如何使用任何而不是在列表中搜索项目时(python)

Question

所以我正在编写一个程序来搜索关键字的职位发布.我已经有了将整个作业描述转换为单个单词列表,删除空格,特殊字符,使所有内容都小写等的代码.

我想做一些事我可以像"打印一些东西,如果这个列表包含python但不打印它,如果有的话python,VBA.这就是我拥有的东西:

def query_job_posting(url, query_list_include, query_list_exclude): 

    soup = create_soup(url)

    ...list formatting functions...

    for i in job_description_list:
        if any(word in i for word in query_list_include) and not any(exclude in i for exclude in query_list_exclude):
            print(url)

job_description list 看起来像这样:

['this',
 'is',
  'a',
  'vba',
  'job',
  'python']

但它似乎没有起作用.

如果query_list_include=['python']和query_list_exclude=[]随后的URL打印.

如果query_list_exclude=['vba']和query_list_include=[]那么URL不打印.

但是如果我python按照包含和vba排除的方式离开,那么网址仍会打印,即使我手动验证了两者vba并且python都在job_descripton_list

我在哪里错了？

Answer 1

问题:

您实际上在列表的每个元素中查找每个单词:

for e in list:
    if any(w in e for w in include) and not any(w in e for w in exclude):
        print(url)

具体如下:

'this'    # do nothing
'is'      # do nothing
'a'       # do nothing
'job'     # do nothing
'python'  # print url

您可以使用以下方式验证它

for e in list:
    if any(w in e for w in include) and not any(w in e for w in exclude):
        print(e, url)

哪个应该打印python <url>.在这种情况下,列表中的"VBA"将不会改变任何内容

解:

从你想要的解释:

url = ...

list = ['this', 'is', 'a', 'job', 'python']
include = ['python']
exclude = ['VBA']

if any(w in list for w in include) and not any(w in list for w in exclude):
    print(url)

Out[]: <url>

它从if声明中评估条件:

'python' in list  --> True
'VBA' not in list --> True

然后执行 print(url)