CODEWITHSUNDEEP
rdataframelapplynamapply
jonnie
jonnie

Reputation: 765

use one list to replace values in another

How can I use one list of vectors to replace the values in another list of data frames? Ultimately, I want a data frame that has certain values replaced by NAs.

list1 <- list(1:2, 3:4, 5:6)
list2 <- as.list(data.frame(mat.or.vec(6, 3)))

desired_ouput <- data.frame(mat.or.vec(6, 3))
desired_ouput[1:2, 1] <- NA
desired_ouput[3:4, 2] <- NA
desired_ouput[5:6, 3] <- NA
desired_ouput

enter image description here

Upvotes: 1

Views: 391

Answers (3)

Zheyuan Li
Zheyuan Li

Reputation: 73325

This may not look obvious, but is rather efficient (fully vectorized, with no for loop or *apply family)

y <- unlist(list2, use.names = FALSE); m <- length(list2[[1]])
x <- unlist(list1) + rep((0:(length(list1)-1)) * m, lengths(list1))
y[x] <- NA; attr(y, "dim") <- c(m, length(y) / m)
as.data.frame(y)

#   V1 V2 V3
# 1 NA  0  0
# 2 NA  0  0
# 3  0 NA  0
# 4  0 NA  0
# 5  0  0 NA
# 6  0  0 NA

Why do this? From OP's statement, I can know that list1 can be ragged, while list2 can not (because later OP wants to turn the updated list2 into a data frame). Therefore, there is a way to use fast indexing to solve this problem.

Thanks to @alexis_laz's comment, that I could use

y <- unlist(list2, use.names = FALSE)

instead of

y <- unname(unlist(list2))

Upvotes: 4

lmo
lmo

Reputation: 38510

Here's a method that uses Map:

myDf <- data.frame(Map(function(x, y) {y[x] <- NA; y}, list1, list2))
names(myDf) <- paste0("X", 1:ncol(myDf))

which returns

myDf
  X1 X2 X3
1 NA  0  0
2 NA  0  0
3  0 NA  0
4  0 NA  0
5  0  0 NA
6  0  0 NA

Map produces a list object, which is wrapped in data.frame to return the proper object. However, it is necessary to provide a nicer name to each column with names.

Upvotes: 2

Bulat
Bulat

Reputation: 6979

Just use for:

list1 <- list(1:2, 3:4, 5:6)
df <- data.frame(mat.or.vec(6, 3))

for(i in 1:length(list1)) {
  df[list1[[i]], i] <- NA
}
# df
#    X1 X2 X3
# 1  NA  0  0
# 2  NA  0  0
# 3  0   NA 0
# 4  0   NA 0
# 5  0   0  NA
# 6  0   0  NA

Upvotes: 3

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