Question

如何在Python中执行以下操作？

以线性数组x开始，以及相关的概率密度函数（pdf）p(x)
按N

p(x)

返回x，删除所选元素（x将更短）
返回N个元素的新数组，按顺序选择

警告：数组x可能有重复的值xi，但它们有不同的概率（p(x)实际上取决于我没有显示的参数），所以我想确定删除所选的“正确”的，而不仅仅是具有相同值xi的元素。

我的代码因此：

import numpy as np

N = 5;
x = np.linspace(0,5,11)
pdf = np.exp(-x**2);
pdf = pdf/np.sum(pdf);

x_rand_sorted = np.random.choice(x,N,replace=False,p = pdf)

print 'x:',x
print 'first N random elements:', x_rand_sorted
print 'x without random elements = ??'

Answer 1

您可以使用布尔掩码：

import numpy as np

N = 5;
x = np.linspace(0,5,11)
pdf = np.exp(-x**2);
pdf = pdf/np.sum(pdf);

index = np.full(x.shape, False, bool)
index[np.random.choice(np.arange(index.shape[0]), N, replace = False,p = pdf)] = True

x_rand_sorted = x[index]
x_rand_remaining = x[~index]

print 'x:',x
print 'first N random elements:', x_rand_sorted
print 'x without random elements: ', x_rand_remaining

Answer 2

输出：

剩余价值：x_remaining
N随机元素：x_rand_vals
随机元素索引：randices

示例：

import numpy as np
N = 9;
n = 4;
x = np.linspace(0,2,N);
pdf = np.exp(-x**2);
pdf = pdf/np.sum(pdf);

#create mask, choose random indices from x according to pdf, set chosen indices to True:
indices = np.full(x.shape, False, bool)
randices = np.random.choice(np.arange(indices.shape[0]), n, replace = False,p = pdf)
indices[randices] = True

x_rand_vals = x[randices]
x_remaining = x[~indices]

np.set_printoptions(precision=3)
print 'original x:\t\t', x
print 'pdf array:\t\t', pdf
print 'random indices:\t\t',randices
print 'random values:\t\t',x_rand_vals
print 'remaining array:\t',x_remaining

（示例输出）：

    original x:         [ 0.   0.2  0.4  0.6  0.8  1.   1.2  1.4  1.6  1.8  2. ]
    random indices:     [4 2 5]
    random values:      [ 0.8  0.4  1. ]
    remaining array:    [ 0.   0.2  0.6  1.2  1.4  1.6  1.8  2. ]

从数组中删除随机选择的元素

2 个答案: