[Solved] Python: return first value by condition of nested dictionary in array [closed]

Question

Your code works fine for me using your (slightly edited) sample data:

data = [{'id': 1, 'name': 'test'}, {'id': 2, 'name': 'test'}, {'id': 3, 'name': 'test'}]
val = [x['id'] for x in data if x['name'] == 'test'][0]

>>> print(val)
1

However, if there is no dictionary containing a name that matches the target string:

>>> val = [x['id'] for x in data if x['name'] == 'blah'][0]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: list index out of range

This is because the list comprehension will create an empty list because there are no dictionaries with d['name'] set to 'blah'. Indexing an empty list results in the IndexError exception. It’s the same as doing this:

>>> [][0]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: list index out of range

A simple way to fix this is to check the list before indexing it:

matches = [x['id'] for x in data if x['name'] == 'test']
val = matches[0] if matches else None

here it is assumed that None can not be used as the value for an id.

Another, more efficient way, again assuming that None is not a valid id is to use next() with a default value:

val = next((x['id'] for x in data if x['name'] == 'test'), None)

This uses a generator expression which avoids generating a whole list containing the matched dictionaries. Instead it will iterate over the data list only until the first match is found, or the data list is exhausted.

Accepted Answer