Question

我有一个格式列表（浮点数，字符串）。如何从列表中删除具有相同浮点值的重复项？

列表按浮点顺序排序。我想保留订单。

[(0.10507038451969995,
  'Deadly stampede in Shanghai - Emergency personnel help victims.'),
 (0.078586381821416265,
  'Deadly stampede in Shanghai - Police and medical staff help injured people after the stampede.'),
 (0.072031446647399661, '- Emergency personnel help victims.'),
 (0.072031446647399661, 'Emergency personnel help victims.')]

看看最后两个。

Answer 1

您可以使用itertools.groupby，因为您已经对这些值进行了排序。这是数据：

>>> lot
[(0.10507038451969995, 'Deadly stampede in Shanghai - Emergency personnel help victims.'), 
(0.07858638182141627, 'Deadly stampede in Shanghai - Police and medical staff help injured people after the stampede.'), 
(0.07203144664739966, '- Emergency personnel help victims.'), 
(0.07203144664739966, 'Emergency personnel help victims.')]

演示：

>>> import itertools
>>> [next(t) for _, t in itertools.groupby(lot, lambda x: x[0])]
[(0.10507038451969995,
  'Deadly stampede in Shanghai - Emergency personnel help victims.'),
 (0.07858638182141627,
  'Deadly stampede in Shanghai - Police and medical staff help injured people after the stampede.'),
 (0.07203144664739966, '- Emergency personnel help victims.')]

这将为您提供组合在一起的第一个值。

Answer 2

如果值不在>>> lst [(0.10507038451969995, 'Deadly stampede in Shanghai - Emergency personnel help victims.'), (0.078586381821416265, 'Deadly stampede in Shanghai - Police and medical staff help injured people after the stampede.'), (0.072031446647399661, '- Emergency personnel help victims.'), (0.072031446647399661, 'Emergency personnel help victims.')] >>> seen = set() >>> result = [] >>> for a, b in lst: ... if not a in seen: ... seen.add(a) ... result.append((a, b)) >>> print result [(0.10507038451969995, 'Deadly stampede in Shanghai - Emergency personnel help victims.'), (0.07858638182141627, 'Deadly stampede in Shanghai - Police and medical staff help injured people after the stampede.'), (0.07203144664739966, '- Emergency personnel help victims.')]中，您可以创建一组看到的值并添加元组：

>>> seen = set()
>>> [(a, b) for a, b in lst if not (a in seen or seen.add(a))]

以下是另一种理解方式：

accessToken

Answer 3

>>> L = [(0.10507038451969995, 'Deadly stampede in Shanghai - Emergency personnel help victims.'),
...  (0.078586381821416265, 'Deadly stampede in Shanghai - Police and medical staff help injured people after the stampede.'),
...  (0.072031446647399661, '- Emergency personnel help victims.'),
...  (0.072031446647399661, 'Emergency personnel help victims.')]

>>> from collections import OrderedDict
>>> OrderedDict(L).items()
[(0.10507038451969995, 'Deadly stampede in Shanghai - Emergency personnel help victims.'),
 (0.07858638182141627, 'Deadly stampede in Shanghai - Police and medical staff help injured people after the stampede.'),
 (0.07203144664739966, 'Emergency personnel help victims.')]

根据值之一从元组列表中删除重复项

3 个答案: