UNIX for Dummies Questions & Answers

I need to know the row number of one column with some value but I don't know how. For example I want to know where is 3 of the first column in:

Code:

2  4
3  5
4  6
5  7

So I should get 2.

I have tried this:

Code:

awk '{if(NF==1){for(i=1;i<=NR;i++){if($i == $3) {print i}}}}' test.ssv

But it is not working

Could someone help me please?

Code:

awk '$1==3{print NR}' infile

OR

Code:

awk '$1==3{print FNR}' infile

From man page

Code:

NR  is the number of input records awk has processed since the beginning of the program's execution. 
NR is set each time a new record is read

FNR is the current record number in the current file. FNR is incremented each time a new record is read . 
It is reinitialized to zero each time a new input file is started.

"NF" is the (total) number of fields, which in your test data is 2 for every record. Therefore "if(NF==1)" will never be true at all.What you want to express (i suppose) is not "if the number of fields equals 1", but "if the field with the number 1 [equals some value]".

The content of the field number x is stored in a variable named "$x": "$1" holds the value of the first field, "$2" the value fo the second field and so on. You might want apply this new insight to your problem.

I hope this helps.

bakunin

Thank you ! You're right

---------- Post updated at 09:59 AM ---------- Previous update was at 08:40 AM ----------

Sorry but I still have a problem because my data are too long and awk is too slow, do you know another way to solve it?

For a value in the first column (assuming space is your column separator):

Code:

grep -n '^3 ' file | cut -d: -f1

for the second column:

Code:

grep -n '^[^ ][^ ]*  *3 ' file | cut -d: -f1

and so on. Admittedly not as flexible as awk. Note there are two spaces between "*" and "*". Or you can:

Code:

cut -d' ' -fN file | awk '...'

But this assumes a SINGLE space between fields.

Please elaborate on "slow": what do you mean by that? What does your real data look like? how big is your file? And what does your real awk script look like (not the one you showed us, the one which you really use)?

There is no point in suggesting anything without understanding the details of the problem - which you completely left out.

@derekludwig: i do not think your solutions will be any faster than awk: because all your suggestions include a pipeline the data have first to be read and then some part of them has to be read again. Given, this part is read from a datastream and not from disk, but still this means some additional work for the machine instead of less.

I hope this helps.

bakunin

This

Code:

sed -n '/^3/=' file

is - on my system - about as fast as the awk above, but a factor of 2 - 3 slower than a grep alone.
If you're sure there's only one occurrence of your pattern, and it is early in the file, this

Code:

awk '$1==3{print NR; exit}' file

will accelerate the awk solution significantly.