List comparison in Python

[Nirmal]

@Vysero : ok let me elaborate . Taking this example :

if " vlan 158 " of list a is present anywhere in list b then "ignore".same for list b (if " vlan 158 " of list b is present anywhere in list a then "ignore"). "vlan 159 " of list b is not available anywhere in list a , so it should return remaining element along with first element.So it should say list a is below missing config . ['vlan 159', ' name SALES', ' mode vpc']

In short if first element ( vlan id : vlan 158 ,vlan 159 etc ) of a list is present in other list then ignore and vice versa , if not then return with remaining element .

[Vysero]

Hrm, so if I have this right then your looking for something like this:

a = [['vlan 158', ' name MARKET', ' mode vpc']] b = [['vlan 158', ' name MARKETING', ' mode vpc'], ['vlan 159', ' name SALES', ' mode vpc']] def check_contents(list_one, list_two): shorter, longer = sorted([list_one, list_two], key=len) for x in longer: if not x: continue for y in shorter: if not y: continue if x[0] != y[0]: print(x) check_contents(a, b)

Output:

['vlan 159', ' name SALES', ' mode vpc']

[Nirmal]

Thanks

[Vysero]

Thanks

Sure no problem!