在R数据框中创建所有可能的列置换 [英] Create all possible column permutations in R dataframe

问题描述

我有一个看起来像这样的df:

I have a df that looks like this:

   code.1 code.2 code.3 code.4
1:     82     93     NA     NA
2:     15     85     93     NA
3:     93     89     NA     NA
4:     81     NA     NA     NA

我想生成一个新的df，其中包含列的所有可能排列，因此例如第5:8行将是"code.2, code.3, code.4, code.1"的值，第9:12行将是"code.3, code.4, code.1, code.2"的值，并且等等.

I'd like to generate a new df that entails all possible permutations of the columns, so for example row 5:8 would be the values of "code.2, code.3, code.4, code.1", the rows 9:12 the values of "code.3, code.4, code.1, code.2", and so on.

我已经尝试过从gtools包中进行排列"，但是我一直遇到"v要么是非原子的，要么太短"的情况.

I've tried "permutations" from the gtools package but I keep running into "v is either non-atomic or too short"

推荐答案

我们可以排列列索引，根据索引选择列，然后组合所有数据框.

We can permutate the column index, select the column based on the indices, and then combine all the data frame.

library(gtools)

per <- permutations(n = 4, r = 4, repeats.allowed = FALSE)

dat2 <- do.call(rbind, lapply(1:nrow(per), function(x){
  dat_temp <- dat[, per[x, ]]
  names(dat_temp) <- names(dat)
  return(dat_temp)
  }))

head(dat2, 12)
#    code.1 code.2 code.3 code.4
# 1      82     93     NA     NA
# 2      15     85     93     NA
# 3      93     89     NA     NA
# 4      81     NA     NA     NA
# 11     82     93     NA     NA
# 21     15     85     NA     93
# 31     93     89     NA     NA
# 41     81     NA     NA     NA
# 12     82     NA     93     NA
# 22     15     93     85     NA
# 32     93     NA     89     NA
# 42     81     NA     NA     NA

数据

dat <- read.table(text = "   code.1 code.2 code.3 code.4
1     82     93     NA     NA
2     15     85     93     NA
3     93     89     NA     NA
4     81     NA     NA     NA",
                  header = TRUE, stringsAsFactors = FALSE)

这篇关于在R数据框中创建所有可能的列置换的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！