在numpy中,可以创建一个矩阵并使用方便的切片表示法
arr=np.array([[1,2,3], [4,5,6], [7, 8, 9], [10,11,12]])
print (arr[2, :])
print (arr[1:2, 2])
这可以扩展到N个维度。
但现在,如果我希望有一个相同的轴,但一个轴不是数字轴,而是基于字符串的轴,该怎么办?所以索引一个元素应该是这样的:
print(arr["cylinder", :, :]) #prints all cylinders
print(arr["sphere", 4, 100]) #prints sphere of 4 radius, 100 bar
print(arr[:, 4, 100]) #prints every shape with 4 radius 100 bar
我可以为每个"组合"(所有形状,特定半径,特定压力…所有形状,所有半径,特定压强…特定形状,特定直径,特定压强)制作。一个唯一的函数,但这是不可行的,那么我该如何创建它呢?
目前,所有内容都存储为字典的字典(尤其是因为只使用半径和压力的值)。如果底层存储可以作为字典的字典保存,但添加切片/索引运算符将是黄金!
当前代码(是的,我确实有想法研究kwargs,使当前代码库更好地添加新点)-这只是为了防止"NP"问题:
class all_measurements(object):
def __init__(self):
self.measurements = {}
def add_measurement(self, measurement):
shape = measurement.shape
size = measurement.size
pressure = measurement.pressure
fname = measurement.filename
if shape in self.measurements:
shape_dict = self.measurements[shape]
else:
shape_dict = {}
self.measurements[shape] = shape_dict
if size in shape_dict:
size_dict = shape_dict[size]
else:
size_dict ={}
shape_dict[size] = size_dict
if pressure in size_dict:
pressure_dict = size_dict[pressure]
else:
pressure_dict = {}
size_dict[pressure] = pressure_dict
if fname in pressure_dict:
print("adding same file twice!")
pressure_dict[fname] = measurement
def get_measurements(self, shape = None, size = None, pressure = None, fname = None):
current_dict = self.measurements
if shape is None:
return current_dict
if shape in current_dict:
current_dict = current_dict[shape]
else:
return None
if size is None:
return current_dict
if size in current_dict:
current_dict = current_dict[size]
else:
return None
if pressure is None:
return current_dict
if pressure in current_dict:
current_dict = current_dict[pressure]
else:
return None
if fname is None:
return current_dict
if fname in current_dict:
return current_dict[fname]
else:
return None
我认为您在寻找结构化数组,请参阅此处。
示例:
>>> import numpy as np
>>> a = np.zeros(10,dtype={'names':['a','b','c'],'formats':['f64','f64','f64']})
# write some data in a
>>> a['a'] = np.arange(10)
>>> a['b'] = np.arange(10,20)
>>> a['c'] = np.arange(20,30)
>>> a
array([(0.0, 10.0, 20.0),
(1.0, 11.0, 21.0),
(2.0, 12.0, 22.0),
(3.0, 13.0, 23.0),
(4.0, 14.0, 24.0),
(5.0, 15.0, 25.0),
(6.0, 16.0, 26.0),
(7.0, 17.0, 27.0),
(8.0, 18.0, 28.0),
(9.0, 19.0, 29.0)],
dtype=[('a', '<f4'), ('b', '<f4'), ('c', '<f4')])
>>> a['a'][2:6]
array([ 2., 3., 4., 5.], dtype=float32)
>>> a[4:8]
array([(4.0, 14.0, 24.0),
(5.0, 15.0, 25.0),
(6.0, 16.0, 26.0),
(7.0, 17.0, 27.0)],
dtype=[('a', '<f4'), ('b', '<f4'), ('c', '<f4')])
重复使用类似的模式
if shape in self.measurements:
shape_dict = self.measurements[shape]
else:
shape_dict = {}
self.measurements[shape] = shape_dict
建议您可以使用collections.defaultdict
获利。
当我用一些measurements
填充你的all_measurements
对象时(使用我自己的简单类),
A = all_measurements()
A.add_measurement(measurement('round',10,20.0,'test0'))
A.add_measurement(measurement('square',10,30.0,'test1'))
A.add_measurement(measurement('round',1,20.0,'test2'))
print(A.measurements)
我有一本字典,看起来像:
{'square': {10: {30.0: {'test1': measurement: square,10,30.0,test1}}},
'round': {1: {20.0: {'test2': measurement: round,1,20.0,test2}},
10: {20.0: {'test0': measurement: round,10,20.0,test0}}}}
我在这里没有看到任何看起来像3d阵列的东西。
我想如果有一套标准的形状、尺寸和压力,例如
shapes = ['round', 'square', 'flat']
sizes = [1,3,10,20]
pressures = [10.0, 20.0, 30.0]
你可以构建一个三维阵列,例如
np.empty((3,4,3))
以及将标签映射到索引的字典或元组列表,例如
sizemap={1:0, 3:1, 10:2, 20:3}
sizelist=[(1,0),(3,1)...]
但是这个数组的值是多少呢?measurement
对象?对象类型的ndarrays
是可能的,但通常与嵌套列表或字典相比没有优势。
我测试了你的get_measurements
。按照现在的结构,你必须选择形状,然后在这些形状中选择大小等。它不能返回具有特定大小值的所有形状。
这个方法,让我使用索引,包括切片,语法来将参数传递给get_measurements
:
def __getitem__(self, key):
print(key)
key = list(key) # comes in a tuple
for i,k in enumerate(key):
if isinstance(k, slice):
# code to interpret a slice goes here
key[i] = None # fall back, do nothing
return self.get_measurements(*key)
pprint(A['round',10])
pprint(A[:,10])
pprint(A['round':'square', 10:30:10])
产生
('round', 10)
{20.0: {'test0': measurement: round,10,20.0,test0}}
(slice(None, None, None), 10)
{'round': ...}
(slice('round', 'square', None), slice(10, 30, 10))
{'round': ...}
你必须决定像这样的物体
slice('round','square', None)
slice(10, 30, 10)
在属性的上下文中的平均值