python - 单轴索引为字符串的 Numpy 数组(矩阵)

Question

在 numpy 中可以创建一个矩阵并使用方便的切片符号

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 维。

但是现在如果我希望拥有相同的东西，但一个轴不是数字轴，而是一个基于字符串的轴怎么办？所以索引一个元素就像:

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、特定压力……特定形状、特定半径、特定压力)制作。一个独特的功能，但这是不可行的，那么我该如何创建它呢？

目前一切都存储为字典的字典(特别是因为只使用半径和压力的值)。如果底层存储可以保留为字典的字典 - 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 以使当前代码库更好地添加新点)——这只是为了防止“NP”问题而添加的:

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

Answer 1

我认为您正在寻找结构化数组，请参阅 here .

例子:

>>> 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')])