Compare dictionaries ignoring specific keys
dictionary, python
Solution
def equal_dicts(d1, d2, ignore_keys):
d1_filtered = {k:v for k,v in d1.items() if k not in ignore_keys}
d2_filtered = {k:v for k,v in d2.items() if k not in ignore_keys}
return d1_filtered == d2_filtered
EDIT: This might be faster and more memory-efficient:
def equal_dicts(d1, d2, ignore_keys):
ignored = set(ignore_keys)
for k1, v1 in d1.iteritems():
if k1 not in ignored and (k1 not in d2 or d2[k1] != v1):
return False
for k2, v2 in d2.iteritems():
if k2 not in ignored and k2 not in d1:
return False
return True
Problem
How can I test if two dictionaries are equal while taking some keys out of consideration. For example, ``` equal_dicts( {'foo':1, 'bar':2, 'x':55, 'y': 77 }, {'foo':1, 'bar':2, 'x':66, 'z': 88 }, ignore_keys=('x', 'y', 'z') ) ``` should return True. UPD: I'm looking for an efficient, fast solution. UPD2. I ended up with this code, which appears to be the fastest: ``` def equal_dicts_1(a, b, ignore_keys): ka = set(a).difference(ignore_keys) kb = set(b).difference(ignore_keys) return ka == kb and all(a[k] == b[k] for k in ka) ``` Timings: https://gist.github.com/2651872