如何使用字典键比较列表值并使用python创建新字典 [英] How to compare list values with dictionary keys and make a new dictionary of it using python

问题描述

我有一个这样的列表:

lis = ['Date', 'Product', 'Price']

我想将其与:

dict = {'Date' : '2013-05-01', 'Salary' : '$5000', 'Product' : 'Toys', 'Price' : '$10', 'Salesman' : 'Smith'}

我想将列表中的每个项目与字典的键进行比较，然后创建一个新的字典.
我尝试过的是:

I want to compare each item of list with keys of dictionary and make a new dictionary.
What I have tried is:

n = {}
for k,v in dict.items():
    for i in lis:
        if i==k:
            n[k] = v

输出:

n = {'Date' : '2013-05-01', 'Product' : 'Toys', 'Price' : '$10'}

这可行，但是我想通过生成器来实现-有人可以帮助我做到这一点吗?

This works but I want to do it through generators - can someone help me do that?

推荐答案

将lis视为一组，因此您可以使用

Treat lis as a set instead, so you can use dictionary views and an intersection:

# python 2.7:
n = {k: d[k] for k in d.viewkeys() & set(lis)}

# python 3:
n = {k: d[k] for k in d.keys() & set(lis)}

或者您可以将简单的dict理解与针对d的in测试一起使用:

Or you could use a simple dict comprehension with a in test against d:

# python 2.6 or older:
n = dict((k, d[k]) for k in lis if k in d)

# python 2.7 and up:
n = {k: d[k] for k in lis if k in d}

这假定不是lis中的所有值都将在d中；如果if k in d测试始终存在，则可以将其删除.

This presumes that not all values in lis are going to be in d; the if k in d test can be dropped if they are always going to be present.

对于您的特定情况，第二种形式要快得多:

For your specific case, the second form is quite a lot faster:

>>> from timeit import timeit
>>> timeit("{k: d[k] for k in d.viewkeys() & s}", 'from __main__ import d, lis; s=set(lis)')
2.156520128250122
>>> timeit("{k: d[k] for k in lis if k in d}", 'from __main__ import d, lis')
0.9401540756225586

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