[R] populating matrix with binary variable after matching data from data frame

[Adrian Johnson]

Hi.
Thank you for your help.
yes, thats exactly right - but the 1211x1211 matrix has some
row/column elements that may not be present in x1.
Is that the reason I get this error?

My matrix row names and column names are identical. I changed the
order in my dput code for representational purpose so that they can
have 1 for conveying question easily.

Thanks


    A  B C D E
A
B
C
D




> for (i in nrow(x1)) {
+   x[x1$V1[i], x1$V2[i]] <- 1;
+ }
Error in `[<-`(`*tmp*`, x1[i, ]$V1, x1[i, ]$V2, value = 1) :
  subscript out of bounds







On Wed, Aug 13, 2014 at 8:28 AM, John McKown
<john.archie.mckown at gmail.com> wrote:
> On Tue, Aug 12, 2014 at 7:14 PM, Adrian Johnson <oriolebaltimore at gmail.com>
> wrote:
>>
>> Hi:
>> sorry I have a basic question.
>>
>> I have a data frame with two columns:
>> > x1
>>       V1       V2
>> 1   AKT3    TCL1A
>> 2  AKTIP    VPS41
>> 3  AKTIP    PDPK1
>> 4  AKTIP   GTF3C1
>> 5  AKTIP    HOOK2
>> 6  AKTIP    POLA2
>> 7  AKTIP KIAA1377
>> 8  AKTIP FAM160A2
>> 9  AKTIP    VPS16
>> 10 AKTIP    VPS18
>>
>>
>> I have a matrix 1211x1211 (using some elements in x1$V1 and some from
>> x1$V2). I want to populate for every match for example AKT3 = TCL1A = 1
>> whereas AKT3 - VPS41 gets 0)
>> How can i map this binary relations in x.
>>
>>
>> >x
>>        TCLA1 VPS41 ABCA13 ABCA4
>> AKT3       0     0      0     0
>> AKTIP      0     0      0     0
>> ABCA13     0     0      0     0
>> ABCA4      0     0      0     0
>>
>
> <snip>
>
> I'm not totally sure that I understand your data structure. So I will
> rephrase a bit so that I can be corrected, if necessary. You have an
> 1211x1121 matrix already. Every cell in the matrix is initialized to 0. It
> has column names such as TCLA1, VPS41, ABCA13, ABCA4, ... and it has row
> names such as AKT3 AKTPI, ABCA13, ABCA4. The list "x1" has columns named V1
> and V2. V1 values are row names in the matrix. V2 values are column names in
> the matrix. The following should do what you want. It is not a _good_
> solution because it is iterative. But it is a start
>
> for (i in nrow(x1)) {
>    x[x1$V1[i], x1$V2[i]] <- 1;
> }
>
>>
>>
>> Thanks
>> Adrian
>>
>>         [[alternative HTML version deleted]]
>>
>
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>
> --
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>
> Maranatha! <><
> John McKown