Question

我正在尝试编写一个完整搜索嵌套列表的函数，以返回包含某个单词的所有列表，但这只返回None

word = "what song?"
def searchSong(mp3_list, word):
    search = input((word))
    match = [i for i in mp3_list if search in i[2]] 
for confirmed in match:
    print(confirmed[0],'\n', confirmed[1],'\n', confirmed[2])

print(searchSong(mp3_list, word))

当我进行比较测试时，match变量仍然不返回任何内容：

mp3_list = [["Eric Clapton","Tears in heaven","Rush"],["Neil Young", "Heart of gold", "Harvest"]]
match = [i for i in mp3_list if 'heaven' in i[2]]
print(match)      #returns []

但是这很有效，尽管语法看起来完全一样：

li = [["0", "20", "ar"], ["20", "40", "asdasd"], ["50", "199", "bar"]]
match = [i for i in li if 'ar' in i[2]]
print(match)     #returns [['0', '20', 'ar'], ['50', '199', 'bar']]

任何帮助都会非常感激:)

Answer 1

＆＃39;天堂＆＃39;存在于["Eric Clapton","Tears in heaven","Rush"]的第二个元素中，但您要检查第三个元素（i[2]）。 Python数组从0开始编号，而不是1.（但你似乎知道这一点，因为你的第二个例子在i[2]中寻找＆＃34; ar＆＃34;这是正确的。）

此外，searchSong()并未返回任何内容，因此默认情况下它始终返回None。

Answer 2

mp3_list = [["Eric Clapton","Tears in heaven","Rush"],["Neil Young", "Heart of gold", "Harvest"]]
match = [i for i in mp3_list for j in i if 'heaven' in j]
match
Out[7]: [['Eric Clapton', 'Tears in heaven', 'Rush']]


li = [["0", "20", "ar"], ["20", "40", "asdasd"], ["50", "199", "bar"]]
match = [i for i in li for j in i if 'ar' in j]
match
Out[10]: [['0', '20', 'ar'], ['50', '199', 'bar']]

列表理解 - 迭代嵌套列表

2 个答案: