CODEWITHSUNDEEP
rdplyrr-haven
JohnDoeVsJoeSchmoe
JohnDoeVsJoeSchmoe

Reputation: 711

How can I check the type of a column if it is an unusual type?

I have some SPSS data from a .sav file and am trying to work with it in R. Many of the variables are of type haven_labelled. I'd like to convert them to double using mutate_if(). How can I create a predicate for mutate_if() that will catch all the columns of type haven_labelled? There is an is.labelled() function in the haven library.

Upvotes: 2

Views: 185

Answers (1)

akrun
akrun

Reputation: 887118

We can use mutate_if to apply the function on columns based on a condition. Here, in the reproducible example below, the labelled attribute is on the 'Species' column, which is converted to factor

library(dplyr)
library(haven)
iris1 <- iris %>%
            mutate_if(is.labelled, factor)

Or another option is to create the logical condition with class

iris1 <- iris %>%
           mutate_if(~ class(.) ==  "haven_labelled", factor)

-checking the structure

str(iris)
#'data.frame':  150 obs. of  5 variables:
# $ Sepal.Length: num  5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
# $ Sepal.Width : num  3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
# $ Petal.Length: num  1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
# $ Petal.Width : num  0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
# $ Species     : 'haven_labelled' chr  "setosa" "setosa" "setosa" "setosa" ...
#  ..- attr(*, "labels")= Named chr  "S" "ve" "vi"
#  .. ..- attr(*, "names")= chr  "setosa" "versicolor" "virginica"

str(iris1)
#'data.frame':  150 obs. of  5 variables:
# $ Sepal.Length: num  5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
# $ Sepal.Width : num  3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
# $ Petal.Length: num  1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
# $ Petal.Width : num  0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
# $ Species     : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 #...

data

data(iris)
iris$Species <- labelled(as.character(iris$Species),
       c("setosa" = "S", "versicolor" = "ve", "virginica" = "vi"))

Upvotes: 3

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