是否有一个“堆栈和地图"?与字典类似? [英] Is there a "stack and map" analogue with dicts?
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问题描述
我最近不得不将字典键映射到评估问题中的值.我从以下内容开始:
I recently had to map dict keys to values in an assessment question. I started with the following:
files=
{'Code.py': 'Stan', 'Output.txt': 'Randy', 'Input.txt': 'Randy'}
然后将文件映射到它们的所有者,为此,我使用了以下内容:
And was to map the files to their owners, for which I used the following:
mapped={
name:[key for key,value in files.items() if value==name]
for name in list(set([value for key,value in files.items()]))
}
在mapped
字典中,我给了我想要的东西:
Which gave me what I wanted in the mapped
dict:
{'Stan': ['Code.py'], 'Randy': ['Output.txt', 'Input.txt']}
我只是想知道是否还有更多类似熊猫的方法来做同样的事情,但是使用简单的字典.
I was just wondering if there is a more Pandas-like way of doing the same thing, but with a plain dictionary.
推荐答案
您可以只使用defaultdict
:
from collections import defaultdict
mapped = defaultdict(list)
for k, v in files.items():
mapped[v].append(k)
mapped
# defaultdict(list, {'Stan': ['Code.py'], 'Randy': ['Output.txt', 'Input.txt']})
或在字典上使用setdefault
方法:
mapped = {}
for k, v in files.items():
mapped.setdefault(v, []).append(k)
mapped
# {'Stan': ['Code.py'], 'Randy': ['Output.txt', 'Input.txt']}
或者,如果您更喜欢pandas
(但是对于此任务而言效率不高):
Or if you prefer pandas
(which would not be as efficient for this task however):
s = pd.Series(files)
s.groupby(s).agg(lambda x: x.index.tolist()).to_dict()
# {'Randy': ['Input.txt', 'Output.txt'], 'Stan': ['Code.py']}
对小样本数据进行计时:
Timing on the small sample data:
%%timeit
from collections import defaultdict
mapped = defaultdict(list)
for k, v in files.items():
mapped[v].append(k)
# 2 µs ± 33.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
%%timeit
s = pd.Series(files)
s.groupby(s).agg(lambda x: x.index.tolist()).to_dict()
# 2.12 ms ± 54.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
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