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Home/ Questions/Q 649267
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T21:55:32+00:00 2026-05-13T21:55:32+00:00

Please have a look at the code below: import string from collections import defaultdict

  • 0

Please have a look at the code below:

import string
from collections import defaultdict



first_complex=open( "residue_a_chain_a_b_backup.txt", "r" )
first_complex_lines=first_complex.readlines()
first_complex_lines=map( string.strip, first_complex_lines )
first_complex.close()

second_complex=open( "residue_a_chain_a_c_backup.txt", "r" )
second_complex_lines=second_complex.readlines()
second_complex_lines=map( string.strip, second_complex_lines )
second_complex.close()
list_1=[]
list_2=[]
for x in first_complex_lines:
    if x[0]!="d":
        list_1.append( x )
for y in second_complex_lines:
    if y[0]!="d":
        list_2.append( y ) 
j=0
list_3=[]      
list_4=[]
for a in list_1:
    pass
    for b in list_2:
        pass
        if a==b:
            list_3.append( a )    

kvmap=defaultdict( int )
for k in list_3:
    kvmap[k]+=1 
print kvmap

Normally I use izip or izip_longest to club two for loops, but this time the length of the files are different. I don’t want a None entry. If I use the above method, the run time becomes incremental and useless. How am I supposed to get the two for loops going?

Cheers,
Chavanak

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T21:55:32+00:00Added an answer on May 13, 2026 at 9:55 pm

    You want to convert list_2 to a set, and check for membership:

    list_1 = ['a', 'big', 'list']
    list_2 = ['another', 'big', 'list']
    
    target_set = set(list_2)
    
    for a in list_1:
        if a in target_set:
             print a
    

    Outputs:

    big
    list
    

    A set gives you the advantage of O(1) access time to determine membership, so you only have to read all the way through list_2 once (when creating the set). Thereafter, each comparison happens in constant time.

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