Question

我的代码遇到问题，我正在编写一个方法来检查下一个元素是否小于前一个元素，如果是，则删除它。

示例：

输入：[1, 20, 10, 30]

所需的输出：[1,20,30]

实际输出：[30]

def findSmaller(s):
    i = -1
    y = []
    while i <= len(s):
        for j in range(len(s)):
            if s[i+1] <= s[i]:
                del s[i + 1]
                y.append(s[i])
            i += 1
            return y

Answer 1

根据您是否需要稍后使用列表进行计算，您可以使用生成器

s = [1, 20, 10, 30]

def find_smaller_generator(l: list):
    last_item = None
    for item in l:
        if last_item is None or item >= last_item: 
            last_item = item
            yield item
def find_smaller_list(l: list):
    return list(find_smaller_generator(l))



print(find_smaller_list(s))

for i in find_smaller_generator(s):
    print(i)

print([i**2 for i in find_smaller_generator(s)])

返回：

[1, 20, 30]

1
20
30

[1, 400, 900]

Answer 2

>>> s =  [5, 20, 10, 15, 30]
>>> max_so_far = s[0]
>>> result = []
>>> for x in s:
        if x >= max_so_far:
            result.append(x)
            max_so_far = x

>>> result
[5, 20, 30]

Answer 3

您可以尝试这样的事情

def findSmaller(s):
    # sets p (previous) as the first value in s
    p = s[0]
    # initializes y to be an array and sets the first value to p
    y = [p]
    # iterates over s, starting with the second element
    for i in s[1::]:
        # checks if i is greater than or equal to the previous element
        if i >= p:
            # if it is, i is appended to the list y
            y.append(i)
            # also set the previous value to i, so the next iteration can check against p
            p = i
    #returns the list
    return y

这样做是迭代s并检查列表中的当前项是否大于或等于列表中的前一个元素。如果是，则将其附加到y，并返回y。

试用代码here。

Answer 4

如果您不确定循环如何工作我建议添加一些打印语句。这样你就可以看到你的循环实际上在做什么，特别是在更复杂的问题中这是有用的。

这样的事情可以解决你的问题。

a = [1,2,3,2,4]
for k in range(0,len(a)-2): #-2 so that one don't go past the loops length
    #print(k)
    y = a
    if(a[k]>a[k+1]):
        del y[k+1] #delete the k+1 element if it is

查找下一个元素是否小于它之前的元素并从列表python中删除它

4 个答案: