R:将数据帧按组和行迭代拆分为列表 [英] R: Split dataframe into list by group and row iteration

问题描述

我有一个像这样的数据框:

I have a dataframe like so:

set.seed(34)
startingframe <-  data.frame(
  group1=factor(rep(c("a","b"),each=3,times=1)),
  time=rep(1:3,each=1,times=2),
  othercolumn=rnorm(1:6)
)

...我正在尝试将其处理成一个列表，该列表按组和time列的每个迭代"划分，结果如下:

...which I am trying to manipulate into a list that splits by group and each 'iteration' of the time column, results looking like so:

 $a1
  group1 time othercolumn
1      a    1    -0.13889

$a2
  group1 time othercolumn
1      a    1   -0.138890
2      a    2    1.199813

$a3
  group1 time othercolumn
1      a    1  -0.1388900
2      a    2   1.1998129
3      a    3  -0.7477224

$b1
  group1 time othercolumn
4      b    1  -0.5752482

$b2
  group1 time othercolumn
4      b    1  -0.5752482
5      b    2  -0.2635815

$b3
  group1 time othercolumn
4      b    1  -0.5752482
5      b    2  -0.2635815
6      b    3  -0.4554921

我想我只需要将迭代位合并到此代码中即可:

I think I just need the iteration bit incorporated into this code:

split_list <- split(startingframe,list(startingframe$group1,startingframe$time))

我可能已经简化了原始问题，但是实际数据集还有其他一些列需要在列表中传递.

I may have oversimplified the original problem, but the real dataset has additional columns that would need to be brought through in the list.

推荐答案

您可以将head用于序列1:nrow(x)中的每个数字，数据的每个子组x

You could use head for every number in the sequence 1:nrow(x), for each sub-group x of your data

out <- lapply(split(df, df$group1), function(x) lapply(1:nrow(x), head, x = x))
out <- unlist(out, recursive = F)

out
# $`a1`
#    group1 time
# 1:      a    1
# 
# $a2
#    group1 time
# 1:      a    1
# 2:      a    2
# 
# $a3
#    group1 time
# 1:      a    1
# 2:      a    2
# 3:      a    3
# 
# $b1
#    group1 time
# 1:      b    1
# 
# $b2
#    group1 time
# 1:      b    1
# 2:      b    2
# 
# $b3
#    group1 time
# 1:      b    1
# 2:      b    2
# 3:      b    3

尽管您可能不需要创建所有这些数据框的列表，而只需创建索引列表并根据需要使用它们即可.

You might not need to create a list of all these dataframes though, you could just create a list of indices and use them as needed.

inds <- 
  lapply(split(seq(nrow(df)), df$group1), function(x) 
    lapply(1:length(x), function(y) x[seq(y)]))

inds
# $`a`
# $`a`[[1]]
# [1] 1
# 
# $`a`[[2]]
# [1] 1 2
# 
# $`a`[[3]]
# [1] 1 2 3
# 
# 
# $b
# $b[[1]]
# [1] 4
# 
# $b[[2]]
# [1] 4 5
# 
# $b[[3]]
# [1] 4 5 6

df[inds$b[[2]]]
#    group1 time
# 1:      b    1
# 2:      b    2

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