UNIX for Dummies Questions & Answers

Hello all,

I have a .csv file with over 100 columns. I would like to grep for a pattern only searching within a range of those fields, and print the entire line. For example: grep a pattern from columns $47-$87, but print fields $0 - $100

Thanks!

grep doesn't understand columns. awk does, however.

Is it an actual csv, as in, comma-separated?

Code:

awk -F"," '{ M=0; for(N=47; (!M) && (N<=87); N++) if($N ~ /regex/) M++; } M' filename

Thank you for such quick reply. Yes it is comma separated. Will this print every column? And I'm assuming 'regex' is where i will put my desired pattern?

It will print every column, and /regex/ is where you put your pattern, yes. The // are what you use instead of quotes, sort of like in sed.

This command is working, but it seems to stop searching after line 34. My file has about 32,000 rows, and only 34 were printed. I know there will be at least 5000 rows containing my pattern.

Please post enough of your data to show the problem, and please show the exact program you ran, including the regex.

Unfortunately I can't really show the data, it has patient information . To be exact, I have 83 columns/fields and 32,224 rows/lines of data - comma seperated delimiter. I'm interested in the regex /hom/, but only as it appears in fields 43-83.

for example, a few lines will look like this:

And I don't want to return the last line, which contains a "hom" in col 1 - 42

The code I used:
This code works, but my output file hom.csv only has 34 lines returned. By visual inspection, this is only the first 34 instances of "hom" in columns 43-83. I know by looking at my file that in line 23743, there is a "hom".

Any reason why this may be happening? Or will you have to see the entire file?

Thanks