r - Choose one cell per row in data frame -

i have vector tells me, each row in date frame, column index value in row should updated.

> set.seed(12008); n <- 10000; d <- data.frame(c1=1:n, c2=2*(1:n), c3=3*(1:n)) > <- sample.int(3, n, replace=true) > head(d); head(i)   c1 c2 c3 1  1  2  3 2  2  4  6 3  3  6  9 4  4  8 12 5  5 10 15 6  6 12 18 [1] 3 2 2 3 2 1 

this means rows 1 , 4, c3 should updated; rows 2, 3 , 5, c2 should updated (among others). cleanest way achieve in r using vectorized operations, i.e, without apply , friends? edit: and, if @ possible, without r loops?

i have thought transforming d matrix , address matrix elements using one-dimensional vector. haven't found clean way compute one-dimensional address row , column indexes.

if willing first convert data.frame matrix, can index elements-to-be-replaced using two-column matrix. (beginning r-2.16.0, possible data.frames directly.) indexing matrix should have row indices in first column , column indices in second column.

here's example:

## create subset of data set.seed(12008); n  <- 6  d  <- data.frame(c1=1:n, c2=2*(1:n), c3=3*(1:n)) <- seq_len(nrow(d))            # vector of row indices j <- sample(3, n, replace=true)  # vector of column indices  ij <- cbind(i, j)                # 2-column matrix index 2-d array                                   # (this extends smoothly higher-d arrays.)    ## convert matrix     dmat <- as.matrix(d)  ## replace elements indexed 'ij' dmat[ij] <- na dmat #      c1 c2 c3 # [1,]  1  2 na # [2,]  2 na  6 # [3,]  3 na  9 # [4,]  4  8 na # [5,]  5 na 15 # [6,] na 12 18 

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beginning r-2.16.0, able use same syntax dataframes (i.e. without having first convert dataframes matrices).

from r-devel news file:

matrix indexing of dataframes 2 column numeric indices supported replacement extraction.

using current r-devel snapshot, here's looks like:

d[ij] <- na d #   c1 c2 c3 # 1  1  2 na # 2  2 na  6 # 3  3 na  9 # 4  4  8 na # 5  5 na 15 # 6 na 12 18