How to skip line with matching pattern in Python?

Question

I have the following lines in test.fa :

#test.fa
>1
AGAGGGAGCTG
CCTCAGGGCTG
CACTCAGGAAA
TTGGGGCGCTG
AGCATGGGGGG
CAGGAGGGGCC

I need to ignore the lines starting with ">" , and concatenate the following lines into one single string. The following script however not only skips lines with ">" , but also the next line before concatenating remaining.

#!/usr/bin/env python
import sys
import re
string = ""
with open("test.fa","rt") as f:
       for line in f:
           if re.match(">",line):
              line = f.next()
           else:
              line = line.rstrip("\n")
              string = string + line
print (string)

Could anyone help fix the script , or suggest better ways to do it? thanks !!

but, I recommend using Biopython [biopython parse fasta](http://biopython.org/wiki/SeqIO#Sequence_Input) — Jose Ricardo Bustos M., Oct 05 '15 at 17:07

score 5 · Accepted Answer · answered Oct 05 '15 at 16:28

The line counter already increments every loop anyway, so you don't actually need to do anything in the if block.

   for line in f:
       if re.match(">",line):
          pass
       else:
          line = line.rstrip("\n")
          string = string + line

Or

   for line in f:
       if not re.match(">",line):
          line = line.rstrip("\n")
          string = string + line

Additional enhancements: you don't need regex to determine what character a string starts with, and accumulating lines in a list is generally recommended over concatenating a string.

lines = []
for line in f:
    if not line.startswith(">"):
        lines.append(line.rstrip("\n"))
string = "".join(lines)

Or, as a one liner:

string = "".join(line.rstrip("\n") for line in f if not line.startswith(">"))

thank you everyone ! all the solutions are similar, Thank you Kevin for the additional enhancements. helpful for any beginner. — harsh, Oct 05 '15 at 16:49

score 1 · Answer 2 · edited Oct 23 '20 at 18:17

1

You are essentially calling line.next() twice since each time you loop, it is getting the next line.. I'd recommend going with this

#!/usr/bin/env python
import sys
import re
string = ""
with open("test.fa","rt") as f:
       for line in f:
           if not re.match(">",line):
              line = line.rstrip("\n")
              string = string + line
print (string)

edited Oct 23 '20 at 18:17

d_kennetz

5,219
5
21
44

answered Oct 05 '15 at 16:29

CollinD

7,304
2
22
45

score 0 · Answer 3 · answered Oct 05 '15 at 16:30

You don't need the

line = f.next()

that happens automatically in the iterator. Just do this:

#!/usr/bin/env python
import sys
import re

string = ""
with open("test.fa","rt") as f:
    for line in f:
        if not re.match(">",line):
            line = line.rstrip("\n")
            string = string + line
print (string)

How to skip line with matching pattern in Python?

3 Answers3