Python array of tuples group by first, store second

问题

So I have an array of tuples something like this

query_results = [("foo", "bar"), ("foo", "qux"), ("baz", "foo")]

I would like to achieve something like:

{
    "foo": ["bar", "qux"],
    "baz": ["foo"]
}

So I have tried using this

from itertools import groupby

grouped_results = {}
for key, y in groupby(query_results, lambda x: x[0]):
    grouped_results[key] = [y[1] for u in list(y)]

The issue I have is although the number of keys are correct, the number of values in each array is dramatically lower than it should be. Can anyone explain why this happens and what I should be doing?

回答1:

You better use a defaultdict for this:

from collections import defaultdict

result = defaultdict(list)

for k,v in query_results:
    result[k].append(v)

Which yields:

>>> result
defaultdict(<class 'list'>, {'baz': ['foo'], 'foo': ['bar', 'qux']})

If you wish to turn it into a vanilla dictionary again, you can - after the for loop - use:

result = dict(result)

this then results in:

>>> dict(result)
{'baz': ['foo'], 'foo': ['bar', 'qux']}

A defaultdict is constructed with a factory, here list. In case the key cannot be found in the dictionary, the factory is called (list() constructs a new empty list). The result is then associated with the key.

So for each key k that is not yet in the dictionary, we will construct a new list first. We then call .append(v) on that list to append values to it.

回答2:

Well why not use a simple for loop?

grouped_results = {}
for key, value in query_results:
    grouped_results.setdefault(key, []).append(value)

Output:

{'foo': ['bar', 'qux'], 'baz': ['foo']}

回答3:

How about using a defaultdict?

d = defaultdict(list)
for pair in query_results:
    d[pair[0]].append(pair[1])

来源：https://stackoverflow.com/questions/45490667/python-array-of-tuples-group-by-first-store-second

标签

python

list

dictionary

group-by