Get column indexes of class character/integer or any type I define?

Question

Given a dataframe, I want to get column indexes (1,4,55,33..) of columns of type integer.

Given df:

col1           col2
<character>    <integer>
a              1 
a              2  
vb             3

I want to get col1 as character column.

I am trying to do this using:

which(sapply(df, function(x) class(x) == "character") == TRUE)

Please advise.

The `== TRUE` is not needed since the inner "==" already created a logical vector. I suspect the `which` is not needed either, but unless you show your use case we cannot know for sure, Almost any function that accepts numeric arguments for column selection will also accept a logical vector. There is a probability approaching unity that this question has been asked multiple times in the past on SO. — IRTFM, Feb 25 '19 at 15:14
You could also do stuff like `grep("ch", sapply(df, class))` or `grep("in", sapply(df, class))` if you are lazy — David Arenburg, Feb 25 '19 at 15:23
Possible duplicate of [Determine the data types of a data frame's columns](https://stackoverflow.com/questions/21125222/determine-the-data-types-of-a-data-frames-columns) — divibisan, Feb 25 '19 at 18:29
^ It's not an exact duplicate, since they aren't asking for the index, but once you have a list it's just `which(classes == 'integer')` — divibisan, Feb 25 '19 at 18:35

tmfmnk · Answer 1 · 2019-02-25T15:27:51.043

2

You can try:

which(sapply(df, function(x) is.numeric(x)))

col2 
   2 

which(sapply(df, function(x) is.character(x)))

col1 
   1

Or as @David Arenburg suggested:

which(sapply(df, is.character))

col1 
   1

Sample data:

df <- read.table(text = "col1           col2
 a              1 
a              2  
vb             3", header = TRUE, stringsAsFactors = FALSE)

edited Feb 25 '19 at 15:27

answered Feb 25 '19 at 15:14

tmfmnk

@David Arenburg updated the post. Thanks for your remark. – tmfmnk Feb 25 '19 at 15:28

NelsonGon · Accepted Answer · 2019-02-25T15:44:18.130

1

We can also use match. The second part of this answer is admittedly overly complicating things.

Purely base R:

 match(names(Filter(is.numeric,df)),names(df))#speed limitations
 #[1] 2

Using Packages:

library(dplyr)
    df1<-df %>% 
      purrr::select_if(is.numeric)

     match(names(df1),names(df))
    #[1] 2

Data:

df <- read.table(text = "col1           col2
 a              1 
a              2  
vb             3", header = TRUE, stringsAsFactors = FALSE)

edited Feb 25 '19 at 15:44

answered Feb 25 '19 at 15:37

NelsonGon

Correct me if I am wrong, Filter(is.numeric, df) checks which column in df is numeric and returns logical vector? This vector is the input for the names functions. – SteveS Feb 26 '19 at 08:18
`Filter(is.numeric,df)` gets all numeric columns. We then get the names of those columns and `match` them to the original names. – NelsonGon Feb 26 '19 at 10:41

2 Answers2