Concise way to find "key" difference between 2 dictionaries?

Question

I needed to compare 2 dictionaries to find the set of keys in one dictionary which was not in the other.

I know that Python set objects support:

set3=set1-set2

but I can't do:

dict3=dict1-dict2

or:

missingKeys=dict1.keys()-dict2.keys()

(I was a little surprised by the last point, because in Java the keys are a Set object.) One solution is:

missingKeys=set(dict1.keys())-set(dict2.keys())

is there a better or more concise way to do this?

I think the last line is sufficiently concise myself, but... I guess a more interesting question is "How to remove all z in Y from X?" where X and Y are lists. This would be useful where duplicate z's exist in X that do appear in Y should be left alone, for instance. — , Apr 22 '12 at 23:04
@pst:Well, it feels a little weird to create set objects, just to leverage the difference() function... — Sam Goldberg, Apr 22 '12 at 23:05
On the other hand, because it *is* a set, it can leverage a better O ... for list differences using comprehension is easy, but a tad more wordy. Still same performance if the "probed" list is converted to a `set`, though. — , Apr 22 '12 at 23:06

score 15 · Answer 1 · answered Apr 22 '12 at 23:07

15

Python 2.7:

>>> d = {1:2, 2:3, 3:4}
>>> d2 = {2:20, 3:30}
>>> set(d)-set(d2)
set([1])

Python 3.2:

>>> d = {1:2, 2:3, 3:4}
>>> d2 = {2:20, 3:30}
>>> d.keys()-d2.keys()
{1}

answered Apr 22 '12 at 23:07

DSM

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So you are saying that Python 3.2 dict.keys() method returns a set (like Java)? Any reason why earlier version didn't do that? – Sam Goldberg Apr 23 '12 at 00:08
2

@SamGoldberg: no, it actually returns a dict_keys object, not a set, but it has many set-like operations. See, e.g. [this question](http://stackoverflow.com/questions/7296716/what-is-dict-keys-dict-items-and-dict-values). – DSM Apr 23 '12 at 00:19
1

`d.keys() - d2` or even `d - d2.keys()` also work :) Just a general note, they're less readable, IMHO. – Niklas B. Apr 23 '12 at 19:21
How about the time complexity? Will `d.keys()-d2.keys()` work in `O(|len(d2)|)` as expected? – Tregoreg Feb 12 '14 at 20:52
Python 2.7 has a backport of Python 3's `.keys()`, so you can also do `d.viewkeys() - d2.viewkeys()`. – rescdsk Nov 14 '14 at 23:39

score 4 · Answer 2 · answered Nov 18 '13 at 07:47

For portable way of doing it I would suggest using dict.viewkeys in Python 2.7 - it is backport of Python 3.x dict.keys and is automatically converted by 2to3.

Example:

>>> left = {1: 2, 2: 3, 3: 4}
>>> right = {2: 20, 3:30}
>>> left.viewkeys() - right.viewkeys()
set([1])

score 3 · Accepted Answer · answered Apr 22 '12 at 23:10

3

Maybe

[x for x in dict1.keys() if x not in dict2.keys()]

answered Apr 22 '12 at 23:10

user1245262

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I think this could be even a little more concise: `[x for x in dict1.keys() if x not in dict2]` since the `x in dict.keys()` is the same as `x in dict` – mgilson Apr 22 '12 at 23:38
@mgilson: as long as you pointed out the `x in dict` is the same as `x in dict.keys()`, then the complete reduction is `[x for x in dict1 if x not in dict2]`. – Sam Goldberg Apr 23 '12 at 00:15
1

@SamGoldberg You're correct. The reason I didn't think of it is because `for x in dict` calls the `__iter__` method on a dict whereas `if x in dict` calls the `__contains__` method. So, in this case, the `in` does two different things -- they just happen to have the same result. Anyway: `[x for x in dict1 if x not in dict2]` is most concise as you pointed out. – mgilson Apr 23 '12 at 00:54
Both answers were good, but accepting this answer because I am using Python 2.7. Being new to Python, this answer also helped me be more conscious of the [for ... in ... if ...] syntax, which is less natural to me than the d1-d2 syntax. – Sam Goldberg May 04 '12 at 17:26
To compute difference between sets of values you may use sets like in the response of Zbyszek Mandziejewicz. – Julien Palard Jul 14 '14 at 20:16

score 1 · Answer 4 · answered Nov 14 '14 at 23:44

1

This should work in Python 2.7 and 3.x:

>>> keys = getattr(dict, 'viewkeys', dict.keys)
>>> left = {1: 2, 2: 3, 3: 4}
>>> right = {2: 20, 3:30} 
>>> list(keys(left) - keys(right))
[1]

answered Nov 14 '14 at 23:44

rescdsk

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Concise way to find "key" difference between 2 dictionaries?

4 Answers4