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

来自timeit documentation：

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

使用给定语句创建Timer实例，   设置代码和计时器功能，并使用数字运行其timeit（）方法   执行。

所以你的代码运行了1,000,000次。将返回值除以10**6，您就可以了。

Answer 2

profile也是您可以考虑的选项。

Answer 3

尝试此操作，捕获当前时间，执行该功能，然后从当前时间中减去先前捕获的时间。

# 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():
    start = time.clock()
    bubble(y)
    elapsed = (time.clock() - start)
    print "Time taken for bubble = ", elapsed
    start = time.clock()
    insertion(x)
    elapsed = (time.clock() - start)
    print "Time taken for Insertion = ", elapsed

if __name__ == '__main__':
    import time
    test()

如何计时泡沫和插入排序？

3 个答案: