Hi community!!!
I have a metaphlan output file from where I want to extract the first two lines (starting with "ID" and "#SampleID") along with all the lines that contain "s__" in that line. And, want to put them in a new file. Can anyone please tell me how can I do that?
I have used grep 's__' > extracted_file.txt. But it does not keep the first two lines in the output. Here is an example of the output:
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dpc
Hi,
Try the following:
grep "#SampleID\|ID\|s__" <target.file> > extracted_file.txt
Grep allows multiple matches. In this case you are saying match #SampleID or (sign - \|) ID or s__.
I hope this helps,
António
i'd add to that and say
grep "^#SampleID\|^ID\|s__" <target.file> > extracted_file.txt
to make sure all matches are at the beginning of the line. you can also use option -P so you don't have to escape the | operator
Thank you @from the mountains!
That makes sense, though for the s__ match I believe that @dpc wants to identify that pattern in the whole line, not necessarily at the beginning.
I did not know that -P option. Thanks for pointing out.
António
i've edited my comment for clarity.
Hi @dcp,
You don't need to add the -P option. The -P option is to deal with escape, according to @from the mountains! (I never tried it myself)
So, the idea is the following: since you want to match several words, i.e., match #SampleID or ID or s__, you need to add the special character | that means or, matching #SampleID|ID|s__, but as this is a special character (used as pipe in shell/bash), the shell deal with it in a different way, therefore you need to escape it, i.e., add the sign \ before | - \|.
What @from the mountains suggested (I did not know that) is that you don't need to escape if you use the -P option before.
So, the command-line should work well without the -P option.
If that solves your problem, please upvote the answer.
I hope this answers your question,
António
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you can try
awk 'NR < 3 || /s__/{print}' stackoverflow.tsvorsed -n '1,2p;/s__/p' stackoverflow.csv