I have an table/df that looks like this:
How can I replace 0 (0.00%)
to 0
? I tried str_replace_all
and it dosen't work. Any suggestion?
df<-structure(list(SUBJECT = c("PE", "PE", "PE", "PE", "PE",
"PE"), Class = c("3rd Grade", "3rd Grade",
"3rd Grade", "3rd Grade", "3rd Grade",
"3rd Grade"), Number_Student = c(7L, 7L, 7L, 7L,
7L, 7L), WGRADE = structure(1:6, .Label = c("Grade 0", "Grade 1",
"Grade 2", "Grade 3", "Grade 4", "Grade 5"), class = "factor"),
`Grade 0` = c("1 (14.29%)", "1 (14.29%)", "0 (0.00%)", "0 (0.00%)",
"0 (0.00%)", "0 (0.00%)"), `Grade 1` = c("0 (0.00%)", "3 (42.86%)",
"1 (14.29%)", "0 (0.00%)", "0 (0.00%)", "0 (0.00%)"), `Grade 2` = c("0 (0.00%)",
"0 (0.00%)", "0 (0.00%)", "1 (14.29%)", "0 (0.00%)", "0 (0.00%)"
), `Grade 3` = c("0 (0.00%)", "0 (0.00%)", "0 (0.00%)", "0 (0.00%)",
"0 (0.00%)", "0 (0.00%)"), `Grade 4` = c("0 (0.00%)", "0 (0.00%)",
"0 (0.00%)", "0 (0.00%)", "0 (0.00%)", "0 (0.00%)"), `Grade 5` = c("0 (0.00%)",
"0 (0.00%)", "0 (0.00%)", "0 (0.00%)", "0 (0.00%)", "0 (0.00%)"
)), row.names = c(NA, -6L), class = c("tbl_df", "tbl", "data.frame"
))
CodePudding user response:
dplyr
solution:
df %>%
mutate(
across(where(is.character), str_replace_all, '0 (0.00%)', '0')
)
CodePudding user response:
R-base
solution:
df[ df == "0 (0.00%)"] <- "0"