Permutations of dictionary in python

Question

I have a dictionary like this -

{'A': 0, 'B': 0, 'C': 0, 'D': 4}

I want to generate a list like this -

[{'A': 1, 'B': 0, 'C': 0, 'D': 4},
 {'A': 0, 'B': 1, 'C': 0, 'D': 4},
 {'A': 0, 'B': 0, 'C': 1, 'D': 4},
 {'A': 0, 'B': 0, 'C': 0, 'D': 5}]

What is the most pythonic way to do this?

What rules do you have? Some of the numbers in your 'permutations' seem to come from no-where. — anon582847382
– anon582847382, Commented Mar 31, 2014 at 6:47
Oh sorry, I should've said what is the most pythonic way to do this? Updated question to reflect this. — Narabhut
– Narabhut, Commented Mar 31, 2014 at 6:47
And the values in the permutations should be increased by 1 from the source one. — Narabhut
– Narabhut, Commented Mar 31, 2014 at 6:48
as @AlexThornton asked, where those 1s and 5 are coming from? Please provide full input and some explanation about rules. — m.wasowski
– m.wasowski, Commented Mar 31, 2014 at 6:49
Permutation means reorder elements, but your values just changed, value of each key in each step. It's not what you call permutation I'm afraid. — Michał
– Michał, Commented Mar 31, 2014 at 6:49

thefourtheye · Accepted Answer · 2014-03-31 06:56:59Z

5

You can use list comprehension and dictionary comprehension together, like this

d = {'A': 0, 'B': 0, 'C': 0, 'D': 4}
print [{key1: d[key1] + (key1 == key) for key1 in d} for key in d]

Output

[{'A': 1, 'B': 0, 'C': 0, 'D': 4},
 {'A': 0, 'B': 0, 'C': 1, 'D': 4},
 {'A': 0, 'B': 1, 'C': 0, 'D': 4},
 {'A': 0, 'B': 0, 'C': 0, 'D': 5}]

The idea is to generate a new dictionary for each key, and when the key matches the key of the dictionary being constructed with dictionary comprehension, then add 1 to it. (key1 == key) will evaluate to 1 only when both the keys match, otherwise it will be zero.

edited Mar 31, 2014 at 6:56

answered Mar 31, 2014 at 6:51

thefourtheye

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1 Comment

thefourtheye Over a year ago

@Tim Thanks Removed that :) I was about write a normal loop based solution, but I got the comprehension idea. So, forgot to remove that :)

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