Python - choose a single element with some property from an array

Question

I have come across this case a few times now: I have an array which contains an element I want and I want to select that element without knowing the index, and instead knowing some desired property (maybe it is a list of dictionaries and I want the dictionary elem such that elem['foo'] == 'bar').

A solution I have is do a filter (or more pythonically, a list comprehension) and then take the first element (often times I know that the filtered list will be a singleton, so taking the first is taking the only).

Ex. Given a list x = [{'foo': 'bar_1'}, {'foo': 'bar_2'}], I want the element whose 'foo' value is 'bar2'. So I do this :

y = [elem for elem in x if elem['foo'] == 'bar_2'][0]

Is there a more standard way of doing this? It seems like a very simple and common use case.

possible duplicate of Python: find first element in a sequence that matches a predicate — Ryan Haining
– Ryan Haining, Commented Aug 28, 2014 at 21:53

unutbu · Accepted Answer · 2014-08-28 20:31:00Z

1

If you run into this problem frequently then consider using a different data structure. Think about whether a list of dicts is really the best data structure for the problem you are solving. Or, if it is, add a dict to keep track of the information you require. In this case, as you build x, also build index:

index = {'bar2':x[1], 'bar1':x[0]}

Dict lookups are O(1), whereas any list-based lookup is O(n).

edited Aug 28, 2014 at 20:31

answered Aug 28, 2014 at 20:20

unutbu

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Comments

Padraic Cunningham · Accepted Answer · 2014-08-28 20:53:48Z

1

You can use a generator calling next on it to get the first match:

l = [{'foo': 'bar_1'}, {'foo': 'bar_2'}]
print next(d for d in l if d["foo"] == "bar_2")
{'foo': 'bar_2'}

d = (d for d in l if d["foo"] == "bar_2")
first = next(d)

You could also you itertools.dropwhile and if you want the first two elements:

In [52]: from itertools import dropwhile,takewhile
In [53]: l = [{'foo': 'bar_1'}, {'foo': 'bar_2'},{'foo': 'bar_2',"goo":"bar"}]
In [54]: d = dropwhile(lambda x: x["foo"] != "bar_2",l)   # drop elements whose where value of key "foo" is not equal to "bar_2"
In [55]: first = next(d)
In [56]: second = next(d)    
In [57]: first
Out[57]: {'foo': 'bar_2'}    
In [58]: second
Out[58]: {'foo': 'bar_2', 'goo': 'bar'}

edited Aug 28, 2014 at 20:53

answered Aug 28, 2014 at 20:24

Padraic Cunningham

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2 Comments

Dunes Over a year ago

Alternatively, item, = (d for d in l if d["foo"] == "bar_2") will assert that there is only one item in the list with the given property (which may, or may not be desired).

Padraic Cunningham Over a year ago

@Dunes, yes true, I am not sure if the OP wants to check if there is only one match or just get the first.

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