独立于列数，在数据帧上按行应用函数 [英] Apply a function by row on a dataframe independently of the number of columns

问题描述

我想在数据框架上按行应用一个函数，以根据行中的值连接列标题。

I'd like to apply a function by rows on a data.frame to concatenate column titles depending on the value in the row.

df 
      A     B
1  TRUE  TRUE
2 FALSE  TRUE
3 FALSE FALSE

      A     B Result
1  TRUE  TRUE A / B
2 FALSE  TRUE   B
3 FALSE FALSE NA

我阅读关于dplyr使用mutate（）和rowwise（），但我不知道如何应用它们，因为列不是常量。

I read about dplyr using mutate() and rowwise(), but I don't know how to apply them since the columns aren't constants.

一行我我会做一些事情：

for a row "i" I would do something like:

paste(names(df)[as.logical(df[i,])], collapse = ' / ')

欢迎任何帮助。

谢谢。

推荐答案

如果数据集不是很大（即数百万/十亿行），我们可以使用应用与 MARGIN = 1 循环遍历行，将名称矢量使用逻辑矢量作为索引和粘贴它们在一起。

If the dataset is not really big (i.e. in millions/billions of rows) we can use apply with MARGIN=1 to loop over the rows, subset the names of the vector using the logical vector as index and paste them together. It is easier to code in a single line.

df$Result <- apply(df, 1, FUN = function(x) paste(names(x)[x], collapse=" / "))

---

但是，如果我们有一个大数据集，另一个选项是创建一个键/值对，并通过匹配替换值，并且比上述解决方案更快。

---

However, if we have a big dataset, another option is to create a key/value pair and replace the values by matching and it is faster than the above solution.

v1 <- do.call(paste, df)
unname(setNames(c("A / B", "B", "A", NA), do.call(paste, 
          expand.grid(rep(list(c(TRUE, FALSE)), 2))))[v1])
#[1] "A / B" "B"     NA   

---

或者我们可以使用算术运算来执行

---

Or we can use arithmetic operation to do this

c(NA, "A", "B", "A / B")[1 + df[,1] + 2 * df[,2]]
#[1] "A / B" "B"     NA  

基准

使用@ DavidArenburg的数据集并包含这里发布的两个解决方案（将df的列名称更改为 A'和'B'）

Benchmarks

Using @DavidArenburg's dataset and including the two solutions posted here (changed the column names of 'df' to 'A' and 'B')

newPaste <- function(df) {
    v1 <- do.call(paste, df)
  unname(setNames(c("A / B", "B", "A", NA), do.call(paste, 
      expand.grid(rep(list(c(TRUE, FALSE)), 2))))[v1])
}

arith <- function(df){
     c(NA, "A", "B", "A / B")[1 + df[,1] + 2 * df[,2]]
}

microbenchmark::microbenchmark(Rowwise(df), Colwise(df), newPaste(df),arith(df))
#Unit: milliseconds
#        expr        min        lq      mean     median         uq       max neval
#  Rowwise(df) 398.024791 453.68129 488.07312 481.051431 523.466771 688.36084   100
#  Colwise(df)  25.361609  28.10300  34.20972  30.952365  35.885061  95.92575   100
# newPaste(df)  65.777304  69.07432  82.08602  71.606890  82.232980 176.66516   100
#   arith(df)   1.790622   1.88339   4.74913   2.027674   4.753279  58.50942   100

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