Question

现在我有这个嵌套列表和字典：

phrases = [['Lionhearted', 'Baby Preach'], ['Lionhearted Baby', 'Preach'], ['Lionhearted', 'Baby', 'Preach']]
artist_info = {u'Baby': u'Baby by Justin Bieber', u'Lionhearted': u'Lionhearted by Porter Robinson', u'Preach': u'Preach by Drake', u'Baby Preach': u:'Baby Preach by some pop singer'}

基本上，我想要做的是尝试使用来自artist_info的值替换短语中的值（如果存在匹配），另外我想省略该特定的嵌套列表。因为有'Lionhearted'，'Baby'，＆＃39;婴儿讲道＆＃39;和'Preach'，输出将读取如下内容：

phrases2 = [['Lionhearted by Porter Robinson','Baby Preach by some pop singer'] ['Lionhearted by Porter Robinson', 'Baby by Justin Bieber', 'Preach by Drake']]

有办法做到这一点吗？我已经在这几个小时了。重要的是，这三个短语导致了“狮子会婴儿讲道”＃34;的顺序。我有点像Python菜鸟，但我还没能找到一个类似的例子。

Answer 1

这是你在找什么？在您的示例中，如果该子列表中的任何项目不在字典phrases中，则您似乎想要省略artist_info中的子列表。在执行您提到的替换之前，下面的代码会这样做。

phrases = [['Lionhearted', 'Baby Preach'], ['Lionhearted Baby', 'Preach'], ['Lionhearted', 'Baby', 'Preach']]
artist_info = {u'Baby': u'Baby by Justin Bieber', u'Lionhearted': u'Lionhearted by Porter Robinson', u'Preach': u'Preach by Drake', u'Baby Preach': u'Baby Preach by some pop singer'}

lst = filter(lambda x: all(y in artist_info for y in x), phrases)
phrases2 = map(lambda x: [artist_info[y] for y in x], lst)
print phrases2

Answer 2

首先，我假设结果列表的排序顺序是任意的。

一步一步：

phrases = [['Lionhearted', 'Baby Preach'], ['Lionhearted Baby', 'Preach'], ['Lionhearted', 'Baby', 'Preach']]
artist_info = {u'Baby': u'Baby by Justin Bieber', u'Lionhearted': u'Lionhearted by Porter Robinson', u'Preach': u'Preach by Drake'}

phrases = [' '.join(item).split() for item in phrases]

phrases2 = set()

for item in phrases:
    for word in item:
        for key in artist_info:
            if word in artist_info.get(key):
                phrases2.add(artist_info.get(key))

phrases2 = list(phrases2)

print(*phrases2, sep='\n')

或者，理解：

phrases = [['Lionhearted', 'Baby Preach'], ['Lionhearted Baby', 'Preach'], ['Lionhearted', 'Baby', 'Preach']]
artist_info = {u'Baby': u'Baby by Justin Bieber', u'Lionhearted': u'Lionhearted by Porter Robinson', u'Preach': u'Preach by Drake'}

phrases2 = list({artist_info.get(key) for item in [' '.join(phrase).split() for phrase in phrases] for word in item for key in artist_info if word in artist_info.get(key)})

具有任一版本的phrases2的内容：

['Baby by Justin Bieber', 'Lionhearted by Porter Robinson', 'Preach by Drake']

使用以下内容打印出来：

print(*phrases2, sep='\n')

根据字典中的值有条件地替换嵌套列表中的项目

2 个答案: