Python字典：获取键列表的值列表

Question

是否有内置/快速方法可以使用字典键列表来获取相应项目列表？

例如我有：

>>> mydict = {'one': 1, 'two': 2, 'three': 3}
>>> mykeys = ['three', 'one']

如何使用mykeys将字典中的相应值作为列表获取？

>>> mydict.WHAT_GOES_HERE(mykeys)
[3, 1]

Answer 1

列表理解似乎是一种很好的方法：

>>> [mydict[x] for x in mykeys]
[3, 1]

Answer 2

除了list-comp之外的其他几种方式：

• 如果找不到密钥，则构建列表并抛出异常：map(mydict.__getitem__, mykeys)
• 如果找不到密钥，则使用None构建列表：map(mydict.get, mykeys)

或者，使用operator.itemgetter可以返回元组：

from operator import itemgetter
myvalues = itemgetter(*mykeys)(mydict)
# use `list(...)` if list is required

注意：在Python3中，map返回迭代器而不是列表。使用list(map(...))获取列表。

Answer 3

一点速度比较：

Python 2.7.11 |Anaconda 2.4.1 (64-bit)| (default, Dec  7 2015, 14:10:42) [MSC v.1500 64 bit (AMD64)] on win32
In[1]: l = [0,1,2,3,2,3,1,2,0]
In[2]: m = {0:10, 1:11, 2:12, 3:13}
In[3]: %timeit [m[_] for _ in l]  # list comprehension
1000000 loops, best of 3: 762 ns per loop
In[4]: %timeit map(lambda _: m[_], l)  # using 'map'
1000000 loops, best of 3: 1.66 µs per loop
In[5]: %timeit list(m[_] for _ in l)  # a generator expression passed to a list constructor.
1000000 loops, best of 3: 1.65 µs per loop
In[6]: %timeit map(m.__getitem__, l)
The slowest run took 4.01 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 853 ns per loop
In[7]: %timeit map(m.get, l)
1000000 loops, best of 3: 908 ns per loop
In[33]: from operator import itemgetter
In[34]: %timeit list(itemgetter(*l)(m))
The slowest run took 9.26 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 739 ns per loop

所以列表理解和itemgetter是最快的方法。

更新： 对于大型随机列表和地图，我的结果有点不同：

Python 2.7.11 |Anaconda 2.4.1 (64-bit)| (default, Dec  7 2015, 14:10:42) [MSC v.1500 64 bit (AMD64)] on win32
In[2]: import numpy.random as nprnd
l = nprnd.randint(1000, size=10000)
m = dict([(_, nprnd.rand()) for _ in range(1000)])
from operator import itemgetter
import operator
f = operator.itemgetter(*l)
%timeit f(m)
%timeit list(itemgetter(*l)(m))
%timeit [m[_] for _ in l]  # list comprehension
%timeit map(m.__getitem__, l)
%timeit list(m[_] for _ in l)  # a generator expression passed to a list constructor.
%timeit map(m.get, l)
%timeit map(lambda _: m[_], l)
1000 loops, best of 3: 1.14 ms per loop
1000 loops, best of 3: 1.68 ms per loop
100 loops, best of 3: 2 ms per loop
100 loops, best of 3: 2.05 ms per loop
100 loops, best of 3: 2.19 ms per loop
100 loops, best of 3: 2.53 ms per loop
100 loops, best of 3: 2.9 ms per loop

因此，在这种情况下，明显的赢家是f = operator.itemgetter(*l); f(m)，而明确的局外人：map(lambda _: m[_], l)。

Python 3.6.4的更新：

import numpy.random as nprnd
l = nprnd.randint(1000, size=10000)
m = dict([(_, nprnd.rand()) for _ in range(1000)])
from operator import itemgetter
import operator
f = operator.itemgetter(*l)
%timeit f(m)
%timeit list(itemgetter(*l)(m))
%timeit [m[_] for _ in l]  # list comprehension
%timeit list(map(m.__getitem__, l))
%timeit list(m[_] for _ in l)  # a generator expression passed to a list constructor.
%timeit list(map(m.get, l))
%timeit list(map(lambda _: m[_], l)
1.66 ms ± 74.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
2.1 ms ± 93.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.58 ms ± 88.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.36 ms ± 60.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.98 ms ± 142 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.7 ms ± 284 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
3.14 ms ± 62.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

因此，Python 3.6.4的结果几乎相同。

Answer 4

试试这个：

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Slingshot.fileRestrictions( "myDefinedDirective", {
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Slingshot.createDirective( "myDefinedDirective", Slingshot.GoogleCloud, {
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    bucket: "img_appengine",
    acl: "public-read",
    authorize: function () {
        return true;
    },
    key: function ( file ) {
        var user = Meteor.users.findOne( this.userId );
        return user.emails[0].address + "/" + file.name;
    }
});

Answer 5

以下是三种方式。

找不到密钥时提升await：

KeyError

缺失密钥的默认值。

result = [mapping[k] for k in iterable]

跳过丢失的钥匙。

result = [mapping.get(k, default_value) for k in iterable]

Answer 6

试试这个：

mydict = {'one': 1, 'two': 2, 'three': 3}
mykeys = ['three', 'one'] # if there are many keys, use a set

[mydict[k] for k in mykeys]
=> [3, 1]

Answer 7

Pandas非常优雅地做到这一点，尽管列表理解总是在技术上更像Pythonic。我现在没有时间进行速度比较（稍后我会回来把它放进去）：

import pandas as pd
mydict = {'one': 1, 'two': 2, 'three': 3}
mykeys = ['three', 'one']
temp_df = pd.DataFrame().append(mydict)
# You can export DataFrames to a number of formats, using a list here. 
temp_df[mykeys].values[0]
# Returns: array([ 3.,  1.])

# If you want a dict then use this instead:
# temp_df[mykeys].to_dict(orient='records')[0]
# Returns: {'one': 1.0, 'three': 3.0}

Answer 8

kmeans

Answer 9

或者只是mydict.keys()这是对字典的内置方法调用。还要探索mydict.values()和mydict.items()。

//啊，OP帖子让我困惑。

Answer 10

关闭Python: efficient way to create a list from dict values with a given order

在不构建列表的情况下检索密钥：

from __future__ import (absolute_import, division, print_function,
                        unicode_literals)

import collections


class DictListProxy(collections.Sequence):
    def __init__(self, klist, kdict, *args, **kwargs):
        super(DictListProxy, self).__init__(*args, **kwargs)
        self.klist = klist
        self.kdict = kdict

    def __len__(self):
        return len(self.klist)

    def __getitem__(self, key):
        return self.kdict[self.klist[key]]


myDict = {'age': 'value1', 'size': 'value2', 'weigth': 'value3'}
order_list = ['age', 'weigth', 'size']

dlp = DictListProxy(order_list, myDict)

print(','.join(dlp))
print()
print(dlp[1])

输出：

value1,value3,value2

value3

哪个匹配列表给出的顺序

Answer 11

reduce(lambda x,y: mydict.get(y) and x.append(mydict[y]) or x, mykeys,[])

如果键中没有键。