I have a dataframe with several vectors of ordered factors that could take the values 1-4. v2 does not have any instances of value 3. Here's a basic example:
myData <- data.frame(v1=factor(c(1, 1, 2, 2, 3, 3, 4, 4),
levels=c(1, 2, 3, 4),
ordered=TRUE),
v2=factor(c(1, 1, 2, 2, 4, 4, 4, 4),
levels=c(1, 2, 3, 4),
ordered=TRUE))
myData
# v1 v2
# 1 1 1
# 2 1 1
# 3 2 2
# 4 2 2
# 5 3 4
# 6 3 4
# 7 4 4
# 8 4 4
levels(myData$v2)
# [1] "1" "2" "3" "4"
I needed to impute missing data, so I dropped unused levels across all columns (e.g., level 3 in v2):
myData <- droplevels(myData)
levels(myData$v2)
# [1] "1" "2" "4"
Now I want to create some Likert plots, but all columns in the dataframe need to have the same number of response options (i.e., the same number of levels).
I've defined the max levels found in the dataframe programmatically in an object called L. So in this case, the result is an object L:
L <- c(1, 2, 3, 4)
I need to make sure that every column in the dataframe has levels L. For v2, this will mean that there are 0 observations with level 3. I have more than just two columns, so I need to run the solution across all columns.
The apply style approaches I've tried just resulted in NAs.
myData[] <- lapply(myData, `levels<-`, L)? - David Arenburglevels<-. That's helpful. However, while this does add the missing level, 4's in1 1 2 2 4 4 4 4become 3's1 1 2 2 3 3 3 3. I need the values to remain the same1 1 2 2 4 4 4 4but have levelsLevels: 1 < 2 < 3 < 4(so adding back unused levels without changing values). - Eric GreenmyData[] <- lapply(myData, factor, levels = L)though it beats me what happening withlevels<-. - David Arenburg