在给定序列中求最大增量

Question

我目前正在研究一个函数，它接受一个序列，并以较高的索引返回从一个元素到另一个元素的最大增量。但是，函数没有返回正确的最大增量。你知道吗

我将for循环放在for循环中，然后尝试从所有差异中返回最大值，但没有成功（它说'int' object is not iterable）

def max_increase(seq):
    i = 0
    maximum_increase = 0
    for i in range(len(seq)):
        difference = 0
        for j in range(i + 1, len(seq)):
            difference = seq[j] - seq[i]
        if 0 <= maximum_increase < difference:
           maximum_increase = difference
    return maximum_increase

对于max_increase([1,2,3,5,0])，它应该返回4，因为从差异列表[1,2,4,-1,1,3,-2,2,-3,-5]，最大值是4。但是，我的函数返回一个负值-1。你知道吗

Answer 1

您可以使用：

my_l = [1,2,3,5,0]

max((e for i in range(len(my_l)) for e in (j - my_l[i] for j in my_l[i + 1:])))

输出：

4

Answer 2

如果您已经收到调试代码的帮助，下面是一个简短的pythonic解决方案：

>>> l=[1,2,3,5,0]
>>> inc = (i-el for p, el in enumerate(l) for i in l[p:])
>>> max(inc)
4

但更好的方法是避免创建不必要的切片（以颠倒顺序为代价）：

import itertools as it


def incs(seq):
    pr = []
    for el in reversed(seq):
        print(f"{el} is compared with {pr}")
        yield (i-el for i in pr)
        pr.append(el)

seq = [1, 2, 3, 5, 0]
print("The max inc is", max(it.chain.from_iterable(incs(seq))))

产生

0 is compared with []
5 is compared with [0]
3 is compared with [0, 5]
2 is compared with [0, 5, 3]
1 is compared with [0, 5, 3, 2]
The max inc is 4

注：如果增加的是两个数字之间的距离，即无论符号如何，始终为正，则进行更改

yield (abs(i-el) for i in pr)

Answer 3

你有缩进问题。这解决了它：

def max_increase(seq):
    i = 0
    maximum_increase = 0
    for i in range(len(seq)):
        difference = 0
        for j in range(i + 1, len(seq)):
            difference = seq[j] - seq[i]
            if 0 <= maximum_increase < difference:
                maximum_increase = difference
    return maximum_increase