view mercurial/dicthelpers.py @ 21810:4b2ebd3187a1

dirstate.status: assign members one by one instead of unpacking the tuple

With this patch, hg status and hg diff regain their previous speed.

The following tests are run against a working copy with over 270,000 files.
Here, 'before' means without this or the previous patch applied.

Note that in this case `hg perfstatus` isn't representative since it doesn't
take dirstate parsing time into account.

$ time hg status  # best of 5
before: 2.03s user 1.25s system 99% cpu 3.290 total
after:  2.01s user 1.25s system 99% cpu 3.261 total

$ time hg diff    # best of 5
before: 1.32s user 0.78s system 99% cpu 2.105 total
after:  1.27s user 0.79s system 99% cpu 2.066 total

# 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