python - 字典：获取键列表的值列表

Question

是否有一种内置/快速的方法来使用字典的键列表来获取相应项目的列表？

例如我有：

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

我如何使用mykeys将字典中的相应值作为列表获取？

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

Answer 1

列表推导似乎是做到这一点的好方法：

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

Answer 2

除了 list-comp 之外的其他几种方法：

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

或者， usingoperator.itemgetter可以返回一个元组：

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

注意：在 Python3 中，map返回一个迭代器而不是一个列表。用于list(map(...))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，我得到了一些不同的结果：

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)
1000 loops, best of 3: 1.14 ms per loop

%timeit list(itemgetter(*l)(m))
1000 loops, best of 3: 1.68 ms per loop

%timeit [m[_] for _ in l]  # list comprehension
100 loops, best of 3: 2 ms per loop

%timeit map(m.__getitem__, l)
100 loops, best of 3: 2.05 ms per loop

%timeit list(m[_] for _ in l)  # a generator expression passed to a list constructor.
100 loops, best of 3: 2.19 ms per loop

%timeit map(m.get, l)
100 loops, best of 3: 2.53 ms per loop

%timeit map(lambda _: m[_], l)
100 loops, best of 3: 2.9 ms per loop

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

Python 3.6.4 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)
1.66 ms ± 74.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit list(itemgetter(*l)(m))
2.1 ms ± 93.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit [m[_] for _ in l]  # list comprehension
2.58 ms ± 88.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit list(map(m.__getitem__, l))
2.36 ms ± 60.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit list(m[_] for _ in l)  # a generator expression passed to a list constructor.
2.98 ms ± 142 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit list(map(m.get, l))
2.7 ms ± 284 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit list(map(lambda _: m[_], l)
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

这里有三种方式。

KeyError未找到密钥时引发：

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

缺失键的默认值。

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

跳过丢失的键。

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

Answer 5

尝试这个：

mydict = {'one': 1, 'two': 2, 'three': 3}
mykeys = ['three', 'one','ten']
newList=[mydict[k] for k in mykeys if k in mydict]
print newList
[3, 1]

Answer 6

尝试这个：

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

new_dict = {x: v for x, v in mydict.items() if x in mykeys}

Answer 8

Pandas 非常优雅地做到了这一点，尽管 ofc 列表推导在技术上总是更加 Pythonic。我现在没有时间进行速度比较（我稍后会回来放入）：

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 9

在 Python 关闭之后：从具有给定顺序的 dict 值创建列表的有效方法

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

与列表给出的顺序相匹配

Answer 10

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

以防字典中没有键。

Answer 11

如果您发现自己经常这样做，您可能希望子类dict化以获取键列表并返回值列表。

>>> d = MyDict(mydict)
>>> d[mykeys]
[3, 1]

这是一个演示实现。

class MyDict(dict):
    def __getitem__(self, key):
        getitem = super().__getitem__
        if isinstance(key, list):
            return [getitem(x) for x in key]
        else:
            return getitem(key)

子类化dict需要更多的工作，而且您可能想要实现.get()、.__setitem__()和.__delitem__()，等等。