List comparison in Python

[Nirmal]

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

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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:

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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:

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

[Nirmal]

Thanks

[Vysero]

Thanks

Sure no problem!