R help - Dec 2015 - 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`
[1] "a" "b"

$`FALSE`
[1] "d"

$OR
[1] "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.

I as not able to make sense of your post. Maybe others can. However,
you have posted here several times, now. You should therefore know
that providing a reproducible example (providing data by e.g. dput() )
**and showing exactly what output you want** would improve your chance
of a helpful response.

'Nuff said.

Cheers,
Bert


Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Tue, Dec 15, 2015 at 7:56 AM, debra ragland via R-help
<r-help at r-project.org> wrote:
> 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`
> [1] "a" "b"
>
> $`FALSE`
> [1] "d"
>
> $OR
> [1] "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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