matrix assignment failing within lapply

问题

I have a data.frame with character data, I want to end up with a matrix with the same column headings but with counts for each value. So far I can get an empty matrix of the dimensions I want, but when I try to populate myMatrix with counts, it doesn't work.

myData <- data.frame(a=LETTERS[5:8], b=LETTERS[6:9], c=rep(LETTERS[5:6],2), d=rep(LETTERS[7],4))
#   a b c d
# 1 E F E G
# 2 F G F G
# 3 G H E G
# 4 H I F G
myValues <- sort(unique(unlist(myData))) # E F G H I
myList <- lapply(myData, table)
myMatrix <- matrix(nrow=length(myValues), ncol=length(myList), dimnames=list(myValues,names(myList)))
#    a  b  c  d
# E NA NA NA NA
# F NA NA NA NA
# G NA NA NA NA
# H NA NA NA NA
# I NA NA NA NA

So far so good. This is the part that doesn't do what I expect:

lapply(seq_along(myList), function(i) {myMatrix[names(myList[[i]]),names(myList[i])] <- myList[[i]]})

It returns the right values, but myMatrix is still full of NAs. Oddly, this one works:

myMatrix[names(myList[[2]]),names(myList[2])] <- myList[[2]]
#    a  b  c  d
# E NA NA NA NA
# F NA  1 NA NA
# G NA  1 NA NA
# H NA  1 NA NA
# I NA  1 NA NA

Why is the assignment to myMatrix failing within lapply and how can I get it to work (without a for loop)?

回答1:

@orizon is correct about why your use of lapply is not working as you expected. You would have to replace <- with <<- for it to work but it is in general considered bad practice for *apply functions to have such side-effects.

Instead, you can use

sapply(lapply(myData, factor, unique(unlist(myData))), table)

#   a b c d
# E 1 0 2 0
# F 1 1 2 0
# G 1 1 0 4
# H 1 1 0 0
# I 0 1 0 0

回答2:

Here is an approach that will return a data.frame

# create table, convert to data.frames then give appropriate column names
myList <- Map(setNames, lapply(lapply(myData, table), data.frame),  Map(c, 'Var', names(myList)))
# merge recursively
Reduce(function(...) merge(..., by = 'Var', all = T), myList)
  Var  a  b  c  d
1   E  1 NA  2 NA
2   F  1  1  2 NA
3   G  1  1 NA  4
4   H  1  1 NA NA
5   I NA  1 NA NA

回答3:

A single call to table can get the desired result, once you collapse everything back to two vectors. 1 vector for the values in the data.frame, 1 vector for the column identifier using col:

table(unlist(myData), colnames(myData)[col(myData)])

Result:

来源：https://stackoverflow.com/questions/13735525/matrix-assignment-failing-within-lapply

标签

lapply