view mercurial/dicthelpers.py @ 21977:4ca4e1572022

run-tests: '--time' option provide more details to Linux users

As our tests execute in child processes, this patch uses os.times()
module in replace of time.time() module to provide additional info like
user time and system time spent by child's processes along with real elapsed
time taken by a process.

There is one limitation of this patch. It can work only for Linux users and
not for Windows.

"os.times" module returns a 5-tuple of a floaing point numbers.
1) User time
2) System time
3) Child's user time
4) Child's system time
5) Ellapsed real time

On Windows, only the first two items are filled, the others are zero.
Therefore, below test cases does not break on Windows but instead gives the
zero value.

# 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