# [R] adding vector values to corresponding components of a list

David L Carlson dcarlson at tamu.edu
Tue Dec 15 18:19:33 CET 2015

```With each post you seem to be changing what you are trying to accomplish. Your WRONG way seems pretty close if I understand you at all. Just tag the names to the numeric vector instead of the logical vector, eg:

> names(dbv) <- letters[1:8]
> split(dbv,logmat1)
\$`FALSE`
a        d        g        h
1.540183 1.172661 1.038135 1.172661

\$`TRUE`
b        c        e        f
1.295202 1.038135 1.540183 1.295202

But the other problem is that you may not realize what is happening. Using logmat1 (a matrix) as a vector in split means that it is converted by columns:

> logmat1
[,1]  [,2] [,3]  [,4]
[1,] FALSE  TRUE TRUE FALSE
[2,]  TRUE FALSE TRUE FALSE

> as.vector(logmat1)
 FALSE  TRUE  TRUE FALSE  TRUE  TRUE FALSE FALSE

Earlier it seemed you wanted to combine the by rows:

c(logmat1[1,], logmat[2,])
 FALSE  TRUE  TRUE FALSE  TRUE FALSE  TRUE FALSE

As you can see, it makes a difference.

-------------------------------------
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352

-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of debra ragland via R-help
Sent: Tuesday, December 15, 2015 9:56 AM
To: R-help
Subject: [R] adding vector values to corresponding components of a list

Hello,

I have tried this 2 ways and I keep coming to a dead end as I am not very proficient in R.

I have a logical matrix, where I would like to generate every row-wise pair of logical values for further testing. That is row1, row2; row 1, row3 etc. Ideally, I would like to assign a pre-generated vector of values to each pair such that each combination of rows may be split into 2 groups with each variable appearing only once.

This is similar to the WRONG way I've been trying to do this;

vector = runif(4, 1.0, 2.0)
logmat<-matrix(c(rep(c(F,T,F),3), rep(c(T,F,T),3), rep(c(T,T,F),3), rep(c(F,F,T),3)), ncol=4)
logmat1 = logmat[1:2,]
dbv=rep(vector, 2)
split(dbv,logmat1)

As you can see there's no proper distinction between what should be absolutely assigned to "TRUE" and what should be assigned to "FALSE". --Note my actual data is named.

To try and correct this mistake on a subset of my data I've been trying to use what I'll call "method 1" where

a = c(TRUE, TRUE, TRUE, FALSE, FALSE, TRUE)
names(a)=c(a,b,c,d,e,f)

b = c(TRUE, TRUE, FALSE, FALSE, TRUE, FALSE)
names(b)=c("a","b","c","d","e","f")

Using this function, I am able to separate which variables are exactly "TRUE" or "FALSE

logical_groups = function (a, b) {
r = list()
s = a + b
r\$'TRUE' = names(subset(s, s == 2))
r\$'FALSE' = names(subset(s, s == 0))
r\$'OR' = names(subset(s, s==1))
return(r)
}

> logical_groups(a,b)
\$`TRUE`
 "a" "b"

\$`FALSE`
 "d"

\$OR
 "c" "e" "f"

The return, as you can see is a list.
I have a vector of values that match the names a-f in a and b above, i.e.
vector=runif(6, 1.0, 2.0)
names(vector) = c("a","b","c","d","e","f")

and I would like to add these values to the corresponding (names in the) list so that I may run the wilcox.test correctly using the 'TRUE' and 'FALSE elements. But I am not sure how to achieve this.

I have also tried "method 2"-- starting with;
t=rbind(a, b, vector) to give;
>t
a        b        c       d        e        f
a      1.000000 1.000000 1.000000 0.00000 0.000000 1.000000
b      1.000000 1.000000 0.000000 0.00000 1.000000 0.000000
vector 1.012097 1.431088 1.832276 1.12801 1.780018 1.682804

where I then try
t2 =t[1,]+t[2,]
t3=rbind(t2, t[3,]) to give;
> t3
a        b        c       d        e        f
t2 2.000000 2.000000 1.000000 0.00000 1.000000 1.000000
1.012097 1.431088 1.832276 1.12801 1.780018 1.682804

But again, I am not sure how to split the t3 matrix where the first row == 2 or == 0.

I have been searching but I think I'm just confusing myself. I am very late in my pregnancy and my brain just won't function fully.

I hope this makes sense. Any help is appreciated.

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