sum cells of certain columns for each row

Question

I would like to calculate sums for certain columns and then apply this summation for every row. Unfortunately, I can only get to the first step. How do I now make it happen for each row? I know that R doesn't need loops; what are good approaches?

My matrix (zscore) looks like this:

   a    b    c    t   y
1  3    4    7    7   4

2  4    56   6    6   4

3  3    3    2    1   7

4  3    88   9    9   9

Now I would want to calculate the row sum for each row, based on some of the columns. For one row it could look like this:

f1 <- sum(zscore[1,1:2], zscore[1,3], zscore[1,5])

How do I do that now for each row?

score 55 · Accepted Answer · answered Nov 20 '12 at 21:37

55

You could do something like this:

summed <- rowSums(zscore[, c(1, 2, 3, 5)])

answered Nov 20 '12 at 21:37

alestanis

21,519
4
48
67

AWESOME, I did it with a loop in the meantime, but your solution is WAY cleaner! – user1807857 Nov 20 '12 at 21:52

score 3 · Answer 2 · answered Oct 26 '20 at 08:23

The summation of all individual rows can also be done using the row-wise operations of dplyr (with col1, col2, col3 defining three selected columns for which the row-wise sum is calculated):

library(tidyverse)

df <- df %>% 
    rowwise() %>% 
    mutate(rowsum = sum(c(col1, col2,col3)))

score 0 · Answer 3 · answered Sep 21 '16 at 16:13

0

If you don't have NA you can apply this

suma.zscore = (zscore$a + zscore$c + zscore$t + zscore$y)

answered Sep 21 '16 at 16:13

Edwin Torres

117
2
6

1

depending on the dataframe or matrix, as.numeric maybe necessary to get the addition to work such as: suma.zscore = (as.numeric(zscore$a) + as.numeric(zscore$c) ... etc) – Phil_T Sep 02 '17 at 16:21

sum cells of certain columns for each row

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