将元组的第二个元素（在元组列表中）获取为字符串

Question

我有一个输出是元组列表。 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]')…

我只需要使用元组的第二部分来应用“拆分”并分别获取句子中的每个单词。

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(' '))

但是我得到：

TypeError: sequence item 0: expected string, tuple found

我尝试使用annot1 [1]，但是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

您可以使用列表推导功能执行以下操作：

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']]

您可以像这样在annot1和annot2中的相应位置计算字符串的交集：

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是一个元组列表。 要从每个元素中获取字符串，您可以执行以下操作

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