我可以创建一个本地的numpy随机种子吗? [英] Can I create a local numpy random seed?

问题描述

有一个功能foo，它使用np.random功能. 我想控制foo使用的种子，但实际上不更改函数本身. 我该怎么做?

There is a function, foo, that uses the np.random functionality. I want to control the seed that foo uses, but without actually changing the function itself. How do I do this?

基本上我想要这样的东西:

Essentially I want something like this:

bar() # should have normal seed
with np.random.seed(0): # Doesn't work
    foo()
bar() # should have normal seed

---

类似的解决方案 此:

---

Solutions like this:

rng = random.Random(42)
number = rng.randint(10, 20)

在这种情况下不起作用，因为我无权访问foo的内部工作原理(或者我错过了什么吗?).

doesn't work in this case, as I don't have access to the inner workings of foo (or am I missing something??).

推荐答案

您可以将全局随机状态保留在一个临时变量中，并在函数完成后将其重置:

You could keep the global random state in a temporary variable and reset it once your function is done:

import contextlib
import numpy as np

@contextlib.contextmanager
def temp_seed(seed):
    state = np.random.get_state()
    np.random.seed(seed)
    try:
        yield
    finally:
        np.random.set_state(state)

演示:

>>> np.random.seed(0)
>>> np.random.randn(3)
array([1.76405235, 0.40015721, 0.97873798])
>>> np.random.randn(3)
array([ 2.2408932 ,  1.86755799, -0.97727788])

>>> np.random.seed(0)
>>> np.random.randn(3)
array([1.76405235, 0.40015721, 0.97873798])
>>> with temp_seed(5):
...     np.random.randn(3)                                                                                        
array([ 0.44122749, -0.33087015,  2.43077119])
>>> np.random.randn(3)
array([ 2.2408932 ,  1.86755799, -0.97727788])

这篇关于我可以创建一个本地的numpy随机种子吗?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！