CODEWITHSUNDEEP
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mhershey3
mhershey3

Reputation: 25

Fill data frame rows with vectors of varying column subsets of all columns in the data frame

Say I initialize an empty data frame like so:

df = data.frame(matrix(ncol = 5, nrow = 0))
colnames(df) <- c("c1","c2","c3","c4","c5")

And wish to populate the data frame rows with vectors such as

v1 =
c2 c3 c4                
6  4  3

v2 =
c1 c3 c5                
2  5  7

So that df will look like

c1 c2 c3 c4 c5
0  6  4  3  0
2  0  5  0  7

What's the most efficient way to go about this? The vectors are originally lists, which I was converting to a matrix and attempting to append df with so that the column labels are maintained throughout. However, I'm struggling to find an automated way to fill the missing element values since the columns themselves don't exist (ie v1 has no "c1" column).

Upvotes: 0

Views: 32

Answers (2)

Jonas
Jonas

Reputation: 1810

Assuming your vectors are still named lists, like

v1 <- as.list(setNames(c(6,4,3),c("c2","c3","c4")))
v2 <- as.list(setNames(c(6,4,3),c("c1","c3","c5")))

(so you don't have to do the step of processing them to vectors), you can use

df = data.frame(matrix(ncol = 5, nrow = 0))
colnames(df) <- c("c1","c2","c3","c4","c5")

listOfNamedValueLists <- list(v1,v2)
df <- do.call(plyr::rbind.fill,c(list(df),lapply(listOfNamedValueLists,data.frame)))
df[is.na(df)] <- 0

Note that appending df to the list as first element ensures you that the names are in right order.

Upvotes: 0

Konrad Rudolph
Konrad Rudolph

Reputation: 545608

The following works:

df[1L, c('c2', 'c3', 'c4')] = v1
df[2L, c('c1', 'c3', 'c5')] = v2
df[is.na(df)] = 0

If your vectors are named, it’s even easier:

df[1L, names(v1)] = v1
df[2L, names(v2)] = v2
df[is.na(df)] = 0

Upvotes: 0

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