Calculate difference in keys contained in two Python dictionaries

Question

Suppose I have two Python dictionaries - dictA and dictB. I need to find out if there are any keys which are present in dictB but not in dictA. What is the fastest way to go about it?

Should I convert the dictionary keys into a set and then go about?

Interested in knowing your thoughts...

Thanks for your responses.

Apologies for not stating my question properly. My scenario is like this - I have a dictA which can be the same as dictB or may have some keys missing as compared to dictB or else the value of some keys might be different which has to be set to that of dictA key's value.

Problem is the dictionary has no standard and can have values which can be dict of dict.

Say

dictA={'key1':a, 'key2':b, 'key3':{'key11':cc, 'key12':dd}, 'key4':{'key111':{....}}}
dictB={'key1':a, 'key2:':newb, 'key3':{'key11':cc, 'key12':newdd, 'key13':ee}.......

So 'key2' value has to be reset to the new value and 'key13' has to be added inside the dict. The key value does not have a fixed format. It can be a simple value or a dict or a dict of dict.

score 177 · Answer 1 · Oct 11 '13 at 17:38

up vote 177 down vote

You can use set operations on the keys:

diff = set(dictb.keys()) - set(dicta.keys())

Here is a class to find all the possibilities: what was added, what was removed, which key-value pairs are the same, and which key-value pairs are changed.

class DictDiffer(object):
    """
    Calculate the difference between two dictionaries as:
    (1) items added
    (2) items removed
    (3) keys same in both but changed values
    (4) keys same in both and unchanged values
    """
    def __init__(self, current_dict, past_dict):
        self.current_dict, self.past_dict = current_dict, past_dict
        self.set_current, self.set_past = set(current_dict.keys()), set(past_dict.keys())
        self.intersect = self.set_current.intersection(self.set_past)
    def added(self):
        return self.set_current - self.intersect 
    def removed(self):
        return self.set_past - self.intersect 
    def changed(self):
        return set(o for o in self.intersect if self.past_dict[o] != self.current_dict[o])
    def unchanged(self):
        return set(o for o in self.intersect if self.past_dict[o] == self.current_dict[o])

Here is some sample output:

>>> a = {'a': 1, 'b': 1, 'c': 0}
>>> b = {'a': 1, 'b': 2, 'd': 0}
>>> d = DictDiffer(b, a)
>>> print "Added:", d.added()
Added: set(['d'])
>>> print "Removed:", d.removed()
Removed: set(['c'])
>>> print "Changed:", d.changed()
Changed: set(['b'])
>>> print "Unchanged:", d.unchanged()
Unchanged: set(['a'])

Available as a github repo: https://github.com/hughdbrown/dictdiffer

edited Oct 11 '13 at 17:38

answered Jul 22 '09 at 14:11

hughdbrown

23.1k 13 62 87

3

Smart solution, thanks! I've made it work with nested dicts by checking whether changed or unchanged values are dict instances and calling a recursive function to check them again using your class. – AJJ Dec 12 '12 at 12:24

@AJJ I'd love to see that implementation. – urschrei Mar 7 '13 at 11:30

1

How about a def update(self, new_dict): self.__init__(new_dict, self.current_dict) or the like so you can do a rolling comparison – Nick T Oct 21 '13 at 20:02

Some remarks : the DictDiffer class is a stateless class and could be a function. The changed andunchanged values could be computed in the same loop. This two functions could return a list instead of a set which is certainly less expensive. For deep comparison, you can take a look at the Unit testing framework: docs.python.org/2/library/unittest.html, just follow the assertDictEqual method in the source code. – Laurent LAPORTE Nov 20 '13 at 6:52

FWIW, set(dictb) is probably better than set(dictb.keys()). – mgilson May 19 '14 at 15:13

show 3 more comments

score 23 · Answer 2 · Mar 10 at 21:54

In case you want the difference recursively, I have written a package for python:https://github.com/seperman/deepdiff

Installation

Install from PyPi:

pip install deepdiff

Example usage

Importing

>>> from deepdiff import DeepDiff
>>> from pprint import pprint
>>> from __future__ import print_function # In case running on Python 2

Same object returns empty

>>> t1 = {1:1, 2:2, 3:3}
>>> t2 = t1
>>> print(DeepDiff(t1, t2))
{}

Type of an item has changed

>>> t1 = {1:1, 2:2, 3:3}
>>> t2 = {1:1, 2:"2", 3:3}
>>> pprint(DeepDiff(t1, t2), indent=2)
{ 'type_changes': { 'root[2]': { 'newtype': <class 'str'>,
                                 'newvalue': '2',
                                 'oldtype': <class 'int'>,
                                 'oldvalue': 2}}}

Value of an item has changed

>>> t1 = {1:1, 2:2, 3:3}
>>> t2 = {1:1, 2:4, 3:3}
>>> pprint(DeepDiff(t1, t2), indent=2)
{'values_changed': {'root[2]': {'newvalue': 4, 'oldvalue': 2}}}

Item added and/or removed

>>> t1 = {1:1, 2:2, 3:3, 4:4}
>>> t2 = {1:1, 2:4, 3:3, 5:5, 6:6}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff)
{'dic_item_added': ['root[5]', 'root[6]'],
 'dic_item_removed': ['root[4]'],
 'values_changed': {'root[2]': {'newvalue': 4, 'oldvalue': 2}}}

String difference

>>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":"world"}}
>>> t2 = {1:1, 2:4, 3:3, 4:{"a":"hello", "b":"world!"}}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
{ 'values_changed': { 'root[2]': {'newvalue': 4, 'oldvalue': 2},
                      "root[4]['b']": { 'newvalue': 'world!',
                                        'oldvalue': 'world'}}}

String difference 2

>>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":"world!\nGoodbye!\n1\n2\nEnd"}}
>>> t2 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":"world\n1\n2\nEnd"}}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
{ 'values_changed': { "root[4]['b']": { 'diff': '--- \n'
                                                '+++ \n'
                                                '@@ -1,5 +1,4 @@\n'
                                                '-world!\n'
                                                '-Goodbye!\n'
                                                '+world\n'
                                                ' 1\n'
                                                ' 2\n'
                                                ' End',
                                        'newvalue': 'world\n1\n2\nEnd',
                                        'oldvalue': 'world!\n'
                                                    'Goodbye!\n'
                                                    '1\n'
                                                    '2\n'
                                                    'End'}}}

>>> 
>>> print (ddiff['values_changed']["root[4]['b']"]["diff"])
--- 
+++ 
@@ -1,5 +1,4 @@
-world!
-Goodbye!
+world
 1
 2
 End

Type change

>>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2, 3]}}
>>> t2 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":"world\n\n\nEnd"}}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
{ 'type_changes': { "root[4]['b']": { 'newtype': <class 'str'>,
                                      'newvalue': 'world\n\n\nEnd',
                                      'oldtype': <class 'list'>,
                                      'oldvalue': [1, 2, 3]}}}

List difference

>>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2, 3, 4]}}
>>> t2 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2]}}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
{'iterable_item_removed': {"root[4]['b'][2]": 3, "root[4]['b'][3]": 4}}

List difference 2:

>>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2, 3]}}
>>> t2 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 3, 2, 3]}}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
{ 'iterable_item_added': {"root[4]['b'][3]": 3},
  'values_changed': { "root[4]['b'][1]": {'newvalue': 3, 'oldvalue': 2},
                      "root[4]['b'][2]": {'newvalue': 2, 'oldvalue': 3}}}

List difference ignoring order or duplicates: (with the same dictionaries as above)

>>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2, 3]}}
>>> t2 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 3, 2, 3]}}
>>> ddiff = DeepDiff(t1, t2, ignore_order=True)
>>> print (ddiff)
{}

List that contains dictionary:

>>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2, {1:1, 2:2}]}}
>>> t2 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2, {1:3}]}}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
{ 'dic_item_removed': ["root[4]['b'][2][2]"],
  'values_changed': {"root[4]['b'][2][1]": {'newvalue': 3, 'oldvalue': 1}}}

Sets:

>>> t1 = {1, 2, 8}
>>> t2 = {1, 2, 3, 5}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (DeepDiff(t1, t2))
{'set_item_added': ['root[3]', 'root[5]'], 'set_item_removed': ['root[8]']}

Named Tuples:

>>> from collections import namedtuple
>>> Point = namedtuple('Point', ['x', 'y'])
>>> t1 = Point(x=11, y=22)
>>> t2 = Point(x=11, y=23)
>>> pprint (DeepDiff(t1, t2))
{'values_changed': {'root.y': {'newvalue': 23, 'oldvalue': 22}}}

Custom objects:

>>> class ClassA(object):
...     a = 1
...     def __init__(self, b):
...         self.b = b
... 
>>> t1 = ClassA(1)
>>> t2 = ClassA(2)
>>> 
>>> pprint(DeepDiff(t1, t2))
{'values_changed': {'root.b': {'newvalue': 2, 'oldvalue': 1}}}

Object attribute added:

>>> t2.c = "new attribute"
>>> pprint(DeepDiff(t1, t2))
{'attribute_added': ['root.c'],
 'values_changed': {'root.b': {'newvalue': 2, 'oldvalue': 1}}}

Thank you for this! Just implemented on my project, works great! — Jun 19 at 21:50

Eli Courtwright 84.9k 44 174 232 · Answer 3 · Jul 22 '09 at 13:52

up vote 14 down vote

not sure whether its "fast" or not, but normally, one can do this

dicta = {"a":1,"b":2,"c":3,"d":4}
dictb = {"a":1,"d":2}
for key in dicta.keys():
    if not key in dictb:
        print key

edited Jul 22 '09 at 13:52

Eli Courtwright

84.9k 44 174 232

answered Jul 22 '09 at 13:48

ghostdog74

144k 26 155 234

You have to swap dicta and dictb since he wants to know those keys of dictb that are not indicta. – Gumbo Jul 22 '09 at 13:51

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Oz123 9,062 12 45 92 · Answer 4 · May 8 '13 at 14:35

As Alex Martelli wrote, if you simply want to check if any key in B is not in A, any(True for k in dictB if k not in dictA) would be the way to go.

To find the keys that are missing:

diff = set(dictB)-set(dictA) #sets

C:\Dokumente und Einstellungen\thc>python -m timeit -s "dictA =    
dict(zip(range(1000),range
(1000))); dictB = dict(zip(range(0,2000,2),range(1000)))" "diff=set(dictB)-set(dictA)"
10000 loops, best of 3: 107 usec per loop

diff = [ k for k in dictB if k not in dictA ] #lc

C:\Dokumente und Einstellungen\thc>python -m timeit -s "dictA = 
dict(zip(range(1000),range
(1000))); dictB = dict(zip(range(0,2000,2),range(1000)))" "diff=[ k for k in dictB if
k not in dictA ]"
10000 loops, best of 3: 95.9 usec per loop

So those two solutions are pretty much the same speed.

This makes more sense: any(k not in dictA for k in dictB) – hughdbrown May 2 '11 at 10:07 — May 2 '11 at 10:07

score 12 · Answer 5 · Jul 23 '09 at 0:23

If you really mean exactly what you say (that you only need to find out IF "there are any keys" in B and not in A, not WHICH ONES might those be if any), the fastest way should be:

if any(True for k in dictB if k not in dictA): ...

If you actually need to find out WHICH KEYS, if any, are in B and not in A, and not just "IF" there are such keys, then existing answers are quite appropriate (but I do suggest more precision in future questions if that's indeed what you mean;-).

softwareplay 518 2 12 29 · Answer 6 · Oct 28 '13 at 11:47

up vote 4 down vote

There is an other question in stackoverflow about this argument and i have to admit that there is a simple solution explained: the datadiff library of python helps printing the difference between two dictionaries.

answered Oct 28 '13 at 11:47

softwareplay

518 2 12 29

add a comment

Joziel Costa 31 1 · Answer 7 · Jul 28 at 20:16

up vote 3 down vote

use set(), documentation in https://docs.python.org/2/library/sets.html

Example:

set(dictA.keys()).intersection(dictB.keys())

answered Jul 28 at 20:16

Joziel Costa

31 1

add a comment

score 2 · Answer 8 · Jul 22 '09 at 17:06

up vote 2 down vote

Here's a way that will work, allows for keys that evaluate to False, and still uses a generator expression to fall out early if possible. It's not exceptionally pretty though.

any(map(lambda x: True, (k for k in b if k not in a)))

EDIT:

THC4k posted a reply to my comment on another answer. Here's a better, prettier way to do the above:

any(True for k in b if k not in a)

Not sure how that never crossed my mind...

edited Jul 22 '09 at 17:06

answered Jul 22 '09 at 16:56

Steve Losh

14.8k 2 41 43

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Oz123 9,062 12 45 92 · Answer 9 · May 8 '13 at 14:27

up vote 2 down vote

If on Python ≥ 2.7:

# update different values in dictB
# I would assume only dictA should be updated,
# but the question specifies otherwise

for k in dictA.viewkeys() & dictB.viewkeys():
    if dictA[k] != dictB[k]:
        dictB[k]= dictA[k]

# add missing keys to dictA

dictA.update( (k,dictB[k]) for k in dictB.viewkeys() - dictA.viewkeys() )

edited May 8 '13 at 14:27

Oz123

9,062 12 45 92

answered Oct 11 '11 at 11:01

tzot

47.4k 15 86 147

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velis 1,748 15 31 · Answer 10 · Apr 17 '14 at 13:40

This is an old question and asks a little bit less than what I needed so this answer actually solves more than this question asks. The answers in this question helped me solve the following:

(asked) Record differences between two dictionaries
Merge differences from #1 into base dictionary
(asked) Merge differences between two dictionaries (treat dictionary #2 as if it were a diff dictionary)
Try to detect item movements as well as changes
(asked) Do all of this recursively

All this combined with JSON makes for a pretty powerful configuration storage support.

The solution (also on github):

from collections import OrderedDict
from pprint import pprint


class izipDestinationMatching(object):
    __slots__ = ("attr", "value", "index")

    def __init__(self, attr, value, index):
        self.attr, self.value, self.index = attr, value, index

    def __repr__(self):
        return "izip_destination_matching: found match by '%s' = '%s' @ %d" % (self.attr, self.value, self.index)


def izip_destination(a, b, attrs, addMarker=True):
    """
    Returns zipped lists, but final size is equal to b with (if shorter) a padded with nulls
    Additionally also tries to find item reallocations by searching child dicts (if they are dicts) for attribute, listed in attrs)
    When addMarker == False (patching), final size will be the longer of a, b
    """
    for idx, item in enumerate(b):
        try:
            attr = next((x for x in attrs if x in item), None)  # See if the item has any of the ID attributes
            match, matchIdx = next(((orgItm, idx) for idx, orgItm in enumerate(a) if attr in orgItm and orgItm[attr] == item[attr]), (None, None)) if attr else (None, None)
            if match and matchIdx != idx and addMarker: item[izipDestinationMatching] = izipDestinationMatching(attr, item[attr], matchIdx)
        except:
            match = None
        yield (match if match else a[idx] if len(a) > idx else None), item
    if not addMarker and len(a) > len(b):
        for item in a[len(b) - len(a):]:
            yield item, item


def dictdiff(a, b, searchAttrs=[]):
    """
    returns a dictionary which represents difference from a to b
    the return dict is as short as possible:
      equal items are removed
      added / changed items are listed
      removed items are listed with value=None
    Also processes list values where the resulting list size will match that of b.
    It can also search said list items (that are dicts) for identity values to detect changed positions.
      In case such identity value is found, it is kept so that it can be re-found during the merge phase
    @param a: original dict
    @param b: new dict
    @param searchAttrs: list of strings (keys to search for in sub-dicts)
    @return: dict / list / whatever input is
    """
    if not (isinstance(a, dict) and isinstance(b, dict)):
        if isinstance(a, list) and isinstance(b, list):
            return [dictdiff(v1, v2, searchAttrs) for v1, v2 in izip_destination(a, b, searchAttrs)]
        return b
    res = OrderedDict()
    if izipDestinationMatching in b:
        keepKey = b[izipDestinationMatching].attr
        del b[izipDestinationMatching]
    else:
        keepKey = izipDestinationMatching
    for key in sorted(set(a.keys() + b.keys())):
        v1 = a.get(key, None)
        v2 = b.get(key, None)
        if keepKey == key or v1 != v2: res[key] = dictdiff(v1, v2, searchAttrs)
    if len(res) <= 1: res = dict(res)  # This is only here for pretty print (OrderedDict doesn't pprint nicely)
    return res


def dictmerge(a, b, searchAttrs=[]):
    """
    Returns a dictionary which merges differences recorded in b to base dictionary a
    Also processes list values where the resulting list size will match that of a
    It can also search said list items (that are dicts) for identity values to detect changed positions
    @param a: original dict
    @param b: diff dict to patch into a
    @param searchAttrs: list of strings (keys to search for in sub-dicts)
    @return: dict / list / whatever input is
    """
    if not (isinstance(a, dict) and isinstance(b, dict)):
        if isinstance(a, list) and isinstance(b, list):
            return [dictmerge(v1, v2, searchAttrs) for v1, v2 in izip_destination(a, b, searchAttrs, False)]
        return b
    res = OrderedDict()
    for key in sorted(set(a.keys() + b.keys())):
        v1 = a.get(key, None)
        v2 = b.get(key, None)
        #print "processing", key, v1, v2, key not in b, dictmerge(v1, v2)
        if v2 is not None: res[key] = dictmerge(v1, v2, searchAttrs)
        elif key not in b: res[key] = v1
    if len(res) <= 1: res = dict(res)  # This is only here for pretty print (OrderedDict doesn't pprint nicely)
    return res

Roei Bahumi 810 8 10 · Answer 11 · Nov 6 '14 at 9:49

Here is a solution for deep comparing 2 dictionaries keys:

def compareDictKeys(dict1, dict2):
  if type(dict1) != dict or type(dict2) != dict:
      return False

  keys1, keys2 = dict1.keys(), dict2.keys()
  diff = set(keys1) - set(keys2) or set(keys2) - set(keys1)

  if not diff:
      for key in keys1:
          if (type(dict1[key]) == dict or type(dict2[key]) == dict) and not compareDictKeys(dict1[key], dict2[key]):
              diff = True
              break

  return not diff

Maxx 25 1 · Answer 12 · Jan 26 '12 at 20:19

what about standart (compare FULL Object)

PyDev->new PyDev Module->Module: unittest

import unittest


class Test(unittest.TestCase):


    def testName(self):
        obj1 = {1:1, 2:2}
        obj2 = {1:1, 2:2}
        self.maxDiff = None # sometimes is usefull
        self.assertDictEqual(d1, d2)

if __name__ == "__main__":
    #import sys;sys.argv = ['', 'Test.testName']

    unittest.main()

Leigh 24.7k 7 36 69 · Answer 13 · May 9 '13 at 1:59

up vote 0 down vote

Not sure if it is still relevant but I came across this problem, my situation i just needed to return a dictionary of the changes for all nested dictionaries etc etc. Could not find a good solution out there but I did end up writing a simple function to do this. Hope this helps,

edited May 9 '13 at 1:59

Leigh

24.7k 7 36 69

answered May 8 '13 at 23:33

Jonathan Mickle

11

2

It would be preferable to have the smallest amount of code that fixes the OP's problem actually in the answer, instead of a link. If the link dies or moves, your answer becomes useless. – George Stocker ♦ May 9 '13 at 12:22

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Brad 1,853 2 18 23 · Answer 14 · Dec 31 '14 at 6:26

up vote 0 down vote

If you want a built-in solution for a full comparison with arbitrary dict structures, @Maxx's answer is a good start.

import unittest

test = unittest.TestCase()
test.assertEqual(dictA, dictB)

answered Dec 31 '14 at 6:26

Brad

1,853 2 18 23

You apparently can't instantiate a test class like that, which is too bad. – Ben Liyanage Oct 6 '15 at 22:59

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j0ker5 1,678 1 8 9 · Answer 15 · Mar 10 '15 at 21:47

here's a solution that can compare more than two dicts:

def diff_dict(dicts, default=None):
    diff_dict = {}
    # add 'list()' around 'd.keys()' for python 3 compatibility
    for k in set(sum([d.keys() for d in dicts], [])):
        # we can just use "values = [d.get(k, default) ..." below if 
        # we don't care that d1[k]=default and d2[k]=missing will
        # be treated as equal
        if any(k not in d for d in dicts):
            diff_dict[k] = [d.get(k, default) for d in dicts]
        else:
            values = [d[k] for d in dicts]
            if any(v != values[0] for v in values):
                diff_dict[k] = values
    return diff_dict

usage example:

import matplotlib.pyplot as plt
diff_dict([plt.rcParams, plt.rcParamsDefault, plt.matplotlib.rcParamsOrig])

normalUser 81 2 3 · Answer 16 · Jun 16 '15 at 7:20

up vote 0 down vote

Based on ghostdog74's answer,

dicta = {"a":1,"d":2}
dictb = {"a":5,"d":2}

for value in dicta.values():
    if not value in dictb.values():
        print value

will print differ value of dicta

answered Jun 16 '15 at 7:20

normalUser

81 2 3

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Testerflorida 19 5 · Answer 17 · Mar 10 at 19:50

Below i created two dictionaries. I need to return the key and value differences between them. I am stuck here. I am not sure which way is correct. I need to know how to get the key value difference. I want to first check if they are the same and if they are not to print key -value differences. I dont want to use deep diff. I dont know to compare if they are the same ?

num_list = [1,2]
val_list = [0,1]
dict1 = dict(zip(num_list,val_list))
print dict1

num_list2= [1,2]
val_list2 = [0,6]
dict2 = dict(zip(num_list2,val_list2))
print dict2
if dict1 == dict2

output : currently {1: 0, 2: 1} {1: 0, 2: 6}

smoke_lp 33 5 · Answer 18 · Jun 15 at 12:44

My recipe of symmetric difference between two dictionaries:

def find_dict_diffs(dict1, dict2):
    unequal_keys = []
    unequal_keys.extend(set(dict1.keys()).symmetric_difference(set(dict2.keys())))
    for k in dict1.keys():
        if dict1.get(k, 'N\A') != dict2.get(k, 'N\A'):
            unequal_keys.append(k)
    if unequal_keys:
        print 'param', 'dict1\t', 'dict2'
        for k in set(unequal_keys):
            print str(k)+'\t'+dict1.get(k, 'N\A')+'\t '+dict2.get(k, 'N\A')
    else:
        print 'Dicts are equal'

dict1 = {1:'a', 2:'b', 3:'c', 4:'d', 5:'e'}
dict2 = {1:'b', 2:'a', 3:'c', 4:'d', 6:'f'}

find_dict_diffs(dict1, dict2)

And result is:

param   dict1   dict2
1       a       b
2       b       a
5       e       N\A
6       N\A     f

Ryan de Kleer 96 8 · Answer 19 · Jul 15 at 21:55

As mentioned in other answers, unittest produces some nice output for comparing dicts, but in this example we don't want to have to build a whole test first.

Scraping the unittest source, it looks like you can get a fair solution with just this:

import difflib
import pprint

def diff_dicts(a, b):
    if a == b:
        return ''
    return '\n'.join(
        difflib.ndiff(pprint.pformat(a, width=30).splitlines(),
                      pprint.pformat(b, width=30).splitlines())
    )

so

dictA = dict(zip(range(7), map(ord, 'python')))
dictB = {0: 112, 1: 'spam', 2: [1,2,3], 3: 104, 4: 111}
print diff_dicts(dictA, dictB)

Results in:

{0: 112,
-  1: 121,
-  2: 116,
+  1: 'spam',
+  2: [1, 2, 3],
   3: 104,
-  4: 111,
?        ^

+  4: 111}
?        ^

-  5: 110}

Where:

'-' indicates key/values in the first but not second dict
'+' indicates key/values in the second but not the first dict

Like in unittest, the only caveat is that the final mapping can be thought to be a diff, due to the trailing comma/bracket.

Diogo Santiago 18 6 · Answer 20 · Jul 21 at 23:40

up vote 0 down vote

Try this to find de intersection, the keys that is in both dictionarie, if you want the keys not found on second dictionarie, just use the not in...

intersect = filter(lambda x, dictB=dictB.keys(): x in dictB, dictA.keys())

answered Jul 21 at 23:40

Diogo Santiago

18 6

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