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最近，我对以下事实感到惊讶，is.logical并且is.factor无法开箱即用apply——至少他们没有提供正确的结果。

这是我的小可复制示例：

# generate a dataset that contains a couple of modes
someDf <- data.frame(fac1=gl(2,3,12),
                 int=1:12,
                 char=letters[1:12],
                 logi=rep(c(T,F),6),
                 fac2=gl(3,2,12))

# hooray, this did work, got factors, int,
# characters and logical
str(someDf)

# I expected this to work, but it didn't
# everything is just FALSE
apply(someDf,2,is.logical)

我没有放弃，并找到了一个偷偷摸摸的方法。

unlist(lapply(names(someDf),function(x) is.logical(someDf[,x])))

虽然这提供了正确的结果，但我想知道为什么它必须如此复杂，以及是否有更简单的解决方案。有任何想法吗？

提示：我已经看到了

apply(someDf,2,class)

提供所有characters。这也是意料之外的。也许有些东西eval可以解决我找不到的窍门。

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1 回答 1

7

由于 data.frame 是 a list，您应该使用lapplyor sapply：

sapply(someDf, is.logical)
 fac1   int  char  logi  fac2 
FALSE FALSE FALSE  TRUE FALSE

您的代码不起作用的原因是因为apply需要一个矩阵作为其参数，并且如果您提供一个数据框，则强制转换为一个矩阵。由于矩阵只能包含单个类的元素，因此您的值将转换为character. 试试看：

as.matrix(someDf)
      fac1 int  char logi    fac2
 [1,] "1"  " 1" "a"  " TRUE" "1" 
 [2,] "1"  " 2" "b"  "FALSE" "1" 
 [3,] "1"  " 3" "c"  " TRUE" "2" 
 [4,] "2"  " 4" "d"  "FALSE" "2" 
 [5,] "2"  " 5" "e"  " TRUE" "3" 
 [6,] "2"  " 6" "f"  "FALSE" "3" 
 [7,] "1"  " 7" "g"  " TRUE" "1" 
 [8,] "1"  " 8" "h"  "FALSE" "1" 
 [9,] "1"  " 9" "i"  " TRUE" "2" 
[10,] "2"  "10" "j"  "FALSE" "2" 
[11,] "2"  "11" "k"  " TRUE" "3" 
[12,] "2"  "12" "l"  "FALSE" "3"
于 2012-09-20T08:15:36.717 回答