如何对矩阵中的列求和并将值粘贴到 r 的最后一个单元格中？

Question

Is there a way to sum up all the values in from a column and paste them into the last "cell"?

Example: I have a matrix like:

  1  2  3
1 0  0  1
2 0  0  2
3 0  0  6

and I want (need) a Matrix like:

  1  2  3
1 0  0  0
2 0  0  0
3 0  0  9

In addition to that I need an iterative Approach, because my matrix dimensions vary.

Thank you in advance!

edit:

the following solution worked:

creating a second zero'd matrix with the same dimensions and the following code 
matrix2[nrow(matrix2),] <- colSums(m1)

Answer 1

We provide several solutions depending on the interpretation of the question. They all assume that a zero matrix is to be returned except for the last row. (1) sums just the last column, (2) sums column(s) i and (2a) sums all columns.

All solutions are one line except for defining nr and/or nc and preserve the original m matrix returning the new matrix.

1) Using m shown reproducibly in the note at the end multiply m by zero and then replace the last cell (which would be at position prod(dim(m)) , with the sum of the last column.

nc <- ncol(m)
replace(0 * is.na(m), prod(dim(m)), sum(m[, nc]))
##   1 2 3
## 1 0 0 0
## 2 0 0 0
## 3 0 0 9

If m is known not have NAs then is.na(m) can be replaced with m in the above code.

2) If the column is not necessarily last then if i is the index of the column to sum (or it can be a vector of the indexes of those columns to be summed should there be more than one) then:

i <- 3
nr <- nrow(m)
replace(0 * is.na(m), i * nr, colSums(m[, i, drop = FALSE]))

2a) If all columns are to be summed then i <- 1:nc so the above becomes:

nc <- ncol(m)
nr <- nrow(m)
replace(0 * is.na(m), (1:nc) * nr, colSums(m, na.rm = TRUE))

If you need to be able to handle matrices with zero columns then use seq_len(nc) instead of 1:nc .

Note

m <- matrix(c(0L, 0L, 0L, 0L, 0L, 0L, 1L, 2L, 6L), 3, dimnames = list(1:3, 1:3))

Answer 2

Suppose your matrix is given by

mat <- matrix(c(0,0,1,0,0,2,0,0,6,0,0,7), nrow=4, byrow = TRUE)

Then

mat_2 <- matrix(0L, nrow=dim(mat)[1], ncol=dim(mat)[2])

gives you matrix with same dimensions and containing just 0s . Now calculate the sum by

mat_2[dim(mat)[1],] <- apply(mat, 2, sum)

Answer 3

Another option is colSums

m1[nrow(m1),] <- colSums(m1, na.rm = TRUE)
m1[-nrow(m1),] <- 0
m1
#  1 2 3
#1 0 0 0
#2 0 0 0
#3 0 0 9

---

Or with addmargins

addmargins(m1, 1)[-1,] * rep(0:1, c(nrow(m1)- 1, 1))

data

m1 <-  matrix(c(0L, 0L, 0L, 0L, 0L, 0L, 1L, 2L, 6L), 3, dimnames = list(1:3, 1:3))