# [R] matrix manipulations

Phil Spector spector at stat.berkeley.edu
Mon Jan 17 21:13:09 CET 2011

Monica -
Perhaps this small example can demonstrate how factors can

> d1 = data.frame(cat=sample(c('cat2','cat5','cat6'),100,replace=TRUE),group=sample(c('land','water'),100,replace=TRUE))
> d2 = data.frame(cat=sample(c('cat1','cat3','cat4'),100,replace=TRUE),group=sample(c('land','water'),100,replace=TRUE))
> d1\$cat = factor(d1\$cat,levels=c('cat1','cat2','cat3','cat4','cat5','cat6'))
> d2\$cat = factor(d2\$cat,levels=c('cat1','cat2','cat3','cat4','cat5','cat6'))
> table(d1\$group,d1\$cat) + table(d2\$group,d2\$cat)

cat1 cat2 cat3 cat4 cat5 cat6
land    14   17   18   22   19   23
water   19   15   16   11   10   16

This works because when you include all possible levels in a factor, R will
automatically put zeroes in the right places when you use table():

> table(d1\$group,d1\$cat)
cat1 cat2 cat3 cat4 cat5 cat6
land     0   17    0    0   19   23
water    0   15    0    0   10   16
> table(d2\$group,d2\$cat)
cat1 cat2 cat3 cat4 cat5 cat6
land    14    0   18   22    0    0
water   19    0   16   11    0    0

Hope this helps.
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spector at stat.berkeley.edu

On Mon, 17 Jan 2011, Monica Pisica wrote:

>
> Hi,
>
> I am having some difficulties with matrix operations. It is a little hard to explain it so please bear with me. I have a very large data set, large enough that it needs to be split in parts in order to deal with. I can work things on these "parts" but the problem lies in adding together these parts for the final answer.
>
> So that been said, let's say that i split the data in 2 parts, 1 and 2. Each part has data belonging to 6 different categories, and each category has 2 different classes, these classes being the same for each category. The classes are called "land" and "water" and each category is labeled "cat1" to "cat6". I am using the command (function) table to tabulate each class for each category, but since i split the data in 2 parts, one part has only some of the 6 categories, and the other some other of the 6 categories (and not necessarily exclusive).
>
> So let's built some results after i used the table function.
>
> m1 <- matrix(c(32, 35, 36, 12, 15, 16), nrow = 2, ncol = 3, byrow = TRUE, dimnames = list(c("land", "water"), c("cat2", "cat5", "cat6")))
>
>> m1
>     cat2 cat5 cat6
> land 32    35   36
> water 12   15   16
>
> m2 <- matrix(c(45, 46, 47, 48, 21, 22, 23, 24), nrow = 2, ncol = 4, byrow = TRUE, dimnames = list(c("land", "water"), c("cat1", "cat2", "cat3", "cat4")))
>
>> m2
>     cat1 cat2 cat3 cat4
> land  45   46   47   48
> water 21   22   23   24
>
> So my end desired result should be a matrix (or a data frame) that has 6 columns called cat1 to cat6 and 2 rows labeled land and water, and for the category that appears in both m1 and m2 the end result will be a sum.
>
> results will be m3:
>
>     cat1 cat2 cat3 cat4 cat5 cat6
> land  45  78   47    48   35   36
> water 21  34   23    24   15   16
>
> To do this i thought in making an empty matrix for each m1 and m2 (called m01 and m02 respectively) with 6 columns and 2 rows, and do a long if else statement in which i match the name of the first column in m1 with the name of the first column in m01 and if they match get the data from m1, if not leave it 0 and so on. Same thing for m2 and m02. This is long and extremely clunky but afterwards i can add m01 with m02 and get my desired result m3. Is there any way i can do this more elegantly? My real data is split in 4 parts, but the problem is the same.
>
> Thanks for all your inputs, and sorry for this long email, but i didn't know how else i could explain what i wanted to do.
>
> Monica
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help