如何使用Python defaultdict将一个值除以另一个？

Question

I have a defaultdict built with keys and values, but I need to be able to divide the first value by the second value if there are two values in the pair.

defaultdict(<type 'list'>, {3: [7567, 6525], 4: [0], 65542: [609, 5245], 13:     
[73585, 84764], 14: [159, 19385], 65552: [1834], 22: [47333], 25: [0, 5320],
65562: [0], 98332: [0], 30: [0, 704249], 32: [5612], 33: [76050]}

So, for 3:, I would need to get 7567/6525, and put that into a new dictionary with the same key. But, for 32, I can't do any division, so I'd need to remove it from the set.

How would I go about doing this, knowing I don't always have 2 values to divide with?

Answer 1

Use a dictionary comprehension and the ternary operator.

Python 3

from operator import truediv

{key: (truediv(*val) if len(val) == 2 else val[0]) for key, val in dic.items()}

Python 2

from operator import truediv

{key: (truediv(*val) if len(val) == 2 else val[0]) for key, val in dic.iteritems()}

I'll continue with Python 3. If you want to remove those entries with less than 2 items

{key: div(*val) for key, val in dic.items() if len(val) == 2}

Update per your comment under the question

Wrapping the results back into lists and getting a list of the "outliers".

new_dic = {key: [div(*val)] for key, val in dic.items() if len(val) == 2}
outliers = [key for key, val in dic.items() if len(val) != 2]

Answer 2

I guess since everyone else is interpreting the question this way I'll post my version as an answer:

{k: reduce(operator.div, map(float, v)) for k, v in original.items()}

Alternatively:

{k: reduce(operator.truediv, v) for k, v in original.items()}

Where original is your dict and you import operator . If you're on Python 3 you'll need to from functools import reduce as well (global in Python 2).

Also, if you're using Python 3, you can drop map(float, v) in favor of just v . That's so you don't get burned by integer division (Floor division) on Python 2. You can use truediv instead of map to avoid this -- I was thinking, incorrectly, this was not portable between versions of Python. I stand corrected.

Note: This will not work if your dict values are not always iterable. You're using a defaultdict with a list default so you'll be fine.

Answer 3

You could do something like this:

for k in mydict.keys():
  if len(mydict[k]) != 2:
    mydict.pop(k)
  else:
    v1, v2 = mydict[k]
    mydict[k] = v1 / v2