List comparison in Python

[Nirmal]

HI

I have two list . i want to compare with each other with the list index[1][2][3] of each list with other list .If its a match then ignore , if not the return the whole list.

a = [['Eth1/1/13', 'Marketing', 'connected', '10', 'full', 'a-1000'], ['Eth1/1/14', 'NETFLOW02', 'connected', '10', 'full', '100']]

b = [['Eth1/1/13', 'NETFLOW02', 'connected', '15', 'full', '100'], ['Eth1/1/14', 'Marketing', 'connected', '10', 'full', 'a-1000']]

Expected Output :

Diff a:

Eth1/1/14  NETFLOW02   connected    10   full    100

Diff b:

Eth1/1/13  NETFLOW02    connected    15   full    100

My Code :

p = [i for i in a if i not in b]
for item in p: 
        print item[0]
print "\n++++++++++++++++++++++++++++++\n"
q = [i for i in b if i not in a]
    for item in q: 
        print item[0]

Any idea ?

[woooee]

p = [i for i in a if i not in b]

Print to see what it is doing

for item in a:
    print(item)
    print(item not in b) 

And you will have to be more specific about what is and is not a match. It looks like a date difference doesn't count but I am not going to guess.

[Nirmal]

i did research and manage to match list index[1] .. could not do for list index[2] and index[3]

c = [o for o in a if o[1] not in [n[1] for n in b]]
print c

can we do similar match for o[2] and o[3] .. so that o[1],o[2],o[3] all should be checked in list b ?

[woooee]

If you aren't going to read what I post, there is no point to posting.

And you will have to be more specific about what is and is not a match. It looks like a date difference doesn't count but I am not going to guess

[nilamo]

list index[1][2][3] of each list with other list .If its a match then ignore , if not the return the whole list.

I don't know why you're trying to force a comprehension to do this. There's no need to build a new list if you're just returning the original one.

>>> def all_match(indices=(), *collections):
...   collections = iter(collections)
...   first = next(collections)
...   val = None
...   sub_list = first
...   for index in indices:
...     sub_list = sub_list[index]
...   value = sub_list
...   for items in collections:
...     sub_value = None
...     for index in indices:
...       items = items[index]
...     sub_value = items
...     if sub_value != value:
...       return False
...   return first
...
>>> a = [['Eth1/1/13', 'Marketing', 'connected', '10', 'full', 'a-1000'], ['Eth1/1/14', 'NETFLOW02', 'connected', '10', 'full', '100']]
>>>
... b = [['Eth1/1/13', 'NETFLOW02', 'connected', '15', 'full', '100'], ['Eth1/1/14', 'Marketing', 'connected', '10', 'full', 'a-1000']]
>>> all_match([1, 2, 3], a, b)
[['Eth1/1/13', 'Marketing', 'connected', '10', 'full', 'a-1000'], ['Eth1/1/14', 'NETFLOW02', 'connected', '10', 'full', '100']]
>>> a[1][2][3]
'n'
>>> b[1][2][3]
'n'
>>> a = [[''], ['', '', 'spam']]
>>> a[1][2][3]
'm'
>>> all_match([1, 2, 3], a, b)
False