Question

我有元组列表。必须根据项目值从列表中排除某些元组。具有2个元组的列表的示例如下：

orig_list = [('mydomain', '20150726', 'e-buyeasy.com', 'www.ujizlhekw.e', '1', 
'1', '1', '1', '100.0', '0.0'), 
('myotherdomain', '20150726', 'floating-share-buttons.com', '', '26',
'26', '26', '26', '100.0', '0.0')]

我创建了一个函数，它使用两个元组项的正则表达式匹配，而不是其他“查找”。名单。该函数返回True或False，具体取决于与查找列表的匹配。

def tupMatch(tup):
    sourceReg="semalt.*","anticrawler.*","best-seo-offer.*","best-seo-solution.*","buttons-for-website.*","buttons-for-your-website.*","7makemoneyonline.*","-musicas*-gratis.*","kambasoft.*","savetubevideo.*","ranksonic.*","medispainstitute.*","offers.bycontext.*","100dollars-seo.*","sitevaluation.*","dailyrank.*","videos-for-your-business.*","videos-for-your-business.*","success-seo.*","4webmasters.*","get-free-traffic-now.*","free-social-buttons.*","trafficmonetizer.*","traffic2money.*","floating-share-buttons.*"

    hostnameReg="mydomain.*","myotherdomain.*"

    sourceReg2 = "(" + ")|(".join(sourceReg) + ")"
    hostnameReg2 = "(" + ")|(".join(hostnameReg) + ")"

    sourceMatch = re.match(sourceReg2, tup[2].lower())
    hostMatch = re.match(hostnameReg2, tup[3].lower())

    if (not sourceMatch) and (hostMatch):
        True
    else:
        False
    return

我有一个列表推导，根据功能结果过滤原始列表。

filtered_list = [tup for tup in orig_list if tupMatch(tup)]

但实际上并没有过滤列表。我期待着＆＃39; tupMatch（tup）＆＃39;将返回True或False，列表推导将只有True的元组。

我在这里缺少什么？

Answer 1

以下行不return（没有return个关键字）;该函数隐式返回None;这被视为虚假价值。

if (not sourceMatch) and (hostMatch):
    True
else:
    False

按以下方式更改行：

if (not sourceMatch) and (hostMatch):
    return True
else:
    return False

或更短时间：

return (not sourceMatch) and (hostMatch)   # OR  return bool(...)

Python列表理解条件过滤器语法

1 个答案: