I'm trying to run the following code , but have error. Please suggest!
df$ppiedu2[df$ppi_mm_educ==12]<- df$education[df$ppi_mm_educ==12]
Error in df$ppiedu2[df$ppi_mm_educ == 12] <- df$education[df$ppi_mm_educ == :
NAs are not allowed in subscripted assignments
What is the result of
any(is.na(df$ppi_mm_educ==12))
If that returns TRUE, then at least one of the subscript values you are passing is NA.
below is my dataset, once ppi_mm_educ==12, replace ppiedu2 with education like df$ppiedu2[df$ppi_mm_educ==12]<- df$education[df$ppi_mm_educ==12]
| exdf= data.frame(ppiedu2=c(1, | 2, | 3, | 4, | 5, | 6, | 7, | 8, | 9, | 5, | 5, | NA, | 5, | NA, | 5, | 1, | NA, | 1, | 6, | 5), |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| education=c(5, | 5, | 5, | 7, | 6, | 6, | 5, | 5, | 7, | 5, | 5, | 6, | 5, | 5, | 5, | 5, | 6, | 5, | 5, | 1), |
| ppi_mm_educ=c(6, | 7, | 6, | 5, | 5, | 5, | 1, | 5, | 6, | 12, | 13, | NA, | NA, | 6, | 12, | 5, | 12, | 12, | 3, | 4)) |
Is this what you are trying to do?
exdf= data.frame(ppiedu2=c(1, 2, 3, 4, 5, 6, 7, 8, 9, 5, 5, NA, 5, NA, 5, 1, NA, 1, 6, 5),
education=c(5, 5, 5, 7, 6, 6, 5, 5, 7, 5, 5, 6, 5, 5, 5, 5, 6, 5, 5, 1),
ppi_mm_educ=c(6, 7, 6, 5, 5, 5, 1, 5, 6, 12, 13, NA, NA, 6, 12, 5, 12, 12, 3, 4))
exdf$ppiedu2 <- ifelse(exdf$ppi_mm_educ == 12, exdf$education, exdf$ppiedu2)
I got the right way without error, and thanks for your help!!
df$ppiedu2[df$ppi_mm_educ==12 & !is.na(df$ppi_mm_educ)]<- df$education[df$ppi_mm_educ==12 & !is.na(df$ppi_mm_educ)]
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