迭代一个字典在python和剥离的空白 [英] Iterating over a dictionary in python and stripping white space

问题描述

我正在使用Web Scratch框架Scrapy，而对于python来说，我是一个noob。所以我想知道如何迭代所有似乎在字典中的被刮的项目，并从每个项目中删除空白。

I am working with the web scraping framework Scrapy and I am a bit of a noob when it comes to python. So I am wondering how do I iterate over all of the scraped items which seem to be in a dictionary and strip the white space from each one.

这是代码我在我的物品管道中玩过：

Here is the code I have been playing with in my item pipeline.:

for info in item:
   info[info].lstrip()

但这段代码不起作用，因为我不能单独选择项目。所以我试图这样做：

But this code does not work, because I cannot select items individually. So I tried to do this:

for key, value item.items():
   value[1].lstrip()

第二种方法在某种程度上起作用，但问题是我不知道如何然后循环所有的值。

This second method works to a degree, but the problem is that I have no idea how then to loop over all of the values.

我知道这可能是一个简单的修复，但我似乎找不到。任何帮助将不胜感激。 ：）

I know this is probably such an easy fix, but I cannot seem to find it. Any help would be greatly appreciated. :)

推荐答案

不是直接回答这个问题，而是建议你看一下 Item Loaders 和输入/输出处理器。很多你的清理可以在这里照顾。

Not a direct answer to the question, but I would suggest you look at Item Loaders and input/output processors. A lot of your cleanup can be take care of here.

剥离每个条目的例子是：

An example which strips each entry would be:

class ItemLoader(ItemLoader):

    default_output_processor = MapCompose(unicode.strip)

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