结合Python中的列表字典 [英] Combining Dictionaries Of Lists In Python
问题描述
示例:
原始列表:(1,2),(1,3) ,(2,3)
结果词典:{1:[2,3],2:[3]}
此外,我想高效地组合这些字典。
示例:
原始字典:{1:[2,3],2:[3]},{1:[4],3:[1]}
结果词典:{1 :[2,3,4],2:[3],3:[1]}
这些操作驻留在内部循环中,所以我更喜欢它们尽可能的快。
提前感谢
如果元组列表被排序,则由@gnibbler建议的 itertools.groupby
不是一个坏的替代方法 defaultdict
,但需要使用diff比他建议的更多:
import itertools
import operator
def lot_to_dict(lot) :
key = operator.itemgetter(0)
#如果很多没有排序,你还需要...:
#lot = sorted(lot,key = key)
#不要在现场lot.sort避免改变它!
grob = itertools.groupby(lot,key)
return dict((k,[v [1] for v in itr])for k,itr in grob
将列表的合并到新的dol ..:
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 key)
我给 []
一个昵称 no
以避免无用地构建大量空列表,因为性能很重要如果dols的键的集合只适度地重叠,则速度会更快:
def merge_dols(dol1,dol2):
result = dict(dol1,** dol2)
result.update((k,dol1 [k] + dol2 [k])
在集合(dol1)中的k .intersection(dol2))
返回结果
因为这仅使用列表连接仅用于重叠的键 - 所以,如果那些很少,会更快。
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.
Example:
Original List: (1, 2), (1, 3), (2, 3)
Resultant Dictionary: {1:[2, 3], 2:[3]}
Furthermore, I would like to efficiently combine these dictionaries.
Example:
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.
Thanks in advance
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:
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...:
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 no
to 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:
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.
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