如何在numpy中随机采样2D矩阵

时间:2017-06-08 23:44:53

标签: python arrays numpy matrix random

我有一个像这样的二维数组/矩阵,我如何从这个2D矩阵中随机选取值,例如得到像[-62, 29.23]这样的值。我查看了numpy.choice,但它是为1d数组构建的。

以下是我的4行8列

示例
Space_Position=[
      [[-62,29.23],[-49.73,29.23],[-31.82,29.23],[-14.2,29.23],[3.51,29.23],[21.21,29.23],[39.04,29.23],[57.1,29.23]],

      [[-62,11.28],[-49.73,11.28],[-31.82,11.28],[-14.2,11.28],[3.51,11.28],[21.21,11.28] ,[39.04,11.28],[57.1,11.8]],

      [[-62,-5.54],[-49.73,-5.54],[-31.82,-5.54] ,[-14.2,-5.54],[3.51,-5.54],[21.21,-5.54],[39.04,-5.54],[57.1,-5.54]],

      [[-62,-23.1],[-49.73,-23.1],[-31.82,-23.1],[-14.2,-23.1],[3.51,-23.1],[21.21,-23.1],[39.04,-23.1] ,[57.1,-23.1]]
      ]

在答案中给出了以下解决方案:

random_index1 = np.random.randint(0, Space_Position.shape[0])
random_index2 = np.random.randint(0, Space_Position.shape[1])
Space_Position[random_index1][random_index2]

这确实可以给我一个样本,像np.choice()那样的多个样本怎么样?

我想的另一种方法是将矩阵转换为数组而不是像

这样的矩阵
Space_Position=[
      [-62,29.23],[-49.73,29.23],[-31.82,29.23],[-14.2,29.23],[3.51,29.23],[21.21,29.23],[39.04,29.23],[57.1,29.23], .....   ]

并且最后使用np.choice(),但我无法找到进行转换的方法,np.flatten()使数组像

Space_Position=[-62,29.23,-49.73,29.2, ....]

2 个答案:

答案 0 :(得分:1)

只需使用随机索引(在您的情况下为2,因为您有3个维度):

import numpy as np

Space_Position = np.array(Space_Position)

random_index1 = np.random.randint(0, Space_Position.shape[0])
random_index2 = np.random.randint(0, Space_Position.shape[1])

Space_Position[random_index1, random_index2]  # get the random element.

另一种方法是实际制作2D:

Space_Position = np.array(Space_Position).reshape(-1, 2)

然后使用一个随机索引:

Space_Position = np.array(Space_Position).reshape(-1, 2)      # make it 2D
random_index = np.random.randint(0, Space_Position.shape[0])  # generate a random index
Space_Position[random_index]                                  # get the random element.

如果您想要替换N个样本:

N = 5

Space_Position = np.array(Space_Position).reshape(-1, 2)                # make it 2D
random_indices = np.random.randint(0, Space_Position.shape[0], size=N)  # generate N random indices
Space_Position[random_indices]  # get N samples with replacement

或无需替换:

Space_Position = np.array(Space_Position).reshape(-1, 2)  # make it 2D
random_indices = np.arange(0, Space_Position.shape[0])    # array of all indices
np.random.shuffle(random_indices)                         # shuffle the array
Space_Position[random_indices[:N]]  # get N samples without replacement

答案 1 :(得分:0)

Space_Position[np.random.randint(0, len(Space_Position))]
[np.random.randint(0, len(Space_Position))]

给你你想要的东西