I have the dataframes df, df1, and the list l1 as follows :
df = data.frame(x = c(1,0,0,0,1,1,1,NA), y = c(2,2,2,2,3,3,2,NA),
z = c(1:7,NA), m = c(1,2,3,1,2,3,1,NA) )
df$x = factor(df$x)
df$y = factor(df$y)
df$m = factor(df$m)
df1 = df%>%select_if(is.factor)
l1 = lapply(df1,table(useNA = 'always'))
when trying to include the useNA
argument in l1 list, it doesn't work. However, when applying the argument for each variable like for example table(df1$x,useNA = 'always')
, it works properly. Would appreciate the help to handle this error.
CodePudding user response:
The proper format for lapply is "lapply(X, FUN, ...)" where the ... are for the optional arguments for the function.
So in this case the argument: useNA = 'always'
becomes:
lapply(df1, table, useNA = 'always')