如果你只是想看看它是否存在，试着把这个列表变成一个词典:

# Generate a list
l = [n*n for n in range(1000)]

# Convert to dict - doesn't matter what you map values to
d = dict((x, 1) for x in l)

count = 0
for n in range(1000000):
    # Compare with "if n in l"
    if n in d:
        count += 1

在我的机器上，“if n in l”需要37秒，而“if n in d”需要0.4秒。

这段代码以递归的方式处理整数列表。寻找最简单的情况，即:列表长度小于2。这意味着答案已经存在，并执行测试以检查正确答案。 如果不正确，则设置中间值并测试其是否正确，如果不正确，则再次调用该函数，但将中间值设置为上限或下限，将其向左或向右移动

def binary_search(intList, intValue, lowValue, highValue):
    if(highValue - lowValue) < 2:
        return intList[lowValue] == intValue or intList[highValue] == intValue
    middleValue = lowValue + ((highValue - lowValue)/2)
    if intList[middleValue] == intValue:
        return True
    if intList[middleValue] > intValue:
        return binary_search(intList, intValue, lowValue, middleValue - 1)
   return binary_search(intList, intValue, middleValue + 1, highValue)

虽然Python中没有显式的二进制搜索算法，但有一个模块- bisect -用于使用二进制搜索在排序列表中找到元素的插入点。这可以被“欺骗”为执行二分搜索。它最大的优势和大多数库代码一样——它是高性能的，经过良好测试的，并且可以正常工作(特别是二进制搜索很难成功实现——特别是在没有仔细考虑边缘情况的情况下)。

基本类型

对于基本类型，如string或int，这是非常简单的-你只需要bisect模块和一个排序列表:

>>> import bisect
>>> names = ['bender', 'fry', 'leela', 'nibbler', 'zoidberg']
>>> bisect.bisect_left(names, 'fry')
1
>>> keyword = 'fry'
>>> x = bisect.bisect_left(names, keyword)
>>> names[x] == keyword
True
>>> keyword = 'arnie'
>>> x = bisect.bisect_left(names, keyword)
>>> names[x] == keyword
False

你也可以用它来查找副本:

...
>>> names = ['bender', 'fry', 'fry', 'fry', 'leela', 'nibbler', 'zoidberg']
>>> keyword = 'fry'
>>> leftIndex = bisect.bisect_left(names, keyword)
>>> rightIndex = bisect.bisect_right(names, keyword)
>>> names[leftIndex:rightIndex]
['fry', 'fry', 'fry']

显然，如果需要，您可以只返回索引，而不是该索引处的值。

对象

对于自定义类型或对象，事情有点棘手:您必须确保实现丰富的比较方法，以便正确地进行比较。

>>> import bisect
>>> class Tag(object):  # a simple wrapper around strings
...     def __init__(self, tag):
...         self.tag = tag
...     def __lt__(self, other):
...         return self.tag < other.tag
...     def __gt__(self, other):
...         return self.tag > other.tag
...
>>> tags = [Tag('bender'), Tag('fry'), Tag('leela'), Tag('nibbler'), Tag('zoidbe
rg')]
>>> key = Tag('fry')
>>> leftIndex = bisect.bisect_left(tags, key)
>>> rightIndex = bisect.bisect_right(tags, key)
>>> print([tag.tag for tag in tags[leftIndex:rightIndex]])
['fry']

这应该至少在Python 2.7 -> 3.3中工作

这有点跑题了(因为Moe的回答似乎完整地回答了OP的问题)，但从头到尾考虑整个过程的复杂性可能是值得的。如果你把东西存储在一个排序的列表中(这是二进制搜索会有帮助的地方)，然后只是检查是否存在，你会遇到(最坏情况，除非指定):

排序的列表

O(n log n)来初始创建列表(如果它是未排序的数据。O(n)，如果是排序的) O(log n)次查找(这是二分查找部分) O(n)插入/删除(可能是O(1)或O(log n)平均情况，这取决于您的模式)

而使用set()则会导致

O(n)来创造 O(1)查找 O(1)插入/删除

一个排序列表真正让你得到的是“下一个”，“前一个”和“范围”(包括插入或删除范围)，它们是O(1)或O(|范围|)，给定一个起始索引。如果你不经常使用这些类型的操作，那么存储为集合，排序显示可能是一个更好的整体交易。Set()在python中只会引起很少的额外开销。

上面有很多好的解决方案，但我还没有看到一个简单的(KISS保持简单(因为我是)愚蠢的使用Python内置/通用平分函数来做二进制搜索。使用bisect函数的一些代码，我想下面有一个示例，其中我测试了一个小字符串数组的所有情况。上面的一些解决方案提到了这一点，但希望下面的简单代码能帮助像我一样困惑的人。

Python bisect is used to indicate where to insert an a new value/search item into a sorted list. The below code which uses bisect_left which will return the index of the hit if the search item in the list/array is found (Note bisect and bisect_right will return the index of the element after the hit or match as the insertion point) If not found, bisect_left will return an index to the next item in the sorted list which will not == the search value. The only other case is where the search item would go at the end of the list where the index returned would be beyond the end of the list/array, and which in the code below the early exit by Python with "and" logic handles. (first condition False Python does not check subsequent conditions)

#Code
from bisect import bisect_left
names=["Adam","Donny","Jalan","Zach","Zayed"]
search=""
lenNames = len(names)
while search !="none":
    search =input("Enter name to search for or 'none' to terminate program:")
    if search == "none":
        break
    i = bisect_left(names,search)
    print(i) # show index returned by Python bisect_left
    if i < (lenNames) and names[i] == search:
        print(names[i],"found") #return True - if function
    else:
        print(search,"not found") #return False – if function
##Exhaustive test cases:
##Enter name to search for or 'none' to terminate program:Zayed
##4
##Zayed found
##Enter name to search for or 'none' to terminate program:Zach
##3
##Zach found
##Enter name to search for or 'none' to terminate program:Jalan
##2
##Jalan found
##Enter name to search for or 'none' to terminate program:Donny
##1
##Donny found
##Enter name to search for or 'none' to terminate program:Adam
##0
##Adam found
##Enter name to search for or 'none' to terminate program:Abie
##0
##Abie not found
##Enter name to search for or 'none' to terminate program:Carla
##1
##Carla not found
##Enter name to search for or 'none' to terminate program:Ed
##2
##Ed not found
##Enter name to search for or 'none' to terminate program:Roger
##3
##Roger not found
##Enter name to search for or 'none' to terminate program:Zap
##4
##Zap not found
##Enter name to search for or 'none' to terminate program:Zyss
##5
##Zyss not found

如果你只是想看看它是否存在，试着把这个列表变成一个词典:

# Generate a list
l = [n*n for n in range(1000)]

# Convert to dict - doesn't matter what you map values to
d = dict((x, 1) for x in l)

count = 0
for n in range(1000000):
    # Compare with "if n in l"
    if n in d:
        count += 1

在我的机器上，“if n in l”需要37秒，而“if n in d”需要0.4秒。