将元组的第二个元素（在元组列表中）作为字符串获取

Question

我有一个输出是一个元组列表。它看起来像这样：

annot1=[(402L, u"[It's very seldom that you're blessed to find your equal]"), 
     (415L, u'[He very seldom has them in this show or his movies]')… 

我需要使用元组的第二部分只应用'split'并分别获取每个单词的句子。

在这一点上，我无法隔离元组的第二部分（文本）。

这是我的代码：

def scope_match(annot1): 
    scope = annot1[1:] 
    scope_string = ‘’.join(scope) 
    scope_set = set(scope_string.split(' ')) 

，但我得到：

TypeError: sequence item 0: expected string, tuple found

我试图用annot1 [1]，但它给了我文本的第二个索引，而不是元组的第二个元素。

Answer 1

你可以做这样的事情跟列表解析：

annot1=[(402L, u"[It's very seldom that you're blessed to find your equal]"), 
     (415L, u'[He very seldom has them in this show or his movies]')] 
print [a[1].strip('[]').encode('utf-8').split() for a in annot1] 

输出：

[["It's", 'very', 'seldom', 'that', "you're", 'blessed', 'to', 'find', 'your', 'equal'], ['He', 'very', 'seldom', 'has', 'them', 'in', 'this', 'show', 'or', 'his', 'movies']] 

你可以像这样计算annot1和annot2中对应位置的串：

for x,y in zip(annot1,annot2): 
    print set(x[1].strip('[]').encode('utf-8').split()).intersection(y[1].strip('[]').encode('utf-8').split()) 

Answer 2

annot1是元组列表。为了从每个元素的字符串，可以做这样的事情

def scope_match(annot1): 
    for pair in annot1: 
     string = pair[1] 
     print string # or whatever you want to do