Normalizing a dataframe - Error : non-numeric argument to binary - R

Question

I went through similar questions but I still could not sort my problem out. Anyway, I have a data-frame (size:36*42) that contains numbers while in the last row (36'th row) it has the dates for each column. I wanted to normalize the columns of this data-frame (the whole data-frame except the last row) using the following function:

normalize<-function(x){return((x-min(x, na.rm=TRUE))/(max(x, na.rm=TRUE)-min(x, na.rm=TRUE)))}

but I always get this error:

Error in x - min(x, na.rm = TRUE) : non-numeric argument to binary operator

The thing that I have tried

as.data.frame(lapply(df[c(1:nrow(df)-1),], normalize))

BTW, when I check the typeof(df) it shows that the df is a list but when I check it using is.list(df)->TRUE and also is.data.frame(df)->TRUE, which it makes me confuse.

Thanks

You have some factors or characters in there. Try `sapply(df,class)`. You should provide sample data with `dput`. Did you type this as is `is.list(df)->TRUE`? — NelsonGon, May 03 '19 at 11:07
regarding your question about your `data.frame` having a `list` type, take a look at [this](https://stackoverflow.com/questions/6258004/types-and-classes-of-variables) post. — patL, May 03 '19 at 11:10
@NelsonGon, "Did you type this as is is.list(df)->TRUE?", Yes — Ester Silva, May 03 '19 at 11:18
If you just want to calculate on the `numeric`s, maybe simply do: `normalize<-function(x){ x<-Filter(is.numeric,x) return( (x-min(x, na.rm=TRUE)) / (max(x, na.rm=TRUE)-min(x, na.rm=TRUE))) }`? — NelsonGon, May 03 '19 at 11:41

MatthewR · Accepted Answer · 2019-05-03T11:35:24.690

1

The issue has to do with variable class. Check the class - class( mtcars$mpg ) . This function will convert your cols to numeric

normalize<-function(x){

x <- as.numeric( as.character( x ))

y <- (x-min(x, na.rm=TRUE))/
    (max(x, na.rm=TRUE)-min(x, na.rm=TRUE))

return(y)

}

It works see: normalize( mtcars$mpg ) normalize( as.character( mtcars$mpg ))

You can add to your dataframe

mtcars$newvar <- normalize( mtcars$mpg )

or do several cols

mtcars[ , 3:7]<- sapply( mtcars[ , 3:7], normalize )

edited May 03 '19 at 11:35

answered May 03 '19 at 11:19

MatthewR

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Thanks @MatthewR. It is working but it removes the name of columns and rows. How should I deal with it? – Ester Silva May 03 '19 at 11:31
mtcars$newvar <- normalize( mtcars$mpg ) – MatthewR May 03 '19 at 11:33

score 1 · Answer 2 · edited May 03 '19 at 11:36

You can parse your normalize function through only numeric variables with mutate_if from dplyr.

 library(dplyr)

normalize<-function(x){
 return( (x-min(x, na.rm=TRUE)) / (max(x, na.rm=TRUE)-min(x, na.rm=TRUE)))
}

df %>%
   mutate_if(is.numeric, normalize)

Since you didn't provide a sample from your data, using iris dataset it would be something like:

iris %>% 
  mutate_if(is.numeric, normalize)

and the output would be (first ten rows):

#Sepal.Length Sepal.Width Petal.Length Petal.Width    Species
#1     0.22222222  0.62500000   0.06779661  0.04166667     setosa
#2     0.16666667  0.41666667   0.06779661  0.04166667     setosa
#3     0.11111111  0.50000000   0.05084746  0.04166667     setosa
#4     0.08333333  0.45833333   0.08474576  0.04166667     setosa
#5     0.19444444  0.66666667   0.06779661  0.04166667     setosa
#6     0.30555556  0.79166667   0.11864407  0.12500000     setosa
#7     0.08333333  0.58333333   0.06779661  0.08333333     setosa
#8     0.19444444  0.58333333   0.08474576  0.04166667     setosa
#9     0.02777778  0.37500000   0.06779661  0.04166667     setosa
#10    0.16666667  0.45833333   0.08474576  0.00000000     setosa

Thanks for the answer. Your answer works if we do not have columns with the same name. — Ester Silva, May 03 '19 at 11:40
This should work on `data.frame`. I'd recommend not using same name on different variables... — patL, May 03 '19 at 11:43

Normalizing a dataframe - Error : non-numeric argument to binary - R

2 Answers2