按字典python列表中的键值属性分组

Question

我有包含dict和text值的元组的orig列表。

orig = [({'x': 28.346, 'y': 19},'Text0'),
 ({'x': 109.726, 'y': 19},'Text1'),
 ({'x': 147.776, 'y': 19},'Text2'),
 ({'x': 153.606, 'y': 24}, 'Text3'),
 ({'x': 452.788, 'y': 24}, 'Text4'),
 ({'x': 504.168, 'y': 34}, 'Text5'),
 ({'x': 527.768, 'y': 34}, 'Text6'),
 ({'x': 533.598, 'y': 45},'Text7'),
 ({'x': 64.291, 'y': 55},'Text8'),
 ({'x': 98.623, 'y': 55},'Text9')]

我想从key='y'中过滤组，这将使我根据y唯一值列出它。 类似于以下内容：

res = [
      [({'x': 28.346, 'y': 19},'Text0'),
         ({'x': 109.726, 'y': 19},'Text1'),
         ({'x': 147.776, 'y': 19},'Text2')],
     [({'x': 153.606, 'y': 24}, 'Text3'),
         ({'x': 452.788, 'y': 24}, 'Text4')],
     [({'x': 504.168, 'y': 34}, 'Text5'),
         ({'x': 527.768, 'y': 34}, 'Text6')],
     [({'x': 533.598, 'y': 45},'Text7')],
     [({'x': 64.291, 'y': 55},'Text8'),
         ({'x': 98.623, 'y': 55},'Text9')]]

Answer 1

如果你使用numpy它会更容易一些。

import numpy as np
orig = [({'x': 28.346, 'y': 19}, 'Text0'),
        ({'x': 109.726, 'y': 19}, 'Text1'),
        ({'x': 147.776, 'y': 19}, 'Text2'),
        ({'x': 153.606, 'y': 24}, 'Text3'),
        ({'x': 452.788, 'y': 24}, 'Text4'),
        ({'x': 504.168, 'y': 34}, 'Text5'),
        ({'x': 527.768, 'y': 34}, 'Text6'),
        ({'x': 533.598, 'y': 45}, 'Text7'),
        ({'x': 64.291, 'y': 55}, 'Text8'),
        ({'x': 98.623, 'y': 55}, 'Text9')]


input_array = np.array([val[0]['y'] for val in orig])
out_array = [np.where(input_array == element)[0].tolist() for element in np.unique(input_array)]
res = [[orig[i] for i in ind_arr] for ind_arr in out_array]
print(res)

输出：

[[({'x': 28.346, 'y': 19}, 'Text0'),
  ({'x': 109.726, 'y': 19}, 'Text1'),
  ({'x': 147.776, 'y': 19}, 'Text2')],
 [({'x': 153.606, 'y': 24}, 'Text3'),
    ({'x': 452.788, 'y': 24}, 'Text4')],
 [({'x': 504.168, 'y': 34}, 'Text5'),
    ({'x': 527.768, 'y': 34}, 'Text6')],
 [({'x': 533.598, 'y': 45}, 'Text7')],
 [({'x': 64.291, 'y': 55}, 'Text8'),
    ({'x': 98.623, 'y': 55}, 'Text9')]]

Answer 2

使用itertools.groupby和list comprehension两行解决方案：

from itertools import groupby
# group by the input orig with a key of dict "y" and then take it in a list of list comprehension
print ([[x for x in v] for k, v in groupby(orig, key= lambda x: x[0]["y"])])

结果：

[[({'x': 28.346, 'y': 19}, 'Text0'), ({'x': 109.726, 'y': 19}, 'Text1'), ({'x': 147.776, 'y': 19}, 'Text2')], [({'x': 153.606, 'y': 24}, 'Text3'), ({'x': 452.788, 'y': 24}, 'Text4')], [({'x': 504.168, 'y': 34}, 'Text5'), ({'x': 527.768, 'y': 34}, 'Text6')], [({'x': 533.598, 'y': 45}, 'Text7')], [({'x': 64.291, 'y': 55}, 'Text8'), ({'x': 98.623, 'y': 55}, 'Text9')]]

我希望这很重要:)