比较两个字典的键/值并放入新字典中

Question

I know variations on this question already exist, but I cannot find one that matches exactly what I am trying to achieve. I have the following code, which included a solution I have taken from a solution to a similar question:

b = {"1":0,"2":0,"3":0,"4":0,"5":0}
c = {"1":1,"4":4,"5":5}

d = [k for k in b if c.get(k, object()) > b[k]]

print d

What I want is to compare all the key and value pairs of the dictionary b with those of c . If a key and value pair is missing from c then the key/pair value of b is retained in the dictionary d , else the values in c are retained in d .

In the example above d should look like this:

d = {"1":1,"2":0,"3":0,"4":4,"5":5}

Can anyone tell me the correct syntax I need for the line d = , please?

Thanks

Answer 1

To answer your actual question

What I want is to compare all the key and value pairs of the dictionary b with those of c. If a key and value pair is missing from c then the key/pair value of b is retained in the dictionary d, else the values in c are retained in d.

>>> b = {"1":0,"2":0,"3":0,"4":0,"5":0}
>>> c = {"1":1,"4":4,"5":5}
>>> d = {k: c.get(k, b[k]) for k in b}
>>> d
{'1': 1, '3': 0, '2': 0, '5': 5, '4': 4}

In Python 3 you should use collections.ChainMap (note this is slightly different in that it will take any key from c , not just the ones in b )

>>> from collections import ChainMap
>>> b = {"1":0,"2":0,"3":0,"4":0,"5":0}
>>> c = {"1":1,"4":4,"5":5}
>>> d = ChainMap(c, b)
>>> d
ChainMap({'1': 1, '5': 5, '4': 4}, {'1': 0, '3': 0, '2': 0, '5': 0, '4': 0})
>>> d['1'], d['2'], d['4']
(1, 0, 4)

Answer 2

You need to use an if-else condition in your comprehension, and also you dont need to use get when every thing is clear :

>>> d={k:c[k] if k in c and c[k]>b[k] else v for k,v in b.items()}
>>> d
{'1': 1, '3': 0, '2': 0, '5': 5, '4': 4}

Answer 3

Or perhaps this:

b = {"1":0,"2":0,"3":0,"4":0,"5":0}
c = {"1":1,"4":4,"5":5}

d = {k:b[k] if not c.get(k)>b[k] else c[k] for k in b}

d
{'1': 1, '2': 0, '3': 0, '4': 4, '5': 5}

Or, if you want to unpack both key/values from b to compare:

{k:v if not c.get(k)>v else c[k] for k, v in b.iteritems()}

The part of k:v is treated as key=k , and value= v only if k doesn't exist in c and c[k] value is > v , otherwise take v .

And since c.get(k) returns None if k doesn't exist in c, and None > v will automatically translate to False .