这是我数据的一小部分。我有超过2万个变量和700个变量。我想要做的是获取变量的子集，并创建一个新变量，该变量具有该子集的变量名称，其值为1. 这些子集中的所有变量都将是字符变量。
id gen16 gen18 gen31 gen33 gen35 gen39 gen45 gen51 gen52 gen56 gen58 gen59 gen66 gen68
5962 1 1 2 2 2 2 2 2 2 1 2 2 2 1 2
6085 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2
6183 3 1 2 2 2 2 2 2 2 2 2 2 2 2 2
6386 4 1 2 2 2 2 2 2 2 2 2 2 2 2 2
6989 5 1 2 1 2 2 2 2 2 2 2 2 2 2 2
7057 6 2 1 1 2 2 2 1 2 2 2 2 2 2 2
7276 7 2 2 2 2 2 2 2 1 1 2 2 2 1 2
7748 8 2 1 2 2 2 2 2 1 2 2 2 2 2 2
7917 9 2 2 2 2 2 2 2 1 2 2 2 2 2 2
8654 10 2 2 2 2 2 2 2 2 2 2 2 1 2 2
所以这就是我要找的。
id gen16 gen18 gen31 gen33 gen35 gen39 gen45 gen51 gen52 gen56 gen58 gen59 gen66 gen68 V1
5962 1 1 2 2 2 2 2 2 2 1 2 2 2 1 2 gen16, gen52, gen66
6085 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 gen58
6183 3 1 2 2 2 2 2 2 2 2 2 2 2 2 2 gen16
6386 4 1 2 2 2 2 2 2 2 2 2 2 2 2 2 gen16
6989 5 1 2 1 2 2 2 2 2 2 2 2 2 2 2 gen16, gen31
7057 6 2 1 1 2 2 2 1 2 2 2 2 2 2 2 gen18, gen31, gen45
7276 7 2 2 2 2 2 2 2 1 1 2 2 2 1 2 gen51, gen52, gen66
7748 8 2 1 2 2 2 2 2 1 2 2 2 2 2 2 gen18, gen51
7917 9 2 2 2 2 2 2 2 1 2 2 2 2 2 2 gen51
8654 10 2 2 2 2 2 2 2 2 2 2 2 1 2 2 gen59
我已经编写了一个执行此操作的for循环，但我想避免循环，因为我的数据集只会变大。我的想法是编写一个适用于一行的函数，然后使用apply函数在整个数据集上迭代它。我很幸运获得了两行不同的功能，但在尝试在apply函数中使用它们时会遇到问题。
这是我写的另外两个函数。
inf.type <- function(x) {
foo <- as.data.frame(x[, c("gen16", "gen18", "gen31", "gen33", "gen35",
"gen39", "gen45", "gen51", "gen52", "gen56", "gen58", "gen59", "gen66", "gen68")] == 1)
gentypes <- paste(names(foo[colSums(foo) == "1"]), collapse = ", ")
return(gentypes)
}
inf.type <- function(x) {
foo <- x[, c("gen16", "gen18", "gen31", "gen33", "gen35", "gen39", "gen45",
"gen51", "gen52", "gen56", "gen58", "gen59", "gen66", "gen68")]
return(paste(names(foo[grep("1", foo)]), collapse = ", "))
}
这两个似乎适用于单行，但不适用于使用apply函数的情况。如果有人可以帮我弄清楚如何让其中一个在apply函数中工作，或者对一个完全不同的方法有更好的建议，我将不胜感激。
答案 0 :(得分：3)
这应该完成工作：
df$V1 = apply(df[,-1], 1, function(x) paste(names(which(x=='1')), collapse = ", "))
查看df[,-1]的每一行（不包括id列），返回与which匹配的索引（x=='1'}），提取names对应于那些索引，paste每行的名称。
还可以写下以下内容（使用@ alistaire＆＃39;建议）：
df$V1 = apply(df[, -1] == 1, 1, function(x) toString(names(x)[x]))
df[, -1] == 1将df[, -1]转换为逻辑矩阵，如果它等于TRUE，则每个单元格评估为1，否则为FALSE。然后，可以为每行提取names个TRUE个单元格，然后将这些名称与toString连接起来。
<强>结果：
id gen16 gen18 gen31 gen33 gen35 gen39 gen45 gen51 gen52 gen56 gen58 gen59
5962 1 1 2 2 2 2 2 2 2 1 2 2 2
6085 2 2 2 2 2 2 2 2 2 2 2 1 2
6183 3 1 2 2 2 2 2 2 2 2 2 2 2
6386 4 1 2 2 2 2 2 2 2 2 2 2 2
6989 5 1 2 1 2 2 2 2 2 2 2 2 2
7057 6 2 1 1 2 2 2 1 2 2 2 2 2
7276 7 2 2 2 2 2 2 2 1 1 2 2 2
7748 8 2 1 2 2 2 2 2 1 2 2 2 2
7917 9 2 2 2 2 2 2 2 1 2 2 2 2
8654 10 2 2 2 2 2 2 2 2 2 2 2 1
gen66 gen68 V1
5962 1 2 gen16, gen52, gen66
6085 2 2 gen58
6183 2 2 gen16
6386 2 2 gen16
6989 2 2 gen16, gen31
7057 2 2 gen18, gen31, gen45
7276 1 2 gen51, gen52, gen66
7748 2 2 gen18, gen51
7917 2 2 gen51
8654 2 2 gen59
数据：
df = structure(list(id = c("1", "2", "3", "4", "5", "6", "7", "8",
"9", "10"), gen16 = c("1", "2", "1", "1", "1", "2", "2", "2",
"2", "2"), gen18 = c("2", "2", "2", "2", "2", "1", "2", "1",
"2", "2"), gen31 = c("2", "2", "2", "2", "1", "1", "2", "2",
"2", "2"), gen33 = c("2", "2", "2", "2", "2", "2", "2", "2",
"2", "2"), gen35 = c("2", "2", "2", "2", "2", "2", "2", "2",
"2", "2"), gen39 = c("2", "2", "2", "2", "2", "2", "2", "2",
"2", "2"), gen45 = c("2", "2", "2", "2", "2", "1", "2", "2",
"2", "2"), gen51 = c("2", "2", "2", "2", "2", "2", "1", "1",
"1", "2"), gen52 = c("1", "2", "2", "2", "2", "2", "1", "2",
"2", "2"), gen56 = c("2", "2", "2", "2", "2", "2", "2", "2",
"2", "2"), gen58 = c("2", "1", "2", "2", "2", "2", "2", "2",
"2", "2"), gen59 = c("2", "2", "2", "2", "2", "2", "2", "2",
"2", "1"), gen66 = c("1", "2", "2", "2", "2", "2", "1", "2",
"2", "2"), gen68 = c("2", "2", "2", "2", "2", "2", "2", "2",
"2", "2")), class = "data.frame", .Names = c("id", "gen16", "gen18",
"gen31", "gen33", "gen35", "gen39", "gen45", "gen51", "gen52",
"gen56", "gen58", "gen59", "gen66", "gen68"), row.names = c(NA,
-10L))