Removing entries from a dictionary based on values

Question

I have a dictionary with character-integer key-value pair. I want to remove all those key value pairs where the value is 0.

For example:

>>> hand
{'a': 0, 'i': 0, 'm': 1, 'l': 1, 'q': 0, 'u': 0}

I want to reduce the same dictionary to this:

>>> hand
{'m': 1, 'l': 1}

Is there an easy way to do that?

Asked By: OneMoreError

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Source

Answer 1

A dict comprehension?

{k: v for k, v in hand.items() if v != 0}

In python 2.6 and earlier:

dict((k, v) for k, v in hand.items() if v != 0)

Answered By: Pavel Anossov

Answer 2

You can use a dict comprehension:

>>> { k:v for k, v in hand.items() if v }
{'m': 1, 'l': 1}

Or, in pre-2.7 Python, the dict constructor in combination with a generator expression:

>>> dict((k, v) for k, v in hand.iteritems() if v)
{'m': 1, 'l': 1}

Answered By: Niklas B.

Answer 3

hand = {k: v for k, v in hand.iteritems() if v != 0}

For Pre-Python 2.7:

hand = dict((k, v) for k, v in hand.iteritems() if v != 0)

In both cases you’re filtering out the keys whose values are 0, and assigning hand to the new dictionary.

Answered By: Volatility

Answer 4

If you don’t want to create a new dictionary, you can use this:

>>> hand = {'a': 0, 'i': 0, 'm': 1, 'l': 1, 'q': 0, 'u': 0}
>>> for key in list(hand.keys()):  ## creates a list of all keys
...     if hand[key] == 0:
...             del hand[key]
... 
>>> hand
{'m': 1, 'l': 1}
>>>

Answered By: Fabian

Removing entries from a dictionary based on values

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