Making pairwise matrix using ChIP-Seq peak binding matrix in R

Question

[Ankit]

Hi everyone,

I have a query about R,

How to convert this

     Protein1 Protein2 Protein3
chr1_1564 1 0 0
chr3_9087 0 1 1
chr4_877671 1 1 0
chr9_90988 0 1 1
chr11_87676 1 0 0
chrX_1546 0 1 1

to this

        Protein1 Protein2 Protein3
Protein1 3 1 0
Protein2 1 4 3
Protein3 0 3 3

using R based functions which can do it quickly?

Further explaination, and relevance to biology and bioinformatics:

On rows are genomic sites and columns are protein names. This is a ChIP-seq data for some of my proteins for which I checked the occupancy on some genomic bins and took the start positon and chr name to give an ID to interval. 0 and 1 shows presence or absence of protein on that genomic sites.

I have applied a for-loop but it is taking a long time for large data. Here is my script as of now which is taking long time for a data of 900000 rows and 500 columns (900000 genomic sites and 500 proteins)

mydf <- data.frame(Protein1= c(1,0,1,0,1,0), Protein2=c(0,1,1,1,0,1), Protein3=c(0,1,0,1,0,1))
mydf <- as.matrix(mydf)
# Create empty matrix to store data
converted_mat <- matrix(0, nrow = ncol(mydf), ncol = ncol(mydf))
rownames(converted_mat)  <- colnames(mydf)
colnames(converted_mat)  <- colnames(mydf)

for (i in 1:ncol(mydf)){
  for  (j in 1:ncol(mydf)){
    converted_mat[i,j] <- sum(ifelse(mydf[,i] == 1 & mydf[,j] == 1, 1,0))
  }
}

Any suggestions?

[Ram]

How is this related to bioinformatics? If it is related, please add necessary context. If not, please delete this question and consult StackOverflow.

[Ankit]

It is a bioinformatics query about binding pattern of proteins across genomic regions.

[jared.andrews07]

If you add more information to your question, we may be able to provide better contextual advice for your actual end goal.

[Ram]

Please edit your post and add as much biological context as it takes for your post to make sense. If not, the post will be removed as off-topic.

[Ankit]

The detailed explanation has been added to the post along with biological context. If I am still missing things let me know,

Answer 1

[jared.andrews07]

Use matrix multiplication:

input_matrix <- matrix(c(1, 0, 0,
                         0, 1, 1,
                         1, 1, 0,
                         0, 1, 1,
                         1, 0, 0,
                         0, 1, 1), nrow = 6, byrow = TRUE)

colnames(input_matrix) <- c("a", "b", "c")

# Compute the co-occurrence matrix
cooccurrence_matrix <- t(input_matrix) %*% input_matrix

# Assign column and row names
colnames(cooccurrence_matrix) <- c("a", "b", "c")
rownames(cooccurrence_matrix) <- c("a", "b", "c")

print(cooccurrence_matrix)
  a b c
a 3 1 0
b 1 4 3
c 0 3 3

Didn't benchmark, but should be much faster.

Answer 2

[bk11]

How about using crossprod function. I have not tested how fast it will be though.

mydf <- data.frame(a= c(1,0,1,0,1,0), b=c(0,1,1,1,0,1), c=c(0,1,0,1,0,1))
mydf <- as.matrix(mydf)
out <- crossprod(mydf)
out
a b c
a 3 1 0
b 1 4 3
c 0 3 3

[Ankit]

Both are good suggestions as they are much faster. Thank you.

[Ram]

^ This is the downside to answering questions where OP has not invested sufficient effort - they take the answer and run and in all probability will come back with a different XY problem where X remains the same.

Apologies to both bk11 and jared.andrews07, but I'll delete this entire post unless OP adds context.

[Ankit]

The detailed explanation has been added to the post along with biological context. If I am still missing things let me know,