Python:虽然，如果，否则计数 [英] Python: While, if, else counting

问题描述

我在使用 while/if 语句时遇到了一些问题.

I'm having a bit of an issue with a while/if statement.

我有一个值列表，通常这些值是字符串，但有时它可以返回 None.这是我的两次尝试:

I have a list of values, normally these values will be strings, but sometimes it can return None. Here are two of my attempts:

x = ['One','Two','Three',None,None]
New = []
count=0
for y in x:
    while isinstance(y,str):
        New.append(y)
        count+=1
        break
    else:
        count+=1
        New.append('New - '+str(count))
print New,count
>>> The list repeats several times

New = []
for y in x:
    count=0
    if y is not None:
        New.append(y)
        count+=1
    else:
        count+=1
        New.append('New - '+str(count))
>>>['One','Two','Three','New - 1','New - 1']

我希望输出是:['One','Two','Three', 'New - 4', 'New - 5']，如果 None 值在中间，则保持列表的顺序.

I would like the output to be: ['One','Two','Three', 'New - 4', 'New - 5'], and to keep the ordering of the list if the None value was somewhere in the middle.

我不确定我哪里出错了，两者都相差甚远.对不起，如果这很简单，我还在学习.我在这个论坛上找了一个类似的问题，有些人有帮助，但我还是想不通.

I'm not sure where I'm going wrong, neither of them are far off. Sorry if this is quite simple i'm still learning. I've looked around this forum for a similar query, some have helped but i still can;t figure it out.

推荐答案

每当计算索引和循环列表时，最好使用 enumerate.如果您不希望它从默认的 0 开始，您也可以指定一个起始编号.这里似乎就是这种情况，因为您似乎想从 1

Whenever counting the index and looping over a list, it's best to use enumerate. You can also specify a start number if you don't want it to start from 0 which is the default. That seems to be the case here, since you appear to want to count starting from 1

还有 while 循环似乎毫无意义.一个简单的 if 就足够了.如果您知道项目将是 None，那么检查它是否是 None 可能比检查 isinstance(item, str)

Also the while loop seems pointless. A simple if would be sufficient. And if you know the items will be None it's probably better to check if it's None rather than checking isinstance(item, str)

所以我相信您正在寻找的解决方案类似于

So I believe the solution you're looking for goes something like

x = ['One', 'Two', 'Three', None, None]
new = []
for index, item in enumerate(x, start=1):
    if item is None:
        new.append('New - {}'.format(index))
    else:
        new.append(item)

print(new)

这应该会产生预期的结果.如果你愿意，这也可以写成列表理解.

This should produce the result that is expected. This could also be written as a list comprehension, if you like.

new = [item if item is not None else 'New - {}'.format(index) for index, item in enumerate(x, start=1)]

输出为

['One', 'Two', 'Three', 'New - 4', 'New - 5']

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