从列中取出变量名并根据条件创建新的列值答案

【问题标题】：Taking variable names out of column and creating a new column value based on condition从列中取出变量名并根据条件创建新的列值
【发布时间】：2021-12-27 18:29:49
【问题描述】：

我有这个矩阵 df，第一列中包含所有单词，第 2-75 列中包含单词所属的不同 LIWC 类别。

这是我所拥有的玩具示例：

word	posemo	certain	insight
certainly	1	0	1
obviously	1	1	1
sure	1	0	0
directly	1	0	1
insight	1	1	0
guarantee	0	1	0
prove	1	0	1

这就是我想要实现的目标：

word	posemo	certain	insight	Categories
certainly	1	0	1	posemo, insight
obviously	1	1	1	posemo, certain, insight
sure	1	0	0	posemo
directly	1	0	1	posemo,insight
insight	1	1	0	posemo, certain
guarantee	0	1	0	certain
prove	1	0	1	posemo, insight

我一直在查看 stackoverflow，但似乎找不到适用于我正在尝试做的事情的东西。这个Taking variable names out of column and creating new columns in R 很接近，但不处理条件。

有什么建议吗？提前致谢

【问题讨论】：

标签： r matrix dplyr

【解决方案1】：

尝试使用apply：

data.frame( dat, Categories=t(
   t( apply( dat[,2:4], 1, function(x) colnames(dat[,2:4])[as.logical(x)] ) ) ))

       word posemo certain insight               Categories
1 certainly      1       0       1          posemo, insight
2 obviously      1       1       1 posemo, certain, insight
3      sure      1       0       0                   posemo
4  directly      1       0       1          posemo, insight
5   insight      1       1       0          posemo, certain
6 guarantee      0       1       0                  certain
7     prove      1       0       1          posemo, insight

数据

dat <- structure(list(word = c("certainly", "obviously", "sure", "directly", 
"insight", "guarantee", "prove"), posemo = c(1L, 1L, 1L, 1L, 
1L, 0L, 1L), certain = c(0L, 1L, 0L, 0L, 1L, 1L, 0L), insight = c(1L, 
1L, 0L, 1L, 0L, 0L, 1L)), class = "data.frame", row.names = c(NA, 
-7L))

编辑速度，尝试预定义关键数据

n <- colnames( dat[,2:4] )
lo <- dat[,2:4] == 1

data.frame( dat, Categories=t(t( apply( lo, 1, function(x) n[x] ) ) ))

【讨论】：

谢谢，这真的很棒！这适用于玩具示例，但需要永远加载具有 60.000 行的整个 df。幸好我今天有时间:)

【解决方案2】：

这就是诀窍：

遍历行并找出哪些行等于 1 (cols <- x[i,] == 1)，然后获取这些列名 (cats <- na.omit(colnames(x)[cols]))，然后将它们作为单个字符串粘贴在一起并替换 categories 的值 (x$categories[i] <- paste(cats, collapse = ", ") )

x <- tibble(word = c("love","hate","sad"),
            happy = c(1,0,0),
            sad = c(0,1,1),
            emotion = c(1,1,1),
            categories = c(NA,NA,NA))

for(i in 1:nrow(x)){
  cols <- x[i,] == 1
  cats <- na.omit(colnames(x)[cols])
  x$categories[i] <- paste(cats, collapse = ", ")
}

【讨论】：