CODEWITHSUNDEEP
pythonpython-2.7for-loopdictionary
vincent
vincent

Reputation: 1390

Refer to dictionary element in 'for' loop

If I have a list of dictionaries, is there a way to refer to a specific element within the for loop declaration?

Something like this:

dict_lst = [
  {'a': 1, 'b': 2, 'c': 3},
  {'a': 1, 'b': 2, 'c': 3},
  {'a': 1, 'b': 2, 'c': 3},
]

for d['a'] as dict_elem in dict_lst:
  print dict_elem

Or something similar?

Upvotes: 0

Views: 139

Answers (4)

Andres
Andres

Reputation: 4501

you can use this:

for element in dict_lst:
    print element['some_key']

Upvotes: 0

wenzul
wenzul

Reputation: 4058

This is one of the fastest and Pythonic methods to do it, lean on Python 3:

for key, val in d.iteritems():
    print key, val

There is also a similar method items which does almost the same. items will create a list of tuples with key and value if you call it in the for loop statement. There is also a similar method keys which only returns you a list of keys.

iteritems will not generate a list of tuples when you call it in the for loop. Each element will be evaluated in each loop cycle. So there is no initial overhead and extensive memory usage. If you break the loop it's better as well.

Upvotes: 0

abarnert
abarnert

Reputation: 365835

No, there's no as clause in for statements.

But it's pretty easy to do this explicitly:

for d in dict_lst:
    dict_elem = d['a']
    print dict_elem

Or, more simply:

for d in dict_list:
    print d['a']

Or, if you want to get fancy:

for dict_elem in (d['a'] for d in dict_lst):
    print dict_elem

Or, just for fun:

for dict_elem in map(operator.itemgetter('a'), dict_lst):
    print dict_elem

Which you can wrap up as a reusable function:

def itemmap(dicts, key):
    for item in dicts:
        yield item[key]

for dict_elem in itemmap(dict_lst, 'a'):
    print dict_elem

As a side note, you actually can do for d['a'] in dict_list:, but that just reassigns d['a'] (assuming d is already a dictionary) to each new dictionary in the list, which is usually a confusing thing to do. (It can be useful for obfuscated code contests…)

Upvotes: 3

user2555451
user2555451

Reputation:

You can use a generator expression like so:

>>> dict_lst = [
...   {'a': 1, 'b': 2, 'c': 3},
...   {'a': 1, 'b': 2, 'c': 3},
...   {'a': 1, 'b': 2, 'c': 3},
... ]
>>> for dict_elem in (d['a'] for d in dict_lst):
...     dict_elem
...
1
1
1
>>>

Upvotes: 2

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