2 
            , 'z',  3
            , 't',  4
            )
#
# I'm going to try various ways
# of implementing spread_to_list()
# on t.


# First, what happens if I call spread()?

s <- spread( t, a, b )
#
# s becomes a one-row tibble:
#         t     x     y     z
#   1     4     1     2     3


# How can I convert s to a named list?

# An obvious way is to call map().
# Let's see what the function argument to
# map() gets passed if I call map() on s.

map( s, show )
#
# Displays
#  [1] 4
#  [1] 1
#  [1] 2
#  [1] 3
# So it gets passed a column of the
# tibble as an atomic vector.

map( s, function(x)x )
#
# Returns a list of these elements:
#   list(t = 4, x = 1, y = 2, z = 3)
# That's because map() is defined to return 
# lists. So the call above uses it merely
# as a type converter.

map( s, identity )
#
# Does the same. identity() is a built-in
# identity function.


# But maybe I can avoid mapping. In my
# experiments with by_row(), 
# http://www.j-paine.org/blog/2017/10/experiments-with-by_row.html ,
# I discovered that as.list() will convert
# a tibble to a named list.

as.list( s )
#
# Also returns 
#   list(t = 4, x = 1, y = 2, z = 3).


# But can I avoid spread() altogether?
# Browsing the discussion groups gave me
# the idea of trying setNames().
# Let's try that, passing t's names as its
# second argunent and t's values as its
# first.

setNames( t[[2]], t[[1]] )
# 
# Gives me an atomic named vector. 

# I need to convert it to a list.
# One way is for the argument to setNames()
# to be a list, because that's specified to
# make it return a list.

setNames( as.list( t[[2]] ), t[[1]] )
#
# Returns the list
#   list(x = 1, y = 2, z = 3, t = 4)

# But I could convert the result instead.

as.list( setNames( t[[2]], t[[1]] ) )
#
# Also returns
#   list(x = 1, y = 2, z = 3, t = 4)


# Let's try these four implementations.
# First, define the functions.

spread_to_list_1 <- function( t )
{
  colname1 <- names( t )[[1]]
  colname2 <- names( t )[[2]]
  t %>% 
    spread( !!as.name(colname1), !!as.name(colname2) ) %>% 
    map( identity )
}

spread_to_list_2 <- function( t )
{
  colname1 <- names( t )[[1]]
  colname2 <- names( t )[[2]]
  t %>% 
    spread( !!as.name(colname1), !!as.name(colname2) ) %>% 
    as.list
}

spread_to_list_3 <- function( t )
{
  setNames( as.list( t[[2]] ), t[[1]] )
}

spread_to_list_4 <- function( t )
{
  as.list( setNames( t[[2]], t[[1]] ) )
}


# Now try them.

s1 <- spread_to_list_1( t )
s2 <- spread_to_list_2( t )
s3 <- spread_to_list_3( t )
s4 <- spread_to_list_4( t )

dput( s1 )
dput( s2 )
dput( s3 )
dput( s4 )

identical( s1, s2 )
identical( s1, s3 )
identical( s1, s4 )


# They all return named lists, but the order
# of elements is different for the spread()-based
# versions than from the as.list()-based ones.
# (That was obvious earlier, actually.)
# So I'll sort the lists, then test that
# they're identical. I'll also microbenchmark
# the functions.

sort_list <- function(l)
{ 
  sort( unlist( l ) )
}

identical( sort_list(s1), sort_list(s2) )%>%show
identical( sort_list(s1), sort_list(s3) )%>%show
identical( sort_list(s1), sort_list(s4) )%>%show

mbres <- microbenchmark( spread_to_list_1( t )
                       , spread_to_list_2( t )
                       , spread_to_list_3( t )
                       , spread_to_list_4( t )
                       )
print( mbres )


# Now let's microbenchmark the functions applied
# to bigger tibbles. I'll generate random name-value
# tibbles of sizes n, where n is defined by the
# vector in the 'for' condition.

for ( n in c(10,30,100,300) ) {

  cat( "Trying ", n, "row tibble\n" )

  names <- replicate( n, str_c(sample(letters,5,replace=FALSE),collapse="") )
  #
  # Generate n random alphabetic strings.
  # From Dirk Eddelbuettel's answer to 
  # https://stackoverflow.com/questions/1439513/creating-a-sequential-list-of-letters-with-r .

  values <- runif( n, 1, 100 )
  #
  # Generate n random values.

  t <- tibble( names=names
             , values=values
             )
  #
  # Use these to make a random tibble with
  # two columns and n rows.

  identical( sort_list(s1), sort_list(s2) )%>%show
  identical( sort_list(s1), sort_list(s3) )%>%show
  identical( sort_list(s1), sort_list(s4) )%>%show

  mbres <- microbenchmark( spread_to_list_1( t )
                         , spread_to_list_2( t )
                         , spread_to_list_3( t )
                         , spread_to_list_4( t )
                         )
  print( mbres )
}


# Here are the microbenchmark results for the
# 300-row tibble:
#   Unit: microseconds
#                  expr       min        lq    
#   spread_to_list_1(t) 17738.631 17928.946 
#   spread_to_list_2(t) 14582.521 14805.888 
#   spread_to_list_3(t)    36.223    40.901   
#   spread_to_list_4(t)    35.317    40.146 
#                  expr        mean    median         uq
#   spread_to_list_1(t) 18668.44595 18221.133 19532.8080
#   spread_to_list_2(t) 15386.05835 15050.685 16355.4180
#   spread_to_list_3(t)    46.67244    47.089    51.4655
#   spread_to_list_4(t)    45.77894    45.882    51.6165
#         max neval
#   21477.003   100
#   17314.838   100
#      64.294   100
#      63.087   100

# So the two spread() versions are much slower.
# Converting the spread() result with mapping is
# slower than with as.list(), probably unsurprisingly.
# The two setNames() versions are much faster. 
# It doesn't seem to matter whether we type-convert
# to list by making setNames()'s first argument
# a list, or by making its result one.

Experiments with by_row()

I’ve been experimenting with by_row() from the R purrrlyr package. It’s a function that maps over the rows of a data frame, which I thought might be handy for processing our economic model’s parameter files. I didn’t find the documentation for by_row() told me everything I wanted to know, so I made up and ran some examples. As these might be useful to others, I’m blogging them here.

purrrlyr was written mainly by Hadley Wickham. Regular readers of my posts will know that Hadley wrote the suite of programs known as the Tidyverse, which can be installed with the command install.packages("tidyverse"). However, purrrlyr is not quite part of this, and has to be installed separately, via install.packages("purrrlyr").

Why am I interested in by_rows()? Our economic model reads what we call parameter files. These are spreadsheets describing the taxes and benefits it has to apply, plus various controls. Each spreadsheet contains several worksheets, most of which start with three columns making up a hierarchical parameter name, followed by one or more values. Here’s a fragment from one such sheet:

        Name1              Name2    Name3 Allowance_Spouse PrivatePension BankAccount NationalSavings
1 IncomeLists      IncomeSupport     <NA>                1              1           0               0               
2 IncomeLists      PensionCredit     Main                1              1           0               0               
3 IncomeLists      PensionCredit  Savings                0              1           0               0               
4 IncomeLists NonContributoryESA   Income                1              1           0               0               
5 IncomeLists NonContributoryESA Earnings                0              0           0               0               
6 IncomeLists              HBCTB     <NA>                1              0           0               0               
And here’s a screen shot of it:

I need to process these sheets a row at a time, converting each row into an entry in a lookup table that maps parameter names to their values. Because by_row() maps over rows, it looked worth learning about. So the code below shows the calls I tried. In it, t is the tibble I passed to by_rows(), and tbr is the result of the calls. I’ve commented each call explaining what it tells me, and also listed the things I wanted to know. One was what the documentation meant by saying that the .collate argument makes by_row() collate by row or by column. “Collate” is a very general verb, and didn’t tell me exactly how by_row()‘s results were arranged. I was also puzzled by the .labels argument, in which if TRUE “the returned data frame is prepended with the labels of the slices (the columns in .d used to define the slices)”.

# try_by_row.R
#
# Some experiments with 
# purrrlyr's by_row() function.
# This may be useful when 
# converting segments of tibble
# rows to lists that I can
# store in a single cell.


library( tidyverse )
library( purrrlyr )


t <- tribble( ~a, ~b, ~c
            ,  1,  2,  3
            ,  4,  5,  6
            ,  7,  8,  9
            , 10, 11, 12
            )
#
# I'll try various calls to by_row on this.

# Let's try sum().

tbr <- by_row( t, sum )
#
# tbr becomes t with a .out column added.
#         a     b     c      .out
#   1     1     2     3 
#   2     4     5     6 
#   3     7     8     9 
#   4    10    11    12 
# Each of the 's is a list whose
# single element is the sum of a, b, and c.

tbr <- by_row( t, sum, .collate="rows" )
tbr <- by_row( t, sum, .collate="cols" )
#
# In both of these, tbr becomes a table where
# each .out cell holds a sum rather than a
# list containing the sum:
#   a     b     c        .out
#   1     1     2     3     6
#   2     4     5     6    15
#   3     7     8     9    24
#   4    10    11    12    33

# Let's try mean().

tbr <- by_row( t, mean )
# Warning messages:
# 1: In mean.default(.d[[i]], ...) :
#   argument is not numeric or logical: returning NA
# 2: In mean.default(.d[[i]], ...) :
#   argument is not numeric or logical: returning NA
# 3: In mean.default(.d[[i]], ...) :
#   argument is not numeric or logical: returning NA
# 4: In mean.default(.d[[i]], ...) :
#   argument is not numeric or logical: returning NA

# Let's find out why sum() works but mean() doesn't.

tbr <- by_row( t, show )
#
# I did this to see what by_row()'s function
# argument gets passed. The command above 
# displays four one-row tibbles, each a row of t.
# So that's what the function gets passed.

sum( t[1,] )
#
# 6. So sum() can accept tibbles.

mean( t[1,] )
#   Warning message:
#   In mean.default(t[1, ]) : argument is not numeric 
#   or logical: returning NA
# But mean() can't.

# The above three results show why by_row gave
# an error with mean() but not sum(). It was
# passing one-row tibbles to both. sum() accepts
# these but mean() doesn't.             
    
# Let's see whether I can create a column 
# containing copies of the original rows.

tbr <- by_row( t, function(row)row )
#
# Shows that tbr$.out is a list where
# each element is one of these one-row tibbles.
# So this copies each original row into 
# the corresponding cell of .out .

tbr <- by_row( t, as.list )
#
# This does as above but converts each row to
# a list. So here, 
#   tbr$.out[[1]] 
# is
#   list(a = 1, b = 2, c = 3) .
# and so on. This will be useful when I want to
# convert rows to look-up tables implemented
# as named lists. And I presume it's more
# efficient than storing an entire tibble.

as.list( t[1,] )
#
# Just to show that conversion of one-row
# tibble to named list is what as.list does,
# and not some quirk of by_row(). You never
# know with R.

as.list( t )
#
# I try this for interest. It converts to a
# named list where each element is a 
# column vector. Thus,  
#   as.list( t )
# is
#   list( a = c(1, 4, 7, 10)
#       , b = c(2, 5, 8, 11)
#       , c = c(3, 6, 9, 12)
#       )

# Let's try the .collate argument, because
# I don't understand exactly what the 
# by_row() documentation means by collation.

tbr <- by_row( t, function(row)c(10,20,30)  )
#
# This makes tbr$.out a list column each of
# whose elements is c(10,20,30). So it just 
# puts the entire result of the function 
# into the .out cells.

tbr <- by_row( t, function(row)c(10,20,30), .collate="rows" )
#
# Makes tbr the table
#         a     b     c  .row  .out
#   1     1     2     3     1    10
#   2     1     2     3     1    20
#   3     1     2     3     1    30
#   4     4     5     6     2    10
#   5     4     5     6     2    20
#   6     4     5     6     2    30
#   7     7     8     9     3    10
# etc. So appends one column. But copies
# each original row as many times as there 
# are components in the function's result. 
# The i'th copy's .out cell holds the i'th 
# component. And we get a .row column holding
# the number of the original row.

tbr <- by_row( t, function(row)c(10,20,30), .collate="cols" )
#
# Makes tbr the table
#         a     b     c .out1 .out2 .out3
#   1     1     2     3    10    20    30
#   2     4     5     6    10    20    30
#   3     7     8     9    10    20    30
#   4    10    11    12    10    20    30
# So appends three columns, where the i'th
# column contains the i'th component of the
# function's result.

# Let's see whether .collate can produce
# "Manhattan arrays", where the rows are 
# different lengths.

tbr <- by_row( t, function(row)1:(sum(row)%%5), .collate="cols" )
#
# This gave an error:
#   Error in by_row(t, function(row) 1:(sum(row)%%5), .collate = "cols") : 
#   .f should return equal length vectors or data frames 
#   for collating on `cols`
# My function was intended to return rows of different
# length. But, as the documentation says, they have to
# be the same length.

tbr <- by_row( t, function(row)1:(sum(row)%%5), .collate="rows" )
#
# However, this one does run, producing differing
# numbers of rows for each original row:
#         a     b     c  .row  .out
#   1     1     2     3     1     1
#   2     4     5     6     2     1
#   3     4     5     6     2     0
#   4     7     8     9     3     1
#   5     7     8     9     3     2
#   6     7     8     9     3     3
#   7     7     8     9     3     4
#   8    10    11    12     4     1
#   9    10    11    12     4     2
#  10    10    11    12     4     3

# Let's see what happens with .collate when 
# the function returns a data frame, 
# because the documentation doesn't really 
# explain that.

df <- tribble( ~X, ~Y, "$", "%", "^", "&" )

tbr <- by_row( t, function(row)df )
#
# Each cell of tbr's .out column becomes a copy of
# df.

tbr <- by_row( t, function(row)df, .collate="rows" )
# 
# tbr becomes a table 
#         a     b     c  .row     X     Y
#   1     1     2     3     1     $     %
#   2     1     2     3     1     ^     &
#   3     4     5     6     2     $     %
#   4     4     5     6     2     ^     &
# etc. So we get one copy of each original row
# for each row in df. We get one column for
# each column in df, plus again a .row column 
# holding the number of the original row.

tbr <- by_row( t, function(row)df, .collate="cols" )
#
# I wasn't expecting that. df's cells get
# spread into a single line, thus, and its
# column names suffixed accordingly:
#         a     b     c    X1    X2    Y1    Y2
#   1     1     2     3     $     ^     %     &
#   2     4     5     6     $     ^     %     &
#   3     7     8     9     $     ^     %     &
#   4    10    11    12     $     ^     %     &

# Let's see what the .labels argument does. The
# documentation says: if TRUE, the returned data 
# frame is prepended with the labels of the slices 
# (the columns in .d used to define the slices). 
# But I don't know what it means by slices.

tbr <- by_row( t, sum, .labels=FALSE )
#
# We just get 
#          .out
#   1 <dbl [1]>
#   2 <dbl [1]>
#   3 <dbl [1]>
#   4 <dbl [1]>

tbr <- by_row( t, function(row)df, .collate="rows", .labels=FALSE )
#
# tbr becomes
#      .row     X     Y
#   1     1     $     %
#   2     1     ^     &
#   3     2     $     %
#   4     2     ^     &
# etc.

tbr <- by_row( t, function(row)df, .collate="cols", .labels=FALSE )
#
#        X1    X2    Y1    Y2
#   1     $     ^     %     &
#   2     $     ^     %     &
#   3     $     ^     %     &
#   4     $     ^     %     &

# So .labels=FALSE gives me what I think
# of as a traditional mapping function,
# where by_row() returns only the new
# columns. Except that if .collate="rows",
# it prepends a .row column holding the
# original row number.

Abstract Data Types and the Uniform Referent Principle II: why Douglas T. Ross would hate nest(), unnest(), gather() and spread()

In “Abstract Data Types and the Uniform Referent Principle I: why Douglas T. Ross would hate nest(), unnest(), gather() and spread()”, I explained why the notation for interfacing to a data structure should be independent of that structure’s representation. R programmers honour this principle in the same way that bricks hang in the sky. All published R code that operates on data frames uses column names. Sometimes these follow the $ operator; sometimes the data frame is implicit via attach() or similar. In the Tidyverse, the column names will often be part of a mutate(), the data frame being piped through a sequence of %>% operators. And this is dreadful software engineering.

Why? Look at the tables below. They represent four different ways of storing my income data.

PersonIncome_TypeIncome_Value
AliceWages37000
AliceBonuses0
AliceBenefits0
BobWages14000
BobBonuses1000
BobBenefits6000
PersonIncome_WagesIncome_BonusesIncome_Benefits
Alice370000 0
Bob 1400010006000
PersonIncome
Alice
TypeValue
Wages37000
Bonuses0
Benefits0
Bob
TypeValue
Wages14000
Bonuses1000
Benefits6000
PersonIncome
Alice
WagesBonusesBenefits
3700000
Bob
WagesBonusesBenefits
1400010006000

Abstractly, the data is the same in each case, and if you’re familiar with nest(), unnest(), gather() and spread(), you will easily see how to transform one table into any of the others. But the tables are implemented in very different ways. If you access their elements with $ or an equivalent, and you then change the implementation, you have to rewrite all those accesses. Which is dreadful software engineering.

Cartoon of experimenter peering into innards of complicated piece of machinery. His colleague is holding a plug coming out of it labelled INTERFACE: 'GET' 'INCOME' 'WAGES' and saying 'Don't worry about how it works. It's the interface that's important.'