##// END OF EJS Templates
test-context: add test for performing a diff on a memctx...
test-context: add test for performing a diff on a memctx We now see the first result of all that refactoring of memctx: we can now diff against a memctx.

File last commit:

r18894:ed46c2b9 default
r21837:61b333b9 default
Show More
dicthelpers.py
55 lines | 1.6 KiB | text/x-python | PythonLexer
# dicthelpers.py - helper routines for Python dicts
#
# Copyright 2013 Facebook
#
# This software may be used and distributed according to the terms of the
# GNU General Public License version 2 or any later version.
def diff(d1, d2, default=None):
'''Return all key-value pairs that are different between d1 and d2.
This includes keys that are present in one dict but not the other, and
keys whose values are different. The return value is a dict with values
being pairs of values from d1 and d2 respectively, and missing values
treated as default, so if a value is missing from one dict and the same as
default in the other, it will not be returned.'''
res = {}
if d1 is d2:
# same dict, so diff is empty
return res
for k1, v1 in d1.iteritems():
v2 = d2.get(k1, default)
if v1 != v2:
res[k1] = (v1, v2)
for k2 in d2:
if k2 not in d1:
v2 = d2[k2]
if v2 != default:
res[k2] = (default, v2)
return res
def join(d1, d2, default=None):
'''Return all key-value pairs from both d1 and d2.
This is akin to an outer join in relational algebra. The return value is a
dict with values being pairs of values from d1 and d2 respectively, and
missing values represented as default.'''
res = {}
for k1, v1 in d1.iteritems():
if k1 in d2:
res[k1] = (v1, d2[k1])
else:
res[k1] = (v1, default)
if d1 is d2:
return res
for k2 in d2:
if k2 not in d1:
res[k2] = (default, d2[k2])
return res