How can I check if key exists in list of dicts in python?

Question

Say I have a list of dicts that looks like this:

[{1: "a"}, {2: "b"}]

What is the pythonic way to indicate if a certain key is in one of the dicts in the list?

question from:https://stackoverflow.com/questions/14790980/how-can-i-check-if-key-exists-in-list-of-dicts-in-python

Answer 1

I'd probably write:

>>> lod = [{1: "a"}, {2: "b"}]
>>> any(1 in d for d in lod)
True
>>> any(3 in d for d in lod)
False

although if there are going to be a lot of dicts in this list you might want to reconsider your data structure.

If you want the index and/or the dictionary where the first match is found, one approach is to use next and enumerate:

>>> next(i for i,d in enumerate(lod) if 1 in d)
0
>>> next(d for i,d in enumerate(lod) if 1 in d)
{1: 'a'}
>>> next((i,d) for i,d in enumerate(lod) if 1 in d)
(0, {1: 'a'})

This will raise StopIteration if it's not there:

>>> next(i for i,d in enumerate(lod) if 3 in d)
Traceback (most recent call last):
  File "<ipython-input-107-1f0737b2eae0>", line 1, in <module>
    next(i for i,d in enumerate(lod) if 3 in d)
StopIteration

If you want to avoid that, you can either catch the exception or pass next a default value like None:

>>> next((i for i,d in enumerate(lod) if 3 in d), None)
>>>

As noted in the comments by @drewk, if you want to get multiple indices returned in the case of multiple values, you can use a list comprehension:

>>> lod = [{1: "a"}, {2: "b"}, {2: "c"}]
>>> [i for i,d in enumerate(lod) if 2 in d]
[1, 2]