基本的和不太广泛的测试，比较五个答案的执行时间:

def numpyIndexValues(a, b):
    na = np.array(a)
    nb = np.array(b)
    out = list(na[nb])
    return out

def mapIndexValues(a, b):
    out = map(a.__getitem__, b)
    return list(out)

def getIndexValues(a, b):
    out = operator.itemgetter(*b)(a)
    return out

def pythonLoopOverlap(a, b):
    c = [ a[i] for i in b]
    return c

multipleListItemValues = lambda searchList, ind: [searchList[i] for i in ind]

使用以下输入:

a = range(0, 10000000)
b = range(500, 500000)

简单的python循环是最快的，lambda操作紧随其后，mapIndexValues和getIndexValues始终非常相似，numpy方法在将列表转换为numpy数组后明显更慢。如果数据已经在numpy数组中，则使用numpy. numpyIndexValues方法。删除数组转换是最快的。

numpyIndexValues -> time:1.38940598 (when converted the lists to numpy arrays)
numpyIndexValues -> time:0.0193445 (using numpy array instead of python list as input, and conversion code removed)
mapIndexValues -> time:0.06477512099999999
getIndexValues -> time:0.06391049500000001
multipleListItemValues -> time:0.043773591
pythonLoopOverlap -> time:0.043021754999999995

列表理解显然是最直接和最容易记住的——除了相当python化!

在任何情况下，在提出的解决方案中，它不是最快的(我已经在Windows上使用Python 3.8.3运行了我的测试):

import timeit
from itertools import compress
import random
from operator import itemgetter
import pandas as pd

__N_TESTS__ = 10_000

vector = [str(x) for x in range(100)]
filter_indeces = sorted(random.sample(range(100), 10))
filter_boolean = random.choices([True, False], k=100)

# Different ways for selecting elements given indeces

# list comprehension
def f1(v, f):
   return [v[i] for i in filter_indeces]

# itemgetter
def f2(v, f):
   return itemgetter(*f)(v)

# using pandas.Series
# this is immensely slow
def f3(v, f):
   return list(pd.Series(v)[f])

# using map and __getitem__
def f4(v, f):
   return list(map(v.__getitem__, f))

# using enumerate!
def f5(v, f):
   return [x for i, x in enumerate(v) if i in f]

# using numpy array
def f6(v, f):
   return list(np.array(v)[f])

print("{:30s}:{:f} secs".format("List comprehension", timeit.timeit(lambda:f1(vector, filter_indeces), number=__N_TESTS__)))
print("{:30s}:{:f} secs".format("Operator.itemgetter", timeit.timeit(lambda:f2(vector, filter_indeces), number=__N_TESTS__)))
print("{:30s}:{:f} secs".format("Using Pandas series", timeit.timeit(lambda:f3(vector, filter_indeces), number=__N_TESTS__)))
print("{:30s}:{:f} secs".format("Using map and __getitem__", timeit.timeit(lambda: f4(vector, filter_indeces), number=__N_TESTS__)))
print("{:30s}:{:f} secs".format("Enumeration (Why anyway?)", timeit.timeit(lambda: f5(vector, filter_indeces), number=__N_TESTS__)))

我的结果是:

列表理解:0.007113秒 操作符。Itemgetter:0.003247秒 使用Pandas系列:2.977286秒 使用map和getitem:0.005029秒 枚举(为什么?):0.135156秒 Numpy:0.157018秒

你可以使用operator.itemgetter:

from operator import itemgetter 
a = [-2, 1, 5, 3, 8, 5, 6]
b = [1, 2, 5]
print(itemgetter(*b)(a))
# Result:
(1, 5, 5)

或者你可以使用numpy:

import numpy as np
a = np.array([-2, 1, 5, 3, 8, 5, 6])
b = [1, 2, 5]
print(list(a[b]))
# Result:
[1, 5, 5]

但说真的，你现在的解决方案很好。这可能是其中最简洁的一个。

选择:

>>> map(a.__getitem__, b)
[1, 5, 5]

>>> import operator
>>> operator.itemgetter(*b)(a)
(1, 5, 5)

静态索引和小列表?

不要忘记，如果列表很小，并且索引没有改变，就像你的例子中，有时最好的方法是使用序列解包:

_,a1,a2,_,_,a3,_ = a

性能大大提高，你还可以节省一行代码:

 %timeit _,a1,b1,_,_,c1,_ = a
10000000 loops, best of 3: 154 ns per loop 
%timeit itemgetter(*b)(a)
1000000 loops, best of 3: 753 ns per loop
 %timeit [ a[i] for i in b]
1000000 loops, best of 3: 777 ns per loop
 %timeit map(a.__getitem__, b)
1000000 loops, best of 3: 1.42 µs per loop

截至2022年6月，最新熊猫==1.4.2的结果如下。

注意，简单的切片不再可能，基准测试结果更快。

import timeit
import pandas as pd
print(pd.__version__)
# 1.4.2

pd.Series([-2, 1, 5, 3, 8, 5, 6])[1, 2, 5]
# KeyError: 'key of type tuple not found and not a MultiIndex'

pd.Series([-2, 1, 5, 3, 8, 5, 6]).iloc[[1, 2, 5]].tolist()
# [1, 5, 5]

def extract_multiple_elements():
    return pd.Series([-2, 1, 5, 3, 8, 5, 6]).iloc[[1, 2, 5]].tolist()

__N_TESTS__ = 10_000
t1 = timeit.timeit(extract_multiple_elements, number=__N_TESTS__)
print(round(t1, 3), 'seconds')
# 1.035 seconds