如何检查列表中的所有字符串？

Question

website = str(raw_input('Website: '))
palavras_chaves = ['joomla', 'Joomla']
palavras_chaves2 = ['wordpress', 'Wordpress', 'WordPress', 'wp']    
abrindo = urllib2.urlopen(website)
abrindo = abrindo.read()
if palavras_chaves[0] and palavras_chaves[1] in abrindo:
    print '\nÉ um joomla!'

elif palavras_chaves2[0] and palavras_chaves2[1] and palavras_chaves2[2] and palavras_chaves2[3] in abrindo:        
    print '\nÉ um wordpress!'

else:
    print '\nÉ um CMS desconhecido!'

我想知道如何在没有指定的情况下检查所有字符串。似乎很简单，但我尝试了一切我想到的，但没有工作。你知道吗

Answer 1

正如@fourtheye所指出的，这里需要内置的^{}函数，您可以使用它和列表理解来完成下面描述的工作。你知道吗

所以试试这个：

if all(i in abrindo for i in palavras_chaves):
    print '\nÉ um joomla!'

elif all(i in abrindo for i in palavras_chaves2):
    print '\nÉ um wordpress!'
else:
    print '\nÉ um CMS desconhecido!'

编辑：

我想如果你手动操作，比如：

if palavras_chaves[0] and palavras_chaves[1] in abrindo:

它之所以起作用是因为palavras_chaves[1]在abrindo中，但是palavras_chaves[0]可能不在abrindo中，但是当您使用all()时，^{>>中的所有字符串都有在abrindo中，万一有一个字符串不在abrindo中，它就失败了！

看这个例子，您尝试这样做，您希望它返回'Works as expected'，但是它返回相反的结果

>>> if 'Hey' and 'Bye' in 'Bye':
...     print 'Not expected right?'
... else:
...     print 'Works as expected'
...     
#Not expected right?

但是使用all()：

if all(i in 'Bye'for i in ['Hey','Bye']):
    print 'Not expected right?'
else:
    print 'Works as expected'
#Works as expected

Answer 2

使用python的内置集，它有一个issubset方法。http://docs.python.org/2/library/stdtypes.html#set