CODEWITHSUNDEEP
pythonlist
user1701820
user1701820

Reputation: 31

python list value in group

i read a file and group it by a keyword. I have also a List, how can i check it if a value from list is in group? Not check only one entry, check all entries in list against group.

Example but not checked list against group. I dont wanna write if any or any blabla.

hosts = ['blabla.dyndns.org','blabla1.dyndns.org']
   with open(testfile,'r') as f:
       for key,group in it.groupby(f,lambda line: line.startswith('[hosts]')):
           if not key:
               group = list(group)
               if any("blabla.dyndns.org" in s for s in group) or any("blabla.dyndns.org" in s for s in group):
                   print 'yes'
               else:
                   print 'no'

Thank you, regards.

I wanna script that i can in example change protocol of some hosts.

import itertools as it
testfile='blabla.txt'

hosts = set(['blabla.dyndns.org','blabla1.dyndns.org'])

with open(testfile,'r') as f:
    for key,group in it.groupby(f,lambda line: line.startswith('[host]')):
        if not key:
            group = set(group)
            if group & hosts: # set intersection
                print '[host]\n'
                for (i, item) in enumerate(group):
                    if 'protocol' in item:
                        group[i] = 'protocol     = udp\n'
                j = ', '.join(group)
                y = j.replace(", ", "")
                print y
            else:
                print '[host]\n'
                j = ', '.join(group)
                y = j.replace(", ", "")
                print y

exit(0)

Does not work.

Upvotes: 1

Views: 255

Answers (1)

nneonneo
nneonneo

Reputation: 179452

Use set intersection:

hosts = set(['blabla.dyndns.org','blabla1.dyndns.org'])
with open(testfile,'r') as f:
    for key,group in it.groupby(f,lambda line: line.startswith('[hosts]')):
        if not key:
            group = set(group)
            if group & hosts: # set intersection
                print 'yes'
            else:
                print 'no'

Upvotes: 2

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