I have a dataframe named "adult"
我有一个名为“成人”的数据框
> str(adult[, 1:2)
'data.frame': 32561 obs. of 15 variables:
$ age : int 39 50 38 53 28 37 49 52 31 42 ...
$ worktp : Factor w/ 9 levels " ?"," Federal-gov",..: 8 7 5 5 5 5 5 7 5 5 ...
> is.factor(adult[,1])
[1] FALSE
> is.factor(adult[,2])
[1] TRUE
Everything works well until I use
一切顺利,直到我使用
> apply(adult[,1:2], 2, function(x) is.factor(x))
age worktp
FALSE FALSE
Why I got FALSE on worktp where is.factor() just gave me TRUE? I really need this apply() function to work on my dataframe. Should I use some other apply related functions?
为什么我在worktp上得到FALSE,其中is.factor()只是给了我真的?我真的需要这个apply()函数来处理我的数据帧。我应该使用其他一些应用相关的功能吗?
Thanks!
谢谢!
1 个解决方案
#1
6
apply
will convert your data into a matrix before processing it (see Details section in ?apply
). The factor information is lost during this step.
apply会在处理之前将您的数据转换为矩阵(请参阅?apply中的详细信息部分)。在此步骤中丢失因子信息。
d <- data.frame(num=1:4, fac=factor(1:4))
d[, 2]
[1] 1 2 3 4
Levels: 1 2 3 4 # levels, hence a factor
m <- as.matrix(d)
m[, 2]
[1] "1" "2" "3" "4" # no levels anymore
apply(d, 2, is.factor)
num fac
FALSE FALSE # no factors as converted to matrix
To get what you want you could use lapply
为了得到你想要的东西你可以使用lapply
lapply(d, is.factor)
$num
[1] FALSE
$fac
[1] TRUE
or sapply
或者是讽刺
sapply(d, is.factor)
num fac
FALSE TRUE
#1
6
apply
will convert your data into a matrix before processing it (see Details section in ?apply
). The factor information is lost during this step.
apply会在处理之前将您的数据转换为矩阵(请参阅?apply中的详细信息部分)。在此步骤中丢失因子信息。
d <- data.frame(num=1:4, fac=factor(1:4))
d[, 2]
[1] 1 2 3 4
Levels: 1 2 3 4 # levels, hence a factor
m <- as.matrix(d)
m[, 2]
[1] "1" "2" "3" "4" # no levels anymore
apply(d, 2, is.factor)
num fac
FALSE FALSE # no factors as converted to matrix
To get what you want you could use lapply
为了得到你想要的东西你可以使用lapply
lapply(d, is.factor)
$num
[1] FALSE
$fac
[1] TRUE
or sapply
或者是讽刺
sapply(d, is.factor)
num fac
FALSE TRUE