Its simple question but looks like it doesn't exist in the stackoverflow
I have a data frame where all columns are factors I want to convert it to decimals.
Var1 Var2 Var3 Var4
1 0.76 0.84 0.76 0.73
2 0.76 0.84 0.76 0.73
3 0.76 0.84 0.76 0.73
4 0.76 0.84 0.76 0.73
5 0.76 0.84 0.76 0.73
6 0.76 0.84 0.76 0.73
I want to convert this without loosing the decimals.
df <- sapply(df, as.numeric)
This doesn't retain the decimals.
If they are truly factors, you need to go through another step:
The reason for as.numeric not working directly is because internally each factor is stored by its levels . you can access that through the levels(factor_var). So when you apply as.numeric to a factor directly, what gets returned is their levels.
Therefore, first make it a character, and then apply as.numeric
df <- sapply(df, as.character)
df <- sapply(df, as.numeric)
Or you can nest them in a function:
convert_func<-function(x){ as.numeric(as.character(x))}
then :df <- sapply(df, convert_func)
I have never tried nesting them in the apply/lapply/sapply without a function, but it might work also. or you can make a loop:
for (col in 1:ncol(df){
df[col]<-as.numeric(as.character(df[col]))
}
sapply(df, function(x) as.numeric(as.character(x)) )This should also work:
df[] <- lapply(df, function(x) ifelse(is.numeric(x), as.numeric(x), x))
We can use dplyr to convert the factor columns to numeric
library(dplyr)
library(magrittr)
df %<>%
mutate_if(is.factor, funs(as.numeric(as.character(.))))
With base R, we can do
df[] <- lapply(df, function(x) if(is.factor(x)) as.numeric(as.character(x)) else x)
df <- structure(list(Var1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L),
.Label = "0.76", class = "factor"),
Var2 = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "0.84", class = "factor"),
Var3 = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "0.76", class = "factor"),
Var4 = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "0.73", class = "factor")),
.Names = c("Var1", "Var2", "Var3", "Var4"), row.names = c("1", "2", "3", "4", "5",
"6"), class = "data.frame")
