检查列表中是否存在值的最快方法

7 in a

最清晰，最快捷的方法。

您也可以考虑使用 ，但从列表中构造该集可能比更快的成员资格测试节省的时间要长。确定的唯一方法是做好基准测试。（这也取决于您需要的操作）set

正如其他人所说，对于大型列表来说可能非常慢。以下是 和 的性能的一些比较。请注意，时间（以秒为单位）以对数刻度为单位。ininsetbisect

测试代码：

import random
import bisect
import matplotlib.pyplot as plt
import math
import time


def method_in(a, b, c):
    start_time = time.time()
    for i, x in enumerate(a):
        if x in b:
            c[i] = 1
    return time.time() - start_time


def method_set_in(a, b, c):
    start_time = time.time()
    s = set(b)
    for i, x in enumerate(a):
        if x in s:
            c[i] = 1
    return time.time() - start_time


def method_bisect(a, b, c):
    start_time = time.time()
    b.sort()
    for i, x in enumerate(a):
        index = bisect.bisect_left(b, x)
        if index < len(a):
            if x == b[index]:
                c[i] = 1
    return time.time() - start_time


def profile():
    time_method_in = []
    time_method_set_in = []
    time_method_bisect = []

    # adjust range down if runtime is too long or up if there are too many zero entries in any of the time_method lists
    Nls = [x for x in range(10000, 30000, 1000)]
    for N in Nls:
        a = [x for x in range(0, N)]
        random.shuffle(a)
        b = [x for x in range(0, N)]
        random.shuffle(b)
        c = [0 for x in range(0, N)]

        time_method_in.append(method_in(a, b, c))
        time_method_set_in.append(method_set_in(a, b, c))
        time_method_bisect.append(method_bisect(a, b, c))

    plt.plot(Nls, time_method_in, marker='o', color='r', linestyle='-', label='in')
    plt.plot(Nls, time_method_set_in, marker='o', color='b', linestyle='-', label='set')
    plt.plot(Nls, time_method_bisect, marker='o', color='g', linestyle='-', label='bisect')
    plt.xlabel('list size', fontsize=18)
    plt.ylabel('log(time)', fontsize=18)
    plt.legend(loc='upper left')
    plt.yscale('log')
    plt.show()


profile()