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根据R中的值将多列折叠为一列

multiple-columns merge function dataframe r

Sylvia Rodriguez · 技术社区 · 2 年前

我有 data.frame df1 :

 df1 <- data.frame(apple = c('0', '0', '0', '1', '0', '1', '0', '0', '0', '1'),
      banana = c('1', '0', '0', '0', '1', '0', '1', '0', '0', '0'),
      cherry = c('0', '1', '0', '0', '0', '0', '0', '1', '0', '0'),
      date = c('0', '0', '1', '0', '0', '0', '0', '0', '1', '0'))
 rownames(df1) <- c('one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten')

如何用列名替换每列中的所有“ones”,然后将该列折叠成一个新列?预期结果为 df2 :

 df2 <- data.frame(fruit = c('banana', 'cherry', 'date', 'apple', 'banana', 
      'apple', 'banana', 'cherry', 'date', 'apple'))
 rownames(df2) <- c('one', 'two', 'three', 'four', 'five', 
      'six', 'seven', 'eight', 'nine', 'ten')

我在这里找到了第二步( r collapsing data from multiple columns into one )但我还没有完全到达那里。

3 回复 | 直到 2 年前

TarJae 2 年前

这是 tidyverse 方式使用 transmute 和 unite :

library(dplyr)
library(tidyr)

df1 %>%
  transmute(across(apple:date, ~case_when(. == 1 ~ cur_column()), .names = 'new_{col}')) %>% 
  unite(fruit, starts_with('new'), na.rm = TRUE, sep = ' ')

       fruit
one   banana
two   cherry
three   date
four   apple
five  banana
six    apple
seven banana
eight cherry
nine    date
ten    apple

user2974951 2 年前

使用apply

> apply(df1==1,1,function(x){colnames(df1)[x]})
     one      two    three     four     five      six    seven    eight     nine      ten 
"banana" "cherry"   "date"  "apple" "banana"  "apple" "banana" "cherry"   "date"  "apple"

另一个选项使用

> df2=as.data.frame(which(df1==1,arr.ind=T))
> df2=df2[order(df2$row),]
> df2$fruit=colnames(df1)[df2$col]
      row col  fruit
one     1   2 banana
two     2   3 cherry
three   3   4   date
four    4   1  apple
five    5   2 banana
six     6   1  apple
seven   7   2 banana
eight   8   3 cherry
nine    9   4   date
ten    10   1  apple

jay.sf 2 年前

使用 max.col 子集 names ,创建 data.frame 用相应的 rownames .

> (res <- data.frame(fruit=names(df1)[max.col(df1)]) |> `rownames<-`(rownames(df1)))
       fruit
one   banana
two   cherry
three   date
four   apple
five  banana
six    apple
seven banana
eight cherry
nine    date
ten    apple
> all.equal(df2, res)
[1] TRUE

基准

> DF1 <- df1[sample.int(nrow(df1), 1e4, replace=TRUE), ]
> microbenchmark::microbenchmark(max.col=data.frame(fruit=names(DF1)[max.col(DF1)]) |> `rownames<-`(rownames(DF1)),
+                                tidy=DF1 %>%
+                                  transmute(across(apple:date, ~case_when(. == 1 ~ cur_column()), .names = 'new_{col}')) %>% 
+                                  unite(fruit, starts_with('new'), na.rm = TRUE, sep = ' '),
+                                check='equivalent')
~$ Rscript --vanilla foo.R
Unit: milliseconds
    expr       min        lq      mean    median        uq        max neval cld
 max.col  4.915768  5.041378  5.213677  5.193255  5.321326   5.821633   100  a 
    tidy 65.275247 67.524774 72.382577 71.061575 72.707830 145.169879   100   b

数据:

> dput(df1)
structure(list(apple = c("0", "0", "0", "1", "0", "1", "0", "0", 
"0", "1"), banana = c("1", "0", "0", "0", "1", "0", "1", "0", 
"0", "0"), cherry = c("0", "1", "0", "0", "0", "0", "0", "1", 
"0", "0"), date = c("0", "0", "1", "0", "0", "0", "0", "0", "1", 
"0")), class = "data.frame", row.names = c("one", "two", "three", 
"four", "five", "six", "seven", "eight", "nine", "ten"))