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I have a data frame. Let's call him bob:

> head(bob)

                 phenotype                         exclusion

GSM399350 3- 4- 8- 25- 44+ 11b- 11c- 19- NK1.1- Gr1- TER119-

GSM399351 3- 4- 8- 25- 44+ 11b- 11c- 19- NK1.1- Gr1- TER119-

GSM399352 3- 4- 8- 25- 44+ 11b- 11c- 19- NK1.1- Gr1- TER119-

GSM399353 3- 4- 8- 25+ 44+ 11b- 11c- 19- NK1.1- Gr1- TER119-

GSM399354 3- 4- 8- 25+ 44+ 11b- 11c- 19- NK1.1- Gr1- TER119-

GSM399355 3- 4- 8- 25+ 44+ 11b- 11c- 19- NK1.1- Gr1- TER119-

I'd like to concatenate the rows of this data frame (this will be another question). But look:

> class(bob$phenotype)

[1] "factor"

Bob's columns are factors. So, for example:

> as.character(head(bob))

[1] "c(3, 3, 3, 6, 6, 6)"       "c(3, 3, 3, 3, 3, 3)"      

[3] "c(29, 29, 29, 30, 30, 30)"

I don't begin to understand this, but I guess these are indices into the levels of the factors of the columns (of the court of king caractacus) of bob? Not what I need.

Strangely I can go through the columns of bob by hand and do

bob$phenotype <- as.character(bob$phenotype)

which works fine. And, after some typing, I can get a data.frame whose columns are characters rather than factors. So my question is: how can I do this automatically? How do I convert a data.frame with factor columns into a data.frame with character columns without having to manually go through each column?

1 Answer

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by (25.3k points)

To convert the columns of a data frame to the character class and then concatenate the rows of the data frame, use the following function:

bob <- data.frame(lapply(bob, as.character), stringsAsFactors=FALSE)

 This recreates the data frame with the lapply() function with character class for all variables.

To only convert factors use the following code:

library(dplyr)

bob %>% mutate_if(is.factor, as.character) -> bob

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