只将第一个字母传入字典。 (python 3.x)

时间:2016-09-12 10:47:56

标签: python dictionary

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这就是我的代码的样子。很奇怪,期望值应为{link:http://wizters.com, 状态:200, 联系方式:联系信息不可用&#34;}

然而,结果是:

    print(single_website)
    print(status_str)
    print(contact_link)

    for link, status_code, contact_info in zip(single_website, status_str, contact_link):
        data = {
            "Link": link,
            "Status": status_code, 
            "Contact": contact_info
        }

仅将第一个字母或数字传递给值。顺便说一句,我用str()来表示状态&#34; 200&#34;到一个字符串。

3 个答案:

答案 0 :(得分:0)

因为字符串是可迭代的,这就是为什么它在你的zip中稍后需要的第一个,你应该把它包装成元组或列表

答案 1 :(得分:0)

>>> single_website = 'http://wizters.com'
>>> status_str = '200'
>>> contact_link = 'Contact Info Not Available'
>>> list(zip(single_website, status_str, contact_link))
[('h', '2', 'C'), ('t', '0', 'o'), ('t', '0', 'n')]
>>> list(zip((single_website, ), (status_str, ), (contact_link, )))
[('http://wizters.com', '200', 'Contact Info Not Available')]

答案 2 :(得分:0)

    management_links = [get_management_link(source, soup, single_website)]
    contact_link = [get_contact_link(source, soup, single_website)]
    status_str = [str(status)]
    single_link = [single_website]
    for link, status_code, management_info, contact_info in zip(single_link, status_str, management_links, contact_link):
        data = {
            "Link": link,
            "Status": status_code,
            "Management": management_info,
            "Contact": contact_info
        }

将这些值放入列表后,它现在可以正常工作了。谢谢你们!