Python。 如何减去2个字典

Question

I have 2 dictionaries, A and B. A has 700000 key-value pairs and B has 560000 key-values pairs. All key-value pairs from B are present in A, but some keys in A are duplicates with different values and some have duplicated values but unique keys. I would like to subtract B from A, so I can get the remaining 140000 key-value pairs. When I subtract key-value pairs based on key identity, I remove lets say 150000 key-value pairs because of the repeated keys. I want to subtract key-value pairs based on the identity of BOTH key AND value for each key-value pair, so I get 140000. Any suggestion would be welcome.

This is an example:

A = {'10':1, '11':1, '12':1, '10':2, '11':2, '11':3}
B = {'11':1, '11':2}

I DO want to get: AB = {'10':1, '12':1, '10':2, '11':3}

I DO NOT want to get:

a) When based on keys:

{'10':1, '12':1, '10':2}

or

b) When based on values:

{'11':3}

Answer 1

To get items in A that are not in B, based just on key:

C = {k:v for k,v in A.items() if k not in B}

To get items in A that are not in B, based on key and value:

C = {k:v for k,v in A.items() if k not in B or v != B[k]}

To update A in place (as in A -= B ) do:

from collections import deque
consume = deque(maxlen=0).extend
consume(A.pop(key, None) for key in B)

(Unlike using map() with A.pop , calling A.pop with a None default will not break if a key from B is not present in A. Also, unlike using all , this iterator consumer will iterate over all values, regardless of truthiness of the popped values.)

Answer 2

一个简单、直观的方法是

dict(set(a.items()) - set(b.items()))

Answer 3

A = {'10':1, '11':1, '12':1, '10':2, '11':2, '11':3}
B = {'11':1, '11':2}

You can't have duplicate keys in Python. If you run the above, it will get reduced to:

A={'11': 3, '10': 2, '12': 1}
B={'11': 2}

But to answer you question, to do A - B (based on dict keys):

all(map( A.pop, B))   # use all() so it works for Python 2 and 3.
print A # {'10': 2, '12': 1}

Answer 4

Another way of using the efficiency of sets. This might be more multipurpose than the answer by @brien . His answer is very nice and concise, so I upvoted it.

diffKeys = set(a.keys()) - set(b.keys())
c = dict()
for key in diffKeys:
  c[key] = a.get(key)

EDIT: There is the assumption here, based on the OP's question, that dict B is a subset of dict A, that the key/val pairs in B are in A. The above code will have unexpected results if you are not working strictly with a key/val subset. Thanks to Steven for pointing this out in his comment.

Answer 5

Since I can not (yet) comment: the accepted answer will fail if there are some keys in B not present in A.

Using dict.pop with a default would circumvent it (borrowed from How to remove a key from a Python dictionary? ):

all(A.pop(k, None) for k in B)

or

tuple(A.pop(k, None) for k in B)

Answer 6

dict-views :

Keys views are set-like since their entries are unique and hashable. If all values are hashable, so that (key, value) pairs are unique and hashable, then the items view is also set-like. (Values views are not treated as set-like since the entries are generally not unique.) For set-like views, all of the operations defined for the abstract base class collections.abc.Set are available (for example, ==, <, or ^).

So you can:

>>> A = {'10':1, '11':1, '12':1, '10':2, '11':2, '11':3}
>>> B = {'11':1, '11':2}
>>> A.items() - B.items()
{('11', 3), ('12', 1), ('10', 2)}
>>> dict(A.items() - B.items())
{'11': 3, '12': 1, '10': 2}

For python 2 use dict.viewitems .

PS You can't have duplicate keys in dict.

>>> A = {'10':1, '11':1, '12':1, '10':2, '11':2, '11':3}
>>> A
{'10': 2, '11': 3, '12': 1}
>>> B = {'11':1, '11':2}
>>> B
{'11': 2}

Answer 7

result = A.copy()
[result.pop(key) for key in B if B[key] == A[key]]

Answer 8

Based on only keys assuming A is a superset of B or B is a subset of A:

Python 3: c = {k:a[k] for k in a.keys() - b.keys()}

Python 2: c = {k:a[k] for k in list(set(a.keys())-set(b.keys()))}

Based on keys and can be used to update a in place as well @PaulMcG answer

Answer 9

For subtracting the dictionaries, you could do :

A.subtract(B)

Note: This will give you negative values in a situation where B has keys that A does not.