Divide each data frame row by vector in R

问题

I'm trying to divide each number within a data frame with 16 columns by a specific number for each column. The numbers are stored as a data frame with 1-16 corresponding to the samples in the larger data frames columns 1-16. There is a single number per column that I need to divide by each number in the larger spreadsheet and print the output to a final spreadsheet.

Here's and example of what I'm starting with. The spreadsheet to be divided.

            X131.478.1 X131.478.2 X131.NSC.1 X131.NSC.2 X166.478.1 X166.478.2
1/2-SBSRNA4          4          2          2          6          7          6
A1BG                93         73         88         86         58         65
A1BG-AS1           123        103         96        128         46         57

The numbers to divide the spreadsheet by

X131.478.1 1.0660880
X131.478.2 0.9104053
X131.NSC.1 0.8642545
X131.NSC.2 0.9611866
X166.478.1 0.9711406
X166.478.2 1.0560121

And the expected results, not necessarily rounded as I did here.

    X131.478.1 X131.478.2 X131.NSC.1 X131.NSC.2 X166.478.1 X166.478.2
1/2-SBSRNA4          3.75          2.19          2.31          6.24          7.20         5.68
A1BG                87.23         80.17         101.82         89.47         59.72         61.55
A1BG-AS1           115.37        113.13         111.07        133.16         47.36         53.97

I tried simply dividing the data frames mx2 = mx/sf with mx being the large data set and sf being the data frame of numbers to divide by. That seemed to divide everything by the first number in the sf data set.

The numbers for division were generated by estimateSizeFactors, part of the DESeq package if that helps.

Any help would be great. Thanks!

回答1:

sweep is useful for these sorts of operations, but it requires a matrix as input. As such, convert your data frame to a matrix, do the operation and then convert back. For example, some dummy data where we divide each element in respective columns of matrix mat by the corresponding value in the vector vec:

mat <- matrix(1:25, ncol = 5)
vec <- seq(2, by = 2, length = 5)

sweep(mat, 2, vec, `/`)

In use we have:

> mat
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    6   11   16   21
[2,]    2    7   12   17   22
[3,]    3    8   13   18   23
[4,]    4    9   14   19   24
[5,]    5   10   15   20   25
> vec
[1]  2  4  6  8 10
> sweep(mat, 2, vec, `/`)
     [,1] [,2]     [,3]  [,4] [,5]
[1,]  0.5 1.50 1.833333 2.000  2.1
[2,]  1.0 1.75 2.000000 2.125  2.2
[3,]  1.5 2.00 2.166667 2.250  2.3
[4,]  2.0 2.25 2.333333 2.375  2.4
[5,]  2.5 2.50 2.500000 2.500  2.5
> mat[,1] / vec[1]
[1] 0.5 1.0 1.5 2.0 2.5

To convert from a data frame use as.matrix(df) or data.matrix(df), and as.data.frame(mat) for the reverse.

回答2:

Suppose we have a dataframe, df:

And we want to divide through each row by the same vector, vec:

> vec <- df[1,]
> vec
  a b   c
1 1 3 100

Then we can use mapply as follows:

> mapply('/', df, vec)
     a        b   c
[1,] 1 1.000000 1.0
[2,] 2 1.333333 1.1

回答3:

Just for variety, you could also use mapply

mx <- structure(list(X131.478.1 = c(4L, 93L, 123L), X131.478.2 = c(2L, 
73L, 103L), X131.NSC.1 = c(2L, 88L, 96L), X131.NSC.2 = c(6L, 
86L, 128L), X166.478.1 = c(7L, 58L, 46L), X166.478.2 = c(6L, 
65L, 57L)), .Names = c("X131.478.1", "X131.478.2", "X131.NSC.1", 
"X131.NSC.2", "X166.478.1", "X166.478.2"), class = "data.frame", row.names = c("1/2-SBSRNA4", 
"A1BG", "A1BG-AS1"))

sf <- structure(list(V1 = c(1.066088, 0.9104053, 0.8642545, 0.9611866, 
0.9711406, 1.0560121)), .Names = "V1", row.names = c("X131.478.1", 
"X131.478.2", "X131.NSC.1", "X131.NSC.2", "X166.478.1", "X166.478.2"
), class = "data.frame")


mapply(function(x, y) x * y, mx, t(sf))


    X131.478.1 X131.478.2 X131.NSC.1 X131.NSC.2 X166.478.1 X166.478.2
[1,]   4.264352   1.820811   1.728509    5.76712   6.797984   6.336073
[2,]  99.146184  66.459587  76.054396   82.66205  56.326155  68.640787
[3,] 131.128824  93.771746  82.968432  123.03188  44.672468  60.192690

But for this I think Josh's answer is better... and Gavin's is even better!

回答4:

This is nothing but element-wise matrix multiplication:

mat <- matrix(c(4,2,2,6,7,6, 93,73,88,86,58,65, 123,103,96,128,46,57), nrow=3, byrow=T)

vec = c(1.0660880,0.9104053,0.8642545,0.9611866,0.9711406,1.0560121)

mat %o% 1/vec

           [,1]      [,2]       [,3]       [,4]      [,5]      [,6]
[1,]   3.752035  2.080761   1.876018   6.242284  6.566062  6.242284
[2,] 102.152305 75.169342  96.660246  88.555663 63.707889 66.931606
[3,] 142.319190 97.536761 111.078392 121.210732 53.225063 53.976654

To do that we used the outer-product approach, since directly trying mat %*% 1/vec gives an error on non-conformable arguments because they have different shapes. Or look at the many posts on https://stackoverflow.com/search?q=%5Br%5D+multiply+matrix+by+vector

回答5:

You could use transform

mx2 <- transform(mx, 
    X131.478.1=X131.478.1/sf["X131.478.1",1],
    X131.478.2=X131.478.2/sf["X131.478.2",1],
    etc
)

Quite a bit to type with 16 columns, but it should work.

来源：https://stackoverflow.com/questions/13830979/divide-each-data-frame-row-by-vector-in-r

标签

vector

dataframe

divide