Here is one simple option in base R, where we replicate the values 8, 9 to make the lengths same and compare with the subset of columns of 'df1', resulting in a logical matrix. Subset the 'df2' and assign those columns to 1
df2[nm1] <- Map(function(x, y, z) replace(x, y == z, 1),
df2[nm1], df1[nm1], c(8, 9))
The if/else loop can be also done, but it is not vectorized i.e. it expects input to be of length 1. If we do a loop, then it can be done (but would be inefficient in R)
vals <- c(8, 9)
for(i in seq_len(nrow(df1))) {
for(j in seq_along(nm1)) {
if(df1[[nm1[j]]][i] == vals[j]) df2[[nm1[j]]][i] <- 1
}
}
Here is one
simpleoption inbase R, where we replicate the values 8, 9 to make the lengths same and compare with the subset of columns of 'df1', resulting in a logical matrix. Subset the 'df2' and assign those columns to 1或者这可以分两步完成
Or using
MapThe
if/elseloop can be also done, but it is not vectorized i.e. it expects input to be of length 1. If we do a loop, then it can be done (but would be inefficient inR)数据
可以使用Base R ifelse语句完成一个简单的两行代码