在 Python 中组合列表字典
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Combining Dictionaries Of Lists In Python
提问by user108088
I have a very large collection of (p, q) tuples that I would like to convert into a dictionary of lists where the first item in each tuple is a key that indexes a list that contains q.
我有一个非常大的 (p, q) 元组集合，我想将它们转换为列表字典，其中每个元组中的第一项是索引包含 q 的列表的键。
Example:
例子：
Original List: (1, 2), (1, 3), (2, 3)
Resultant Dictionary: {1:[2, 3], 2:[3]}
Furthermore, I would like to efficiently combine these dictionaries.
此外，我想有效地组合这些词典。
Example:
例子：
Original Dictionaries: {1:[2, 3], 2:[3]}, {1:[4], 3:[1]}
Resultant Dictionary: {1:[2, 3, 4], 2:[3], 3:[1]}
These operations reside within an inner loop, so I would prefer that they be as fast as possible.
这些操作驻留在内部循环中，所以我希望它们尽可能快。
Thanks in advance
提前致谢
采纳答案by Alex Martelli
If the list of tuples is sorted, itertools.groupby, as suggested by @gnibbler, is not a bad alternative to defaultdict, but it needs to be used differently than he suggested:
如果元组列表已排序，itertools.groupby正如@gnibbler 所建议的那样， 不是一个糟糕的替代方案defaultdict，但它的使用方式需要与他建议的不同：
import itertools
import operator
def lot_to_dict(lot):
key = operator.itemgetter(0)
# if lot's not sorted, you also need...:
# lot = sorted(lot, key=key)
# NOT in-place lot.sort to avoid changing it!
grob = itertools.groupby(lot, key)
return dict((k, [v[1] for v in itr]) for k, itr in grob)
For "merging" dicts of lists into a new d.o.l...:
将列表的字典“合并”到一个新的 dol.. 中：
def merge_dols(dol1, dol2):
keys = set(dol1).union(dol2)
no = []
return dict((k, dol1.get(k, no) + dol2.get(k, no)) for k in keys)
I'm giving []a nickname noto avoid uselessly constructing a lot of empty lists, given that performance is important. If the sets of the dols' keys overlap only modestly, faster would be:
鉴于性能很重要，我给出[]了一个昵称，no以避免无用地构建大量空列表。如果 dols 的密钥集仅适度重叠，则更快：
def merge_dols(dol1, dol2):
result = dict(dol1, **dol2)
result.update((k, dol1[k] + dol2[k])
for k in set(dol1).intersection(dol2))
return result
since this uses list-catenation only for overlapping keys -- so, if those are few, it will be faster.
因为这仅对重叠的键使用列表连接 - 所以，如果这些键很少，它会更快。
回答by SilentGhost
collections.defaultdictworks like this:
collections.defaultdict像这样工作：
from collections import defaultdict
dic = defaultdict(list)
for i, j in tuples:
dic[i].append(j)
similar for the dicts:
类似于 dicts：
a, b = {1:[2, 3], 2:[3]}, {1:[4], 3:[1]}
de = defaultdict(list, a)
for i, j in b.items():
de[i].extend(j)
回答by Triptych
defaltdict to the rescue (as usual)
defaltdict 救援（像往常一样）
from collections import defaultdict
my_dict = defaultdict(list)
for key,value in original_list:
my_dict[key].append(value)
Combining the two dicts can be done like this (note that duplicates will be preserved):
可以像这样组合两个字典（注意重复项将被保留）：
for key,value in orig_dict:
new_dict[key].extend(value)
回答by John La Rooy
Here is the iterator style of doing it
这是这样做的迭代器风格
>>> mylist=[(1, 2), (1, 3), (2, 3)] >>> from itertools import groupby >>> from operator import itemgetter >>> mylist=[(1, 2), (1, 3), (2, 3)] >>> groupby(mylist,itemgetter(0)) >>> list(_) [(1, <itertools._grouper object at 0xb7d402ec>), (2, <itertools._grouper object at 0xb7c716ec>)]
回答by Tarnay Kálmán
I wanted these done in one linejust for fun:
我希望这些在一行中完成只是为了好玩：
>>> from itertools import groupby
>>> t=(1, 2), (1, 3), (2, 3)
>>> [(i,[x for _,x in list(f)]) for i,f in groupby(sorted(t),lambda t: t[0])]
[(1, [2, 3]), (2, [3])]
>>> b={1:[2, 3], 2:[3]}, {1:[4], 3:[1]}
>>> dict([(key,sum([i[1::][0] for i in elements],[])) for key,elements in groupby(sorted(b[0].items()+b[1].items()),lambda t: t[0])])
{1: [2, 3, 4], 2: [3], 3: [1]}
