Thank you Jim, but I obtain:
```
'data.frame': 5 obs. of 3 variables:
$ region : Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
$ sales : num 13 16 22 27 34
$ country: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
df1<-df[,!unlist(lapply(df,is.factor))]
str(df1)
[1] 13 16 22 27 34
```
I was expecting
```
str(df)
'data.frame': 5 obs. of 3 variables:
$ region : char "A","B","C","D",..: 1 2 3 4 5
$ sales : num 13 16 22 27 34
$ country: char "a","b","c","d",..: 1 2 3 4 5
```
On Sun, Sep 19, 2021 at 11:37 AM Jim Lemon <drjimlemon at gmail.com> wrote:
Hi Luigi,
It's easy:
df1<-df[,!unlist(lapply(df,is.factor))]
_except_ when there is only one column left, as in your example. In
that case, you will have to coerce the resulting vector back into a
one column data frame.
Jim
On Sun, Sep 19, 2021 at 6:18 PM Luigi Marongiu <marongiu.luigi at gmail.com>
Hello,
I woul dlike to remove factors from all the columns of a dataframe.
I can do it n a column at the time with
```
df <- data.frame(region=factor(c('A', 'B', 'C', 'D', 'E')),
sales = c(13, 16, 22, 27, 34), country=factor(c('a',
'b', 'c', 'd', 'e')))
new_df$region <- droplevels(new_df$region)
```
What is the syntax to remove all factors at once (from all columns)?
For this does not work:
```
'data.frame': 5 obs. of 3 variables:
$ region : Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
$ sales : num 13 16 22 27 34
$ country: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
df = droplevels(df)
str(df)
'data.frame': 5 obs. of 3 variables:
$ region : Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
$ sales : num 13 16 22 27 34
$ country: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
```
Thank you