[英]Finding matching subset of "row" in a numpy structured array
I have data stored in a NumPy structured array where part of the information identifies various cases.我将数据存储在 NumPy 结构化数组中,其中部分信息标识了各种情况。 I would like to find the row that matches a given case.我想找到与给定案例匹配的行。 Eg, let's say I'm storing the name of a building, room number, and the number of chairs and tables in the room in a (2,) array.例如,假设我将建筑物的名称、房间号以及房间中的椅子和桌子的数量存储在一个 (2,) 数组中。 This would then look something like this:这看起来像这样:
import numpy as np
my_dtype = [('building', '<U5'), ('room', '<i8'), ('seating', '<i8', (2,))]
room_info = np.array([('BLDG0', 12, [24, 6]),
('BLDG1', 34, [32, 10]),
('BLDG0', 14, [10, 20])],
dtype=my_dtype)
Now say that I want to find the row for building 'BLDG0'
, room 14
.现在说我想找到建筑物'BLDG0'
的行,房间14
。 Based on the answer to Finding a matching row in a numpy matrix , I tried根据Finding a matching row in a numpy matrix的答案,我尝试了
sub_fields = ['building', 'room']
matching_index, = np.where(room_info[sub_fields] == ('BLDG0', 14))
which would ideally result in [2]
.理想情况下会导致[2]
。 However, this results in the following warning:但是,这会导致以下警告:
FutureWarning: elementwise == comparison failed and returning scalar instead; this will raise an error or perform elementwise comparison in the future.
and returns an empty array.并返回一个空数组。 Is there a way to find the matching sub- row for a large set of data other than comparing each column separately and then finding the matching indices?除了分别比较每一列然后找到匹配的索引之外,有没有办法为大量数据找到匹配的子行?
I am using NumPy version 1.18.5 through miniconda and it doesn't look like I can safely update to a newer version within this environment.我正在通过 miniconda 使用 NumPy 版本 1.18.5,看起来我无法在此环境中安全地更新到更新的版本。 (Though I'm not sure if newer versions support this type of comparison) (虽然我不确定新版本是否支持这种类型的比较)
In [243]: my_dtype = [('building', '<U5'), ('room', '<i8'), ('seating', '<i8', (
...: 2,))]
...: room_info = np.array([('BLDG0', 12, [24, 6]),
...: ('BLDG1', 34, [32, 10]),
...: ('BLDG0', 14, [10, 20])],
...: dtype=my_dtype)
In [244]: room_info
Out[244]:
array([('BLDG0', 12, [24, 6]), ('BLDG1', 34, [32, 10]),
('BLDG0', 14, [10, 20])],
dtype=[('building', '<U5'), ('room', '<i8'), ('seating', '<i8', (2,))])
In [246]: room_info['building']
Out[246]: array(['BLDG0', 'BLDG1', 'BLDG0'], dtype='<U5')
In [247]: room_info['building']=='BLDG0'
Out[247]: array([ True, False, True])
In [248]: room_info['room']==14
Out[248]: array([False, False, True])
combine the two:结合两者:
In [249]: Out[247] & Out[248]
Out[249]: array([False, False, True])
Use that as a boolean mask:将其用作 boolean 掩码:
In [250]: room_info[_]
Out[250]:
array([('BLDG0', 14, [10, 20])],
dtype=[('building', '<U5'), ('room', '<i8'), ('seating', '<i8', (2,))])
and getting the index:并获取索引:
In [251]: np.nonzero(Out[247]&Out[248])
Out[251]: (array([2]),)
Looks like we can test both fields, using a properly constructed structured array:看起来我们可以使用正确构造的结构化数组来测试这两个字段:
In [254]: test=np.array(('BLDG0',14),dtype=my_dtype[:2])
In [255]: room_info[['building','room']]
Out[255]:
array([('BLDG0', 12), ('BLDG1', 34), ('BLDG0', 14)],
dtype={'names':['building','room'], 'formats':['<U5','<i8'], 'offsets':[0,20], 'itemsize':44})
In [256]: room_info[['building','room']]==test
Out[256]: array([False, False, True])
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