Using lapply to group list of data frames by column - r

Question

I have a list that contains multiple data frames. I would like to sort the data by Category (A) and sum the Frequencies (B) using the lapply-command.
The data is df_list
df_list
$`df.1`
A B
1 Apples 2
2 Pears 5
3 Apples 6
4 Pears 1
5 Apples 3
$`df.2`
A B
1 Oranges 2
2 Pineapples 5
3 Oranges 6
4 Pineapples 1
5 Oranges 3
The desired outcome df_list_2 looks like this:
df_list_2
$`df.1`
A B
1 Apples 11
2 Pears 6
$`df.2`
A B
1 Oranges 11
2 Pineapples 6
I have tried the following code based on lapply:
df_list_2<-df_list[, lapply(B, sum), by = A]
However, I get an error code, saying that A was not found.
Either I mistake how the lapply command works in this case or my understating of how it should work is flawed.
Any help much appreciated.

Answer 1

You need to aggregate in lapply
lapply(df_list, function(x) aggregate(B~A, x, sum))
#[[1]]
# A B
#1 Apples 11
#2 Pears 6
#[[2]]
# A B
#1 Oranges 11
#2 Pineapples 6
Using map from purrr and dplyr it would be
library(dplyr)
purrr::map(df_list, ~.x %>% group_by(A) %>% summarise(sum = sum(B)))
data
df_list <- list(structure(list(A = structure(c(1L, 2L, 1L, 2L, 1L),
.Label = c("Apples", "Pears"), class = "factor"), B = c(2L, 5L, 6L, 1L, 3L)),
class = "data.frame", row.names = c("1", "2", "3", "4", "5")),
structure(list(A = structure(c(1L, 2L, 1L, 2L, 1L), .Label = c("Oranges",
"Pineapples"), class = "factor"), B = c(2L, 5L, 6L, 1L, 3L)), class = "data.frame",
row.names = c("1", "2", "3", "4", "5")))

Answer 2

I fear you might not have a clear idea of lapply nor the extract operator ([). Remember lapply(list, function) applies the specified function you give it to each element of the list you give it. Extract gives you the element you specify:
x <- c('a', 'b', 'c')
x[2]
## "b"
I would imagine that somewhere in your R workspace you have an object names B which is why you didn't get an error along the lines of
## Error in lapply(B, sum) : object 'B' not found
Conversely if you had (accidentally or intentionally) defined both A and B you would see the error
## Error in df_list[, lapply(B, sum), by = A] : incorrect number of dimensions
because that's not at all how to use [; remember, you just pass indexes or booleans to [ along with the occasional optional argument, but by is not one of those.
So without further adieu, here's how I would do this (in base R):
# make some data
a <- c(1, 2, 1, 2, 1)
b <- c(2, 5, 6, 1, 3)
df_list <- list(df.1 = data.frame(A = c('Apples', 'Pears')[a], B = b),
df.2 = data.frame(A = c('Oranges', 'Pineapples')[a], B = b))
# simplify it
df_list_2 <- lapply(df_list, function(x) {
aggregate(list(B = x$B), list(A = x$A), sum)
})
# the desired result
df_list_2
## $df.1
## A B
## 1 Apples 11
## 2 Pears 6
##
## $df.2
## A B
## 1 Oranges 11
## 2 Pineapples 6
You can take advantage of the fact that a data.frame is just a list and shorten up your code like this:
df_list_2 <- lapply(df_list, function(x) {
aggregate(x['B'], x['A'], sum)
})
but the first way of writing it should help make more clear what we're doing

Answer 3

The data.table syntax in OP's post can changed to
library(data.table)
lapply(df_list, function(x) as.data.table(x)[, .(B = sum(B)), by = A])
#$df.1
# A B
#1: Apples 11
#2: Pears 6
#$df.2
# A B
#1: Oranges 11
#2: Pineapples 6
data
df_list <- list(df.1 = structure(list(A = structure(c(1L, 2L, 1L, 2L, 1L
), .Label = c("Apples", "Pears"), class = "factor"), B = c(2L,
5L, 6L, 1L, 3L)), class = "data.frame", row.names = c("1", "2",
"3", "4", "5")), df.2 = structure(list(A = structure(c(1L, 2L,
1L, 2L, 1L), .Label = c("Oranges", "Pineapples"), class = "factor"),
B = c(2L, 5L, 6L, 1L, 3L)), class = "data.frame", row.names = c("1",
"2", "3", "4", "5")))