Python字典：获取键列表的值列表

Python dictionary: Get list of values for list of keys

提问by FazJaxton

Is there a built-in/quick way to use a list of keys to a dictionary to get a list of corresponding items?

是否有内置/快速的方法来使用字典的键列表来获取相应项目的列表？

For instance I have:

例如我有：

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

How can I use mykeysto get the corresponding values in the dictionary as a list?

如何使用mykeys以列表形式获取字典中的相应值？

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

采纳答案by FazJaxton

A list comprehension seems to be a good way to do this:

列表理解似乎是一个很好的方法：

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

回答by Edgar Aroutiounian

Or just mydict.keys()That's a builtin method call for dictionaries. Also explore mydict.values()and mydict.items().

或者mydict.keys()这只是对字典的内置方法调用。还探索mydict.values()和mydict.items()。

//Ah, OP post confused me.

//啊，OP帖子让我很困惑。

回答by óscar López

Try this:

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]

回答by Jon Clements

A couple of other ways than list-comp:

除了 list-comp 之外，还有其他几种方式：

• Build list and throw exception if key not found: map(mydict.__getitem__, mykeys)
• Build list with Noneif key not found: map(mydict.get, mykeys)

• 如果未找到密钥，则构建列表并抛出异常： map(mydict.__getitem__, mykeys)
• 使用Noneif key not found构建列表：map(mydict.get, mykeys)

Alternatively, using operator.itemgettercan return a tuple:

或者，使用operator.itemgetter可以返回一个元组：

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

Note: in Python3, mapreturns an iterator rather than a list. Use list(map(...))for a list.

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

回答by yupbank

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

incase there are keys not in dict.

以防字典中没有键。

回答by OdraEncoded

Here are three ways.

这里有三种方法。

Raising KeyErrorwhen key is not found:

KeyError找不到密钥时引发：

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

Default values for missing keys.

缺失键的默认值。

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

Skipping missing keys.

跳过丢失的键。

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

回答by Sklavit

A little speed comparison:

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

So list comprehension and itemgetter are the fastest ways to do this.

所以列表理解和 itemgetter 是最快的方法。

UPDATE: For large random lists and maps I had a bit different results:

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)
%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

So in this case the clear winner is f = operator.itemgetter(*l); f(m), and clear outsider: map(lambda _: m[_], l).

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

UPDATE for Python 3.6.4:

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)
%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)

So, results for Python 3.6.4 is almost the same.

因此，Python 3.6.4 的结果几乎相同。

回答by mementum

Following closure of Python: efficient way to create a list from dict values with a given order

Python关闭后 ：从具有给定顺序的 dict 值创建列表的有效方法

Retrieving the keys without building the list:

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])

The output:

输出：

value1,value3,value2

value3

Which matches the order given by the list

与列表给出的顺序匹配

回答by Vikram Singh Chandel

Try This:

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]

回答by abby sobh

Pandas does this very elegantly, though ofc list comprehensions will always be more technically Pythonic. I don't have time to put in a speed comparison right now (I'll come back later and put it in):

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

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}