mercurial/dicthelpers.py
author Mads Kiilerich <madski@unity3d.com>
Fri, 25 Oct 2013 02:33:59 +0800
branchstable
changeset 19954 427ce5633c1c
parent 18894 ed46c2b98b0d
permissions -rw-r--r--
largefiles: don't prompt for normal/largefile changes when doing plain updates We used to get like: $ hg up -r 2 foo has been turned into a normal file keep as (l)argefile or use (n)ormal file? l getting changed largefiles 0 largefiles updated, 0 removed 0 files updated, 0 files merged, 2 files removed, 0 files unresolved $ cat foo cat: foo: No such file or directory [1] - which both asked the wrong question and did the wrong thing. Instead, skip this conflict resolution when the local conflicting file has been scheduled for removal and there thus is no conflict.

# 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