如何使用R将列表的列拆分为几列 [英] How to split a column of list into several columns using R

问题描述

原始数据框:

     sg                                dt               time
2099     C 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0 2014-07-24 16:23:55.2
2100     C 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0 2014-07-24 16:23:55.4
2101     C 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0 2014-07-24 16:23:55.5
2103     C 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0 2014-07-24 16:23:56.4
2104     C 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0 2014-07-24 16:23:56.5
2102     C 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1 2014-07-24 16:23:56.7

有一列名为"dt"的

> z$dt
[[1]]
 [1] "0" "0" "0" "0" "0" "1" "0" "0" "0" "0" "0" "0" "0"

[[2]]
 [1] "0" "0" "0" "0" "0" "1" "0" "0" "0" "0" "0" "0" "0"

[[3]]
 [1] "0" "0" "1" "0" "0" "1" "0" "0" "0" "0" "0" "0" "0"

[[4]]
 [1] "0" "0" "0" "1" "0" "1" "0" "0" "0" "0" "1" "0" "0"

[[5]]
 [1] "0" "0" "0" "0" "0" "1" "0" "0" "0" "0" "1" "0" "0"

[[6]]
 [1] "0" "0" "0" "0" "0" "1" "0" "0" "0" "0" "0" "0" "1"

我想将"dt"列转换为多个列，例如:

I want convert the column "dt" into multiple columns like:

           sg  A  B  C  D  E  F  G  H  I  G  K  L  M             time
    2099     C 0  0  0  0  0  1  0  0  0  0  0  0  0 2014-07-24 16:23:55.2
    2100     C 0  0  0  0  0  1  0  0  0  0  0  0  0 2014-07-24 16:23:55.4
    2101     C 0  0  1  0  0  1  0  0  0  0  0  0  0 2014-07-24 16:23:55.5
    2103     C 0  0  0  1  0  1  0  0  0  0  1  0  0 2014-07-24 16:23:56.4
    2104     C 0  0  0  0  0  1  0  0  0  0  1  0  0 2014-07-24 16:23:56.5
    2102     C 0  0  0  0  0  1  0  0  0  0  0  0  1 2014-07-24 16:23:56.7

我该怎么办?

推荐答案

如果z$dt中的数据都具有相同的长度，则以下内容应该起作用:

The following should work if the data in z$dt all have the same length:

x <- do.call(rbind, z$dt)
colnames(x) <- LETTERS[1:ncol(x)]
cbind(z[c("sg", "time")], x)

这篇关于如何使用R将列表的列拆分为几列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！