mercurial/dicthelpers.py
author Augie Fackler <raf@durin42.com>
Wed, 12 Mar 2014 13:21:17 -0400
changeset 20685 56b1f39dd0c1
parent 18894 ed46c2b98b0d
permissions -rw-r--r--
test-url: move from dict() construction to {} literals The latter are both faster and more consistent across Python 2 and 3.
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# dicthelpers.py - helper routines for Python dicts
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#
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# Copyright 2013 Facebook
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#
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# This software may be used and distributed according to the terms of the
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# GNU General Public License version 2 or any later version.
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def diff(d1, d2, default=None):
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    '''Return all key-value pairs that are different between d1 and d2.
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    This includes keys that are present in one dict but not the other, and
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    keys whose values are different. The return value is a dict with values
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    being pairs of values from d1 and d2 respectively, and missing values
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    treated as default, so if a value is missing from one dict and the same as
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    default in the other, it will not be returned.'''
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    res = {}
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    if d1 is d2:
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        # same dict, so diff is empty
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        return res
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    for k1, v1 in d1.iteritems():
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        v2 = d2.get(k1, default)
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        if v1 != v2:
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            res[k1] = (v1, v2)
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    for k2 in d2:
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        if k2 not in d1:
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            v2 = d2[k2]
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            if v2 != default:
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                res[k2] = (default, v2)
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    return res
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def join(d1, d2, default=None):
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    '''Return all key-value pairs from both d1 and d2.
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    This is akin to an outer join in relational algebra. The return value is a
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    dict with values being pairs of values from d1 and d2 respectively, and
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    missing values represented as default.'''
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    res = {}
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    for k1, v1 in d1.iteritems():
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        if k1 in d2:
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            res[k1] = (v1, d2[k1])
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        else:
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            res[k1] = (v1, default)
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    if d1 is d2:
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        return res
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    for k2 in d2:
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        if k2 not in d1:
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            res[k2] = (default, d2[k2])
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    return res