Benchmark of merging dicts in python.

benchmark-dict-merge.py

#!/usr/bin/env python
import timeit


def d():
    d1= {
        'title': u'xyx',
        'description': u'xyx',
        'checkin_time': u'xyx',
        'checkout_time': u'xyx',
        'active': u'xyx',
        'booking_from': u'xyx',
        'booking_to': u'xyx',
        'publish_from': u'xyx',
        'publish_to': u'xyx',
        'min_beds': u'xyx',
        'max_beds': u'xyx'
    };
    d2={
        'accept_price': u'xyx',
        'reservation_price': u'xyx',
        'start_price': u'xyx',
        'auction_from': u'xyx',
        'auction_to': u'xyx'
    };
    d3={
        'day-price': u'xyx',
        'days': {
            '0': {'from_date': u'xyx', 'to_date': u'xyx'}
        },
        'checkin_weekday': u'',
        'checkout_weekday': u'',
        'min_day_stay': u'xyx',
        'max_day_stay': u'xyx'
    }
    return d1, d2, d3

statements = [
    'd4 = d1.copy(); d4.update(d2), d4.update(d3)',
    'd4 = dict(d1, **d2); d4.update(d3)',
    'd4 = {}; d4.update(d1); d4.update(d2), d4.update(d3)',
    'd4 = dict(d1, **d2); d5=dict(d4, **d3)',
    'd4 = reduce(lambda x,y: dict(x, **y), (d1, d2, d3))',
    'd4 = {}\nfor d in (d1, d2, d3): d4.update(d)',
    'd4 = dict(d1)\nfor d in (d2, d3): d4.update(d)',
    'd4 = dict(chain(*[d.iteritems() for d in (d1, d2, d3)]))',
    'd4 = dict(d1.items() + d2.items() + d3.items())',
    'd4 = dict(chain.from_iterable(d.iteritems() for d in (d1, d2, d3)))',
    'd4 = dict(chain(d1.items(), d2.items(), d3.items()))',
]


if __name__ == '__main__':
    res = []
    for s in statements:
        t = timeit.timeit(
            s,
            setup="from itertools import chain;from __main__ import d; d1, d2, d3 = d()"
        )
        st = "time %s --- %s " % (t, s.replace('\n', ' '))
        print st
        res.append(st)

    print "------- In time order -------"
    for x in sorted(res):
        print x

Result.txt

time 1.6452050209  --- d4 = d1.copy(); d4.update(d2), d4.update(d3) 
time 1.69228887558 --- d4 = dict(d1, **d2); d4.update(d3) 
time 1.78650903702 --- d4 = {}; d4.update(d1); d4.update(d2), d4.update(d3) 
time 1.92449307442 --- d4 = {} for d in (d1, d2, d3): d4.update(d) 
time 2.00036215782 --- d4 = dict(d1, **d2); d5=dict(d4, **d3) 
time 2.04427695274 --- d4 = dict(d1) for d in (d2, d3): d4.update(d) 
time 2.68033099174 --- d4 = reduce(lambda x,y: dict(x, **y), (d1, d2, d3)) 
time 3.72226691246 --- d4 = dict(chain(*[d.iteritems() for d in (d1, d2, d3)])) 
time 4.01872420311 --- d4 = dict(d1.items() + d2.items() + d3.items()) 
time 4.07779598236 --- d4 = dict(chain.from_iterable(d.iteritems() for d in (d1, d2, d3))) 
time 4.23182702065 --- d4 = dict(chain(d1.items(), d2.items(), d3.items()))