我有orig
个元组列表,其中包含dict
和text
个值。
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
中的唯一值列出该组。如下所示:
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')]]
答案 0 :(得分:1)
如果您使用SELECT *
FROM myTable
WHERE (
SELECT CONVERT(date, myDateTimeColumn)
) = '2019-01-01'
,会更容易。
SELECT *
FROM myTable
WHERE date(myDateTimeColumn) > '2019-01-01'
AND
SELECT *
FROM myTable
WHERE date(myDateTimeColumn) < '2019-01-02'
输出:
numpy
答案 1 :(得分:1)
使用itertools.groupby
和list comprehension
的两层解决方案:
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:)