我使用列表的列表。这些列表中的每一个都是相同的 - 它们包含标题,url和一些额外的统计信息(始终以相同的顺序)。我想创建一个函数 ,它采用所需的标题并返回整个列表(带有标题,url和统计信息)。这是我的尝试find_titledef find_title(title, ls): return(list(filter(lambda x: x[0] == title, ls)))但是它不起作用,它不返回任何内容。这可能是因为仅表示大列表中的第一个元素。如何修复?x[0]编辑。这是ls的一部分:[['Der Vagabund und das Kind (1921)', 'http://www.imdb.com/title/tt0012349/', 0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], ['Goldrausch (1925)', 'http://www.imdb.com/title/tt0015864/', 0,0,1,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], ['Metropolis (1927)', 'http://www.imdb.com/title/tt0017136/', 0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0]]
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明月笑刀无情
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您可以直接从数据帧中提取信息,如下所示:
import pandas as pd
df = pd.DataFrame([['Der Vagabund und das Kind (1921)',
'http://www.imdb.com/title/tt0012349/',
0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
['Goldrausch (1925)',
'http://www.imdb.com/title/tt0015864/',
0,0,1,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
['Metropolis (1927)',
'http://www.imdb.com/title/tt0017136/',
0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0]])
print(df[df[0] =='Goldrausch (1925)'])
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