通过识别具有相同键但具有不同值的集合来比较Python中的两个词典

Question

I'm trying to compare two dictionaries by comparing the keys, if two keys in the two seperate dictionaries are the same the program should check if the values are also the same, if they are not the same the program should identify that.

This is the code I have written:

def compare(firstdict,seconddict):
    shared_items = set(firstdict()) & set(seconddict())
    length = len(shared_items)
    if length > 0:
        return shared_items
    if length < 1:
        return None
print(compare(firstdict,seconddict))

('firstdict' and 'seconddict' are two dictionaries which have been made in previous functions).

When the code is run it prints out all the keys that are the same without their values, even if their values are different.

For example if:

firstdict = {'cat' : 'animal', 'blue' : 'colour', 'sun' : 'star'}

seconddict = {'cat' : 'pet', 'blue' : 'colour', 'earth' : 'star'}   

it would print out:

'cat', 'blue'

whereas I'm tring to get it to print out:

'cat pet (animal)'

in that exact format.

Any advice on how to edit my code to do this is appreciated :)

Answer 1

You can use set intersection on the dictionaries keys() . Then loop over those and check if the values corresponding to those keys are identical. If not, you can print them out with format .

def compare(first, second):
    sharedKeys = set(first.keys()).intersection(second.keys())
    for key in sharedKeys:
        if first[key] != second[key]:
            print('Key: {}, Value 1: {}, Value 2: {}'.format(key, first[key], second[key]))

>>> compare(firstdict, seconddict)
Key: cat, Value 1: animal, Value 2: pet

And for another example

>>> firstdict = {'cat' : 'animal', 'blue' : 'colour', 'sun' : 'star', 'name': 'bob', 'shape': 'circle'}
>>> seconddict = {'cat' : 'pet', 'blue' : 'colour', 'earth' : 'star', 'name': 'steve', 'shape': 'square'}

>>> compare(firstdict, seconddict)
Key: shape, Value 1: circle, Value 2: square
Key: cat, Value 1: animal, Value 2: pet
Key: name, Value 1: bob, Value 2: steve

Answer 2

If you values are hashable also you can use items to get the common key/value pairings:

firstdict = {'cat' : 'animal', 'blue' : 'colour', 'sun' : 'star'}

seconddict = {'cat' : 'pet', 'blue' : 'colour', 'earth' : 'star'}

common = set(firstdict.iteritems()).intersection(seconddict.iteritems())

for k,v in common:
    print("Key: {}, Value: {}".format(k,v))
Key: blue, Value: colour

To check if both dicts are the same, check the len of each:

print(len(common)) == len(firstdict)

To find common keys with different value:

for k,v in firstdict.iteritems():
    if k in seconddict and seconddict[k] != v:
        print("Key: {}, Value: {}".format(k, seconddict[k]))
Key: cat, Value: pet

Answer 3

The use of sets in your example code makes it less efficient than simply looping over the keys, which is also, perhaps, more readable:

firstdict = {'cat' : 'animal', 'blue' : 'colour', 'sun' : 'star'}
seconddict = {'cat' : 'pet', 'blue' : 'colour', 'earth' : 'star'}

def compare(firstdict, seconddict):
    for key in firstdict:
        if key in seconddict:
            first, second = firstdict[key], seconddict[key]
            if first != second:
                print('%s %s (%s)' % (key, first, second))

compare(firstdict, seconddict)

output:

cat animal (pet)

Answer 4

Well Padraic's methods seem better than this one but it is another option. So this is not the best way but it will get the job done.

Step through each element and compare them manually.

def compare(dictOne,dictTwo):
    for keyOne in dictOne:
        for keyTwo in dictTwo:
            if keyTwo == keyOne:
                if dictOne[keyOne] != dictTwo[keyTwo]:
                    print(keyOne+" "+dictOne[keyOne]+" ("+dictTwo[keyTwo]+")")

compare(firstdict,seconddict)

Your question said it was suppose to print out, and did not mention you want them returned in array so the above code will step through both dicts comparing them one by one and print out in the format the one that do not match.