CODEWITHSUNDEEP
pythonlistdictionary
Amby
Amby

Reputation: 467

Python : filter list items from list of dictionary

 names = ['vapp1', 'vapp3', 'vapp4', 'vapp2']

vapps = [{'name':'vapp2', 'ip': '11.21.18.24', 'obj': 'obj523'}, {'name':'vapp3', 'ip': '11.21.18.27', 'obj': 'obj234'}, {'name':'vapp5', 'ip': '11.21.18.25', 'obj': 'obj246'}]


result = [vapp for vapp in vapps if vapp['name'] in names]
print result

Using this list/dict comprehension I am getting what I want in result. But I also want to print that vapp1 & vapp4 are not there in vapps .

What is the most efficient way ? or How to avoid extra looping to achieve all of this so that I will get a filtered list of dictionary whose names are common in the list names. And also I can print those names which are not there.

Upvotes: 0

Views: 954

Answers (2)

kid
kid

Reputation: 153

You could do following

just using some set properties we can get it

vapps_names = [vapp['name'] for vapp in vapps]

now Values not in vapps but in names

not_in_vapps = set(names) - set(vapps_names)

{'vapp1', 'vapp4'}

Values in names not in vapps

not_in_names = set(vapps_names) - set(names)

{'vapp5'}

Upvotes: 1

ch3ka
ch3ka

Reputation: 12178

You could abuse the short-circuiting of and like so:

>>> result = [vapp for vapp in vapps if
...           vapp['name'] in names and
...           (names.remove(vapp['name']) or 1)]
>>> names # now contains names not found in vapps
['vapp1', 'vapp4']

This will yield the same result list as before, and modifies names so it removes found vapp names as a side effect.

This works because the and clause is only evaluated when the first part of the statement (vapp['name'] in names) is True. The or 1 part is needed because .remove() yields None, which is False in a boolean context.

List comprehensions with side effects are generally discouraged as bad style, though - and your names list will get modified, so better save a copy if you need it again.

Generally, do not worry about performance, and just use two loops - or write it out in a readable, classic loop.

Upvotes: 1

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