S是一个列表。 Binary (s, 0, len(s) - 1, find)是初始调用。 函数返回查询项的索引。如果没有这样的项，则返回-1。 def二进制(s, p, q,找到): 如果发现= = s ((p + q) / 2): 返回(p + q) / 2 Elif p==q-1或p==q: 如果发现= = s[问]: 回来问 其他: 返回1 find < s[(p+q)/2]: 返回二进制(s p (p + q) / 2,找到) Elif find > s[(p+q)/2]: 返回二进制(s (p + q) / 2 + 1, q,找到)

为什么不看看bisect_left/right的代码并调整它以适合您的目的呢?

是这样的:

def binary_search(a, x, lo=0, hi=None):
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        midval = a[mid]
        if midval < x:
            lo = mid+1
        elif midval > x: 
            hi = mid
        else:
            return mid
    return -1

戴夫·亚伯拉罕斯的解决方案很好。虽然我会把它做得极简:

def binary_search(L, x):
    i = bisect.bisect_left(L, x)
    if i == len(L) or L[i] != x:
        return -1
    return i

我想使用bisect_left，然后检查是否在那项 position等于我要搜索的内容，但这看起来很麻烦 (我还需要做边界检查，如果数字可以更大 而不是我列表中最大的数字)。如果有更好的方法，我会 我想了解一下。

避免边界检查或相等性检查的一种方法是同时运行bisect_left()和bisect_right():

def find(data, target):
    start = bisect_left(data, target)
    end = bisect_right(data, target)
    return -1 if start == end else start

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。

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