Question

可能重复：
Converting a for loop to a while loop

def splitList(myList, option):
    snappyList = []
    for i in myList:
        if option == 0:
            if i > 0:
                snappyList.append(i)
        if option == 1:
            if i < 0:
                snappyList.append(i)
    return (snappyList)

嗨，我有这个代码在for循环下工作得很好。它根据用户输入的内容返回正面或负面元素。我需要在while循环中使用它，并且我不确定如何在没有被while循环捕获的情况下使它工作。

任何想法或提示都将不胜感激，谢谢！

Answer 1

尝试以下方法：

def splitList(myList, option):
    snappyList = []
    i = 0
    while i < len(myList):
        if option == 0:
            if myList[i] > 0:
                snappyList.append(myList[i])
        if option == 1:
            if myList[i] < 0:
                snappyList.append(myList[i])
        i+=1
    return (snappyList)

Answer 2

由于没有严格遵守你的问题而有可能吸引downvotes，Python比其他更传统的语言有更好的（简单）循环设施。（我也意识到，根据今天早上有非常类似的问题，这可能是家庭作业）。学习while循环工作的方式显然有一些价值，但是在Python中这样做会掩盖其他工具。例如，您的示例使用单个列表解析：

def splitList2(myList, option):
    return [v for v in myList if (1-2*option)*v > 0]

print(splitList2([1,2,3,-10,4,5,6], 0))
print(splitList2([1,2,3,-10,4,5,6], 1))

输出：

[1, 2, 3, 4, 5, 6]
[-10]
>>>

理解中条件的语法看起来很复杂，因为option到效果的映射很差。在Python中，与许多其他动态和函数语言一样，您可以直接传递比较函数：

def splitList3(myList, condition):
    return [v for v in myList if condition(v)]

print(splitList3([1,2,3,-10,4,5,6], lambda v: v>0))
print(splitList3([1,2,3,-10,4,5,6], lambda v: v<0))
print(splitList3([1,2,3,-10,4,5,6], lambda v: v%2==0))

[1, 2, 3, 4, 5, 6]
[-10]
[2, -10, 4, 6]
>>>

请注意，这是多么灵活：将代码调整为完全不同的过滤条件变得微不足道。

Answer 3

def splitList(myList, option):
    snappyList = []
    myListCpy=list(myList[:]) #copy the list, in case the caller cares about it being changed, and convert it to a list (in case it was a tuple or similar)
    while myListCpy:  #There's at least one element in the list.
        i=myListCpy.pop(0)  #Remove the first element, the rest continues as before.
        if option == 0:
            if i > 0:
                snappyList.append(i)
        if option == 1:
            if i < 0:
                snappyList.append(i)
    return (snappyList)

Answer 4

def splitList(myList, option):
    snappyList = []
    while len(myList) > 0:
        i = myList.pop()
        if option == 0:
            if i > 0:
                snappyList.append(i)
        if option == 1:
            if i < 0:
                snappyList.append(i)
    return snappyList

Answer 5

def splitList(myList, option):
    snappyList = []
    myList_iter = iter(myList)
    sentinel = object()
    while True:
        i = next(myList_iter, sentinel)
        if i == sentinel:
            break
        if option == 0:
            if i > 0:
                snappyList.append(i)
        if option == 1:
            if i < 0:
                snappyList.append(i)
    return (snappyList)

或者，您可以使用异常处理程序而不是sentinel

def splitList(myList, option):
    snappyList = []
    myList_iter = iter(myList)
    while True:
        try:
            i = next(myList_iter)
        except StopIteration:
            break
        if option == 0:
            if i > 0:
                snappyList.append(i)
        if option == 1:
            if i < 0:
                snappyList.append(i)
    return (snappyList)

Answer 6

这是一种实际拆分列表的简洁方法，而不仅仅是过滤它：

from operator import ge,lt
def splitlist(input, test=ge, pivot=0):
    left = []
    right = []
    for target, val in (((left if test(n, pivot) else right), n) for n in input):
        target.append(val)

    return left, right

你会注意到它使用for循环，因为它是正确的做事方式。

如何将此for循环并使其成为while循环python

6 个答案: