python - 类型错误: 'float' 对象在内置 max 函数的列表上不可迭代

Question

我正在尝试使用 max 函数及其关键参数在给定实际电影标题的情况下找到与近似电影标题最接近的匹配。如果我定义一个示例列表并测试它的功能...

from difflib import SequenceMatcher as SM
movies = ['fake movie title', 'faker movie title', 'shaun died']
approx_title = 'Shaun of the Dead.'
max(movies, key = lambda title: SM(None, approx_title, title).ratio())
'shaun died'

但是我试图匹配单独数据框中的整个列，所以我尝试将 Pandas Series 转换为列表并运行相同的函数，但我得到了 type_error，即使我已经检查了数据类型movie 和 movie_lst 都是列表。

Old id  New id  Title   Year    Critics Score   Audience Score  Rating
NaN     21736.0 Peter Pan   1999.0  NaN 70.0    PG nothing objectionable
NaN     771471359.0 Dragonheart Battle for the Heartfire    2017.0  NaN 50.0    PG13
NaN     770725090.0 The Nude Vampire Vampire nue, La    1974.0  NaN 24.0    NR
2281.0  19887.0 Beyond the Clouds   1995.0  65.0    67.0    NR
10913.0 11286.0 Wild America    1997.0  27.0    59.0    PG violence

movie_lst = rt_info['Title'].tolist()
 ['Peter Pan',
 'Dragonheart Battle for the Heartfire',
 'The Nude Vampire Vampire nue, La',
 'Beyond the Clouds',
 'Wild America',
 'Sexual Dependency',
 'Body Slam',
 'Hatchet II',
 'Lion of the Desert Omar Mukhtar',
 'Imagine That',
 'Harold',
 'A United Kingdom',
 'Violent City The FamilyCitt violenta',
 'Ratchet  Clank',
 'Wes Craven Presents Carnival of Souls',
 'The Adventures of Ociee Nash',
 'Blackfish',
 'For Petes Sake',
 'Daybreakers',
 'The Big One',
 'Godzilla vs Megaguirus',
 'In a Lonely Place',
 'Case 39', ...
]

max(movie_lst, key = lambda title: SM(None, approx_title, title).ratio())

TypeError                                 Traceback (most recent call last)
<ipython-input-88-0022a3c1bdb9> in <module>()
----> 1 max(movie_lst, key = lambda title: SM(None, approx_title, title).ratio())

<ipython-input-88-0022a3c1bdb9> in <lambda>(title)
----> 1 max(movie_lst, key = lambda title: SM(None, approx_title, title).ratio())

/usr/lib/python3.4/difflib.py in __init__(self, isjunk, a, b, autojunk)
    211         self.a = self.b = None
    212         self.autojunk = autojunk
--> 213         self.set_seqs(a, b)
    214 
    215     def set_seqs(self, a, b):

/usr/lib/python3.4/difflib.py in set_seqs(self, a, b)
    223 
    224         self.set_seq1(a)
--> 225         self.set_seq2(b)
    226 
    227     def set_seq1(self, a):

/usr/lib/python3.4/difflib.py in set_seq2(self, b)
    277         self.matching_blocks = self.opcodes = None
    278         self.fullbcount = None
--> 279         self.__chain_b()
    280 
    281     # For each element x in b, set b2j[x] to a list of the indices in

/usr/lib/python3.4/difflib.py in __chain_b(self)
    309         self.b2j = b2j = {}
    310 
--> 311         for i, elt in enumerate(b):
    312             indices = b2j.setdefault(elt, [])
    313             indices.append(i)

TypeError: 'float' object is not iterable

我很困惑为什么 - 任何帮助将不胜感激!

Answer 1

不是 pandas 专家，无法重现，但取决于文件的读取方式，因为有些标题(例如法国电影 11.6)与 float 匹配，因此某些数据可能是float而不是字符串(你的问题证明它是可能的:))

一个好的解决方法是将数据强制为字符串，如下所示:

movie_lst = [str(x) for x in movie_lst]

如果字符串已经是字符串 ( Should I avoid converting to a string if a value is already a string? )，它不会创建字符串的副本，因此它非常高效，并且您肯定只会获得字符串。

请注意，您可以通过打印找到违规者:

[x for x in movie_lst if not isinstance(x,str)]