List comparison in Python

[Nirmal]

Hi
I have 2 list .I want to compare and get the difference.

a = [['vlan 158', ' name MARKET', ' mode vpc'], []]

b = [['vlan 158', ' name MARKETING', ' mode vpc'], ['vlan 159', ' name SALES', ' mode vpc']]

Expected Output :

Missing in "a"

vlan 159 name SALES mode vpc

please suggest how can it be done?

Note : list "a" has Vlan 158 so as list "b" .So its a MATCH ,though "name" is different.not worried about that...

[Larz60+]

Is this a homework assignment?
look into using set to do comparisons.
by the way, there's no way the 'code' you show would have a chance of running.
Make an attempt.
hint, use a logical and on two sets

[Nirmal]

this is the last part of my script and i am stuck...

[Vysero]

this is the last part of my script and i am stuck...

Assuming this is a hw assignment I won't give you the answer but to get you started:

a = [['vlan 158', 'name MARKET', 'mode vpc']] b = [['vlan 158', 'name MARKETING', 'mode vpc'], ['vlan 159', 'name SALES', 'mode vpc']] x = a.pop() y = b.pop() for i in range(len(y)): for j in range(i + 1, len(x)): if not(x[i] == y[j]): print(x[i],y[j])

[Nirmal]

@Vysero : hey no this is no hw assignment .This i am doing as corporate task.The thing is i cant "pop" as the list content can be extended .It needs to be a generic one . So the first element of "list a" ( e:g vlan 158" in this case ) has to be checked in "list b" and vice versa and if it is found then fine. If it is not found ( e:g vlan 159 ) is not available in "list a" then it should return everything i.e "list a " has below missing

vlan 159 name SALES mode vpc

[wavic]

If there are no repeating elements in the lists you can turn them into sets. Here you can see what you can do with sets.

[Nirmal]

i tried

print ("+++++++++++++++++ The missing configuration is++++++++++++++\n") p = [item for index, item[1] in enumerate(list2) if [] != [it for it in item if it not in list1[index]]] print('\n'.join(['\n'.join(item[1]) for item in p])) print ("+++++++++++++++++ The missing configuration is++++++++++++++\n") q = [item for index, item[1] in enumerate(list1) if [] != [it for it in item if it not in list2[index]]] print('\n'.join(['\n'.join(item[1]) for item in q]))

output:

+++++++++++++++++ The missing configuration is++++++++++++++ vlan 159 name SALES mode vpc vlan 159 name SALES mode vpc +++++++++++++++++ The missing configuration is++++++++++++++

repeated ,.....i am missing something here..please suggest.

[Larz60+]

Another approach:

>>> a = [['vlan 158', ' name MARKET', ' mode vpc'], []] >>> b = [['vlan 158', ' name MARKETING', ' mode vpc'], ['vlan 159', ' name SALES', ' mode vpc']] >>> c = [] >>> d = [] >>> for item in a: ... c.append(set(item)) ... >>> for item in b: ... d.append(set(item)) ... >>> for n in range(len(c)): ... print(c[n] & d[n]) ... {' mode vpc', 'vlan 158'}

[Nirmal]

This does not give the desired output !!

[Vysero]

So the first element of "list a" ( e:g vlan 158" in this case ) has to be checked in "list b" and vice versa and if it is found then fine. If it is not found ( e:g vlan 159 ) is not available in "list a" then it should return everything i.e "list a " has below

When you say: "has to be checked in "list b"" are you meaning the first element of "list a" needs to be checked against the first element of each list in "list b"? S/T each matching instance can be ignored but none matching instances will result in a 'print' of the remaining elements in that list? For instance the program would for:

a = [['vlan 158', ' name MARKET', ' mode vpc'], []] b = [['vlan 158', ' name MARKETING', ' mode vpc'], ['vlan 159', ' name SALES', ' mode vpc']]

produce the output:

#first element of list a matches first element of list b thus ignore name SALES, mode vpc #first element of list a does not match second element of list b thus print name MARKETING, mode vpc #second element in list a is an empty list thus will not match first element in list b thus print name SALES, mode vpc #second element in list a is an empty list thus will not match second element in list b thus print