Compare Dictionaries for close enough match

Easiest way I could think of:

keylist = dict_1.keys()array_1 = numpy.array([dict_1[key] for key in keylist])array_2 = numpy.array([dict_2[key] for key in keylist])if numpy.allclose(array_1, array_2):    print('Equal')else:    print('Not equal')

There is a numpy-function, used in unit testing to compare 2 numbers. You can set the desired precision via significant.

import numpy as npdict_1 = {("a","a"):0.01, ("a","b"): 0.02, ("a","c"): 0.00015}dict_2 = {("a","a"):0.01, ("a","b"): 0.018, ("a","c"): 0.00014}def compare_dicts(dict_1, dict_2, significant=2):    for k in dict_1:        try:            np.testing.assert_approx_equal(dict_1[k], dict_2[k],                 significant=significant)        except AssertionError:            return False        return Truecompare_dicts(dict_1, dict_2)

This will take the keys from one dict and compare each item in both dicts.

See the details for the numpy-function used here.

You can use OrderedDict to get the values in default order and numpy.allclose :

>>> import numpy>>> from collections import OrderedDict>>> dict_1 = {("a","a"):0.01, ("a","b"): 0.02, ("a","c"): 0.00015}>>> dict_2 = {("a","a"):0.01, ("a","b"): 0.018, ("a","c"): 0.00014}>>> d2=OrderedDict(sorted(dict_2.items(), key=lambda t: t)) #d2.values() -> [0.01, 0.018, 0.00014]>>> d1=OrderedDict(sorted(dict_1.items(), key=lambda t: t))>>> atol=0.0001 # as example>>> numpy.allclose(d1.values(),d2.values())False