如何计时泡沫和插入排序？

Question

我正在尝试计算泡沫和插入排序.排序适用于它们两个,但它输出需要15秒,这显然是不正确的.有任何建议如何解决这个问题？

# insertion sort
x = [7, 2, 3, 5, 9, 1]

def insertion(list):
    for index in range(1,len(list)):
        value = list[index]
        i = index - 1
        while i>=0 and (value < list[i]):
            list[i+1] = list[i] # shift number in slot i right to slot i+1
            list[i] = value # shift value left into slot i
            i = i - 1

# bubble sort
y = [7, 2, 3, 5, 9, 1]

def bubble(unsorted_list):
    length = len(unsorted_list) - 1
    sorted = False

    while not sorted:
        sorted = True
        for i in range(length):
            if unsorted_list[i] > unsorted_list[i+1]:
                sorted = False
                unsorted_list[i], unsorted_list[i+1] = unsorted_list[i+1], unsorted_list[i]

def test():
    bubble(y)
    insertion(x)

if __name__ == '__main__':
    import timeit
    print(timeit.timeit("test()", setup="from __main__ import test"))

Answer 1

从timeit文档:

timeit.timeit(stmt ='pass',setup ='pass',timer =,number = 1000000)

使用给定的语句,设置代码和计时器函数创建一个Timer实例,并使用数字执行运行其timeit()方法.

所以你的代码运行了1,000,000次.除以返回值10**6,您就可以了.