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I have a dataframe named "adult"
> 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
> 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?
Thanks!
453
In R Programming Language to apply a function to every integer type value in a data frame, we can use lapply function from dplyr package. And if the datatype of values is string then we can use paste() with lapply.
The apply() function lets us apply a function to the rows or columns of a matrix or data frame. This function takes matrix or data frame as an argument along with function and whether it has to be applied by row or column and returns the result in the form of a vector or array or list of values obtained.
We can create a data frame using the data. frame() function. For example, the above shown data frame can be created as follows. Notice above that the third column, Name is of type factor, instead of a character vector.
apply will convert your data into a matrix before processing it (see Details section in ?apply). The factor information is lost during this step.
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
lapply(d, is.factor)
$num
[1] FALSE
$fac
[1] TRUE
or sapply
sapply(d, is.factor)
num fac
FALSE TRUE
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