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May be your question gets a little clearer if I reformat the data:

NP_041954.1 NP_848263.1 NP_041955.1 NP_041956.1
NP_041956.1 NP_041955.1 
NP_041957.1 NP_848264.1
[download]

Nope, it does not get clearer. The first row (row[0]) has four elements/columns; the second row (row[1]) and third row (row[2]) have two elements each.

Perhaps you mean the structure to be:

NP_041954.1 NP_848263.1
NP_041955.1 NP_041956.1
NP_041956.1 NP_041955.1 
NP_041957.1 NP_848264.1
[download]

That makes a bit more sense.

You need to print out the columns where column[0] of one row is equal to column[1] of another row and vice versa. The result should be NP_041955.1 and NP_041956.1, but that doesn't look right as these are not columns but elements. OK, we will assume you meant elements.

Assuming you are only interested in these elements, then the best and fastest way to do this is to store each column in its own hash and check each element of the first hash against the second hash and vice-versa and print out the matches.

CountZero

A program should be light and agile, its subroutines connected like a string of pearls. The spirit and intent of the program should be retained throughout. There should be neither too little or too much, neither needless loops nor useless variables, neither lack of structure nor overwhelming rigidity." - The Tao of Programming, 4.1 - Geoffrey James

In reply to Re: how to compare column 1 to column 2 and vice versa from multiple rows. by CountZero
in thread how to compare column 1 to column 2 and vice versa from multiple rows. by BhariD

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