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

using list/dict comprehension getting want in result. want print vapp1 & vapp4 not there in vapps .

what efficient way ? or how avoid looping achieve of filtered list of dictionary names common in list names. , can print names not there.

you could abuse short-circuiting of and so:

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

this yield same result list before, , modifies names removes found vapp names side effect.

this works because and clause evaluated when first part of statement (vapp['name'] in names) true. or 1 part needed because .remove() yields none, false in boolean context.

list comprehensions side effects discouraged bad style, though - , names list modified, better save copy if need again.

generally, not worry performance, , use 2 loops - or write out in readable, classic loop.