出现错误:在Python中使用列表理解和条件语句时“没有足够的值要解压" [英] Getting Error: 'not enough values to unpack' when using list comprehension together with conditional statement in Python

问题描述

目标是创建一个输出两个值的列表理解.

The objective is to create a list comprehension that outputs two values.

for循环如下所示

    paper_href_scopus = []
    paper_title = []
    for litag in all_td.find_all('a', {'class': 'ddmDocTitle'}):
        paper_href_scopus.append(litag['href'])
        paper_title.append(litag.text)

根据 OP 的建议，可以通过

    paper_href_scopus, paper_title = zip(*[(litag['href'], litag.text) for litag in all_td.find_all('a', {'class': 'ddmDocTitle'})])

但是，在某些情况下，all_td.find_all('a', {'class': 'ddmDocTitle'})返回empty并且编译器返回错误:

However, there is an instances where the all_td.find_all('a', {'class': 'ddmDocTitle'}) returns empty and the compiler returns an error:

ValueError:没有足够的值可解包(预期2，得到0)

ValueError: not enough values to unpack (expected 2, got 0)

基于此线程中的讨论，似乎上面的代码可以修改为

Based on the discussion in this thread, it seems the above code can be modified as

     paper_href_scopus, paper_title = zip(
                        *((litag['href'], litag.text) for litag in all_td.find_all('a', {'class': 'ddmDocTitle'}) \
                          if all_td.find_all('a', {'class': 'ddmDocTitle'}
                          ))

但是，编译器仍然返回错误

But still, the compiler returns an error

ValueError:没有足够的值可解包(预期2，得到0)

ValueError: not enough values to unpack (expected 2, got 0)

尽管如此，以下代码仍然有效，尽管在某些情况下all_td.find_all('a', {'class': 'ddmDocTitle'})返回empty

Nevertheless, the following code works despite the fact that on some occasions the all_td.find_all('a', {'class': 'ddmDocTitle'}) returns empty

    [(paper_href_scopus.append(litag['href']), paper_title.append(litag.text)) \
                     for litag in all_td.find_all('a', {'class': 'ddmDocTitle'})]

但是，由于要事先初始化paper_href_scopus=[]和paper_title=[]，因此我想避免使用append.

But, I would like to avoid using append as there is requirement to initialize paper_href_scopus=[] and paper_title=[] beforehand.

我可以知道如何解决该代码吗?

May I know, what can I do to fix the code?

    paper_href_scopus, paper_title = zip(
                        *((litag['href'], litag.text) for litag in all_td.find_all('a', {'class': 'ddmDocTitle'}) \
                          if all_td.find_all('a', {'class': 'ddmDocTitle'}
                          ))

推荐答案

首先，带有if的版本基本上等同于:

Firstly, the version with the if is basically equivalent to:

tmp = []
for litag in all_td.find_all('a', {'class': 'ddmDocTitle'}):
    if all_td.find_all('a', {'class': 'ddmDocTitle'}):
        tmp.append((litag['href'], litag.text))

paper_href_scopus, paper_title = zip(*tmp)

如果您的文档有100个匹配元素，它将进行101次搜索.

which, if your document has 100 matching elements, it does 101 searches.

这是我的建议:忘记zip.相反，请分拆一下代码:

Here's my proposal: forget about the zip. Instead, split out the code a bit:

litags = all_td.find_all('a', {'class': 'ddmDocTitle'})
paper_href_scopus = [litag['href'] for litag in litags]
paper_title = [litag.text for litag in litags]

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