I want to replace the NA value in dfABy from the column A, with the value from the column B, based on the year of column year. For example, my df is:
Perhaps the easiest to read/understand answer in R lexicon is to use ifelse. So borrowing Richard's dataframe we could do:
df <- structure(list(A = c(56L, NA, NA, 67L, NA),
B = c(75L, 45L, 77L, 41L, 65L),
Year = c(1921L, 1921L, 1922L, 1923L, 1923L)),.Names = c("A",
"B", "Year"), class = "data.frame", row.names = c(NA, -5L))
df$A <- ifelse(is.na(df$A), df$B, df$A)
The new dplyr function, coalesce, can really simplify these situations.
library(dplyr)
dfABy %>%
coalesce(A,B)
You could use simple replacement with [<-, subsetting for the NA elements.
df$A[is.na(df$A)] <- df$B[is.na(df$A)]
Or alternatively, within()
within(df, A[is.na(A)] <- B[is.na(A)])
Both give
A B Year
1 56 75 1921
2 45 45 1921
3 77 77 1922
4 67 41 1923
5 65 65 1923
Data:
df <- structure(list(A = c(56L, NA, NA, 67L, NA), B = c(75L, 45L, 77L,
41L, 65L), Year = c(1921L, 1921L, 1922L, 1923L, 1923L)), .Names = c("A",
"B", "Year"), class = "data.frame", row.names = c(NA, -5L))
Easy
library(dplyr)
dfABy %>%
mutate(A_new =
A %>%
is.na %>%
ifelse(B, A) )
The solution provided by GGAnderson did return an error message. Using it inside mutate() however worked fine.
df <- structure(list(A = c(56L, NA, NA, 67L, NA),
B = c(75L, 45L, 77L, 41L, 65L),
Year = c(1921L, 1921L, 1922L, 1923L, 1923L)),
.Names = c("A", "B", "Year"),
class = "data.frame",
row.names = c(NA, -5L))
df
df%>%
coalesce(A,B) #returns error
df %>%
mutate(A = coalesce(A,B)) #works
(I am new to Stackoverflow; My low reputation does not allow to comment on GGAnderson´s answer directly)