计算python中字典之间键重复的次数

Question

How do I check to see how many times the keys in one dict1 exist in dict2 . If the keys of dict1 exist in dict2 a variable, val , with an initial value of 4 should be subtracted based on how many times the keys are found.

For example dict1 looks like this

print dict1
{(2, 0): 3, (3, 1): 0, (1, 1): 2, (2, 2): 1} 

and dict2 looks like this

print `dict2`
{(2, 0): 323, (3, 1): 32, (10, 10): 21, (20, 2): 100} 

Since there are two repeat keys between the dicts, val should be equal to 2 .

if dict2 looks identical to dict1 , then val should be 0 .

Also, dict1 will always be the same size, but dict2 can get quite large, so a fast lookup method would be ideal. Lastly, the values of dicts here don't really mean anything.

Answer 1

Using set intersection:

d1 = {(2, 0): 3, (3, 1): 0, (1, 1): 2, (2, 2): 1} 
d2 = {(2, 0): 323, (3, 1): 32, (10, 10): 21, (20, 2): 100} 

sd1 = set(d1.keys())
sd2 = set(d2.keys())
len(sd1.intersection(sd2))

Edit: Because key views are already set-like, so you can do d1.keys() & d2.keys() directly. Note that .keys() is a very cheap call because it simply provides an alternative interface to the existing dict structure (Credits to @PM2RING in the comments).

Answer 2

Since dict_keys are already set-like , you can simply use

len(dict1.keys() & dict2.keys())

That is for Python 3. In Python 2, the equivalent view object is dict.viewkeys() , which you'd use similarly.

len(dict1.viewkeys() & dict2.viewkeys())

Answer 3

Make sets out of the list of keys in each dict. Find the intersection and union of those sets. union - intersection gives you the set of differences. If that's 0, then you return 0; otherwise, return the size of the intersection.

Answer 4

This will work in Python 2 and 3: len(set(dict1).intersection(dict2))

In [1]: dict1 = {(2, 0): 3, (3, 1): 0, (1, 1): 2, (2, 2): 1}

In [2]: dict2 = {(2, 0): 323, (3, 1): 32, (10, 10): 21, (20, 2): 100}

In [3]: len(set(dict1).intersection(dict2))
Out[3]: 2