Parse lines with specific pattern out of file

Question

I have a file that looks roughly like this :

[25]:0.00843832,469:0.0109533):0.00657864,((((872:0.00120503,((980:0.0001
[29]:((962:0.000580339,930:0.000580339):0.00543993 ((758:0.000598847,726:0.000598847)
position:
sites: 5 4 2 1 3 4 543 5  67 657  78 67 8  5645 6 
01010010101010101010101010101011111100011
1111010010010101010101010111101000100000
00000000000000011001100101010010101011111

Now I would like to extract only those lines which start with [numeric]: from the file. It is not always only the first two, it could also be the first 7 or 8 or whatever. How would I read in this file and output a file only containing the lines with [numeric]:?

Answer 1

Using grep :

$ grep "^\[[0-9]\+\]:" file.txt 
[25]:0.00843832,469:0.0109533):0.00657864,((((872:0.00120503,((980:0.0001
[29]:((962:0.000580339,930:0.000580339):0.00543993 ((758:0.000598847,726:0.000598847)

To save the output in a file (output.txt):

grep "^\[[0-9]\+\]:" file.txt > output.txt

Using python:

#!/usr/bin/env python2
import re
with open('/path/to/file.txt') as f:
  print '\n'.join([line.rstrip() for line in f if re.search(r'^\[\d+\]:', line)])

Answer 2

The perl way:

perl -ne 'print "$1\n" if /^(\[[0-9]*\]:.*)/' testdata > out

The awk way:

awk 'match($0, /^\[[0-9]*\]:/)' testdata > out

---

Output for both commands

[25]:0.00843832,469:0.0109533):0.00657864,((((872:0.00120503,((980:0.0001
[29]:((962:0.000580339,930:0.000580339):0.00543993 ((758:0.000598847,726:0.000598847)

Answer 3

This task is perfectly suited for grep, because you're just checking which lines contain a match for a pattern and printing the lines that do.

heemayl's way is excellent. Here's another that's similar but uses Perl regular expression syntax (which GNU grep supports, with -P), for a shorter and slightly simpler pattern:

grep -P '\[\d+\]:' infile

That just prints the output, but you can redirect it to outfile:

grep -P '\[\d+\]:' infile > outfile

In Perl regular expressions, \d matches any single digit, same as [0-9] or [[:digit:]].

---

In case you're interested, here's a sed way:

sed -nr '/^\[[0-9]+\]:/p' infile

sed -nr '/^\[[0-9]+\]:/p' infile > outfile

That checks each line to see if it matches ^\[[0-9]+\]:. If it does, the sed command p is used to print the line. The -n flag prevents any lines from being printed except as provided for explicitly by the sed script.

Answer 4

If there is no posibility of a [non-numeric] at the beginning of your file lines, then simply grep -E '^\[' will do the trick, viz.:

$ cat /tmp/tmp.tmp
[25]:0.00843832,469:0.0109533):0.00657864,((((872:0.00120503,((980:0.0001
[29]:((962:0.000580339,930:0.000580339):0.00543993 ((758:0.000598847,726:0.000598847)
position:
sites: 5 4 2 1 3 4 543 5  67 657  78 67 8  5645 6
01010010101010101010101010101011111100011
1111010010010101010101010111101000100000
00000000000000011001100101010010101011111

$ grep -E '^\[' /tmp/tmp.tmp
[25]:0.00843832,469:0.0109533):0.00657864,((((872:0.00120503,((980:0.0001
[29]:((962:0.000580339,930:0.000580339):0.00543993 ((758:0.000598847,726:0.000598847)

$