为什么我的二进制搜索要比线性搜索慢？

Question

我曾尝试对二进制搜索和线性搜索进行编码，但看到二进制搜索比线性搜索要慢两倍甚至两倍，我感到震惊。请帮我。这是我的代码。

二进制搜索代码：

def binary_search(array, target, n=0):
    l = len(array)-1
    i = l//2
    try:
        ai = array[i]
    except:
        return False
    if ai == target:
        n += i
        return (True, n)

    elif target >= ai:
        array = array[i+1:l+1]
        n += i + 1
        return binary_search(array, target, n)

    elif target <= ai:
        array = array[0: i]
        return binary_search(array, target, n)

线性搜索代码

def linear_search(array, target):
    for i, num in enumerate(array):
        if num == target:
            return True, i

    return False

测试用例代码：

import random 
import time
n = 10000000
num = sorted([random.randint(0, n) for x in range(n)])

start = time.time()
print(linear_search(num, 1000000))
print(f'Linear Search: {time.time() - start}')

start_new = time.time()
print(binary_search(num, 1000000))
print(f'Binary Search: {time.time() - start_new}')

Answer 1

如@khelwood所说，您的代码将更快地进行，而无需切片。

def binary_search_no_slice(array, target, low, high):
    if low > high:
        return False
    mid = (low + high) // 2
    if array[mid] == target:
        return True
    elif array[mid] > target:
        return binary_search_no_slice(array, target, low, mid - 1)
    else:
        return binary_search_no_slice(array, target, mid + 1, high)

以下添加到您的测试代码中。

start_new2 = time.time()
print(binary_search_no_slice(num, 1000000, 0, len(num) - 1))
print(f'Binary Search no slice: {time.time() - start_new2}')

这是我机器上的结果（macOS Catalina，2.8GHz Corei7，8GB RAM）

False
Linear Search: 2.172485113143921
False
Binary Search: 0.56640625
False
Binary Search no slice: 2.8133392333984375e-05