如何在 PyTorch 中获得 [r1,r2] 范围内的均匀分布？

Question

我想在 PyTorch 中获得一个大小为[a,b]的二维torch.Tensor ，其中填充了来自均匀分布（在[r1,r2]范围内）的值。

Answer 1

If U is a random variable uniformly distributed on [0, 1], then (r1 - r2) * U + r2 is uniformly distributed on [r1, r2].

Thus, you just need:

(r1 - r2) * torch.rand(a, b) + r2

Alternatively, you can simply use:

torch.FloatTensor(a, b).uniform_(r1, r2)

---

To fully explain this formulation, let's look at some concrete numbers:

r1 = 2 # Create uniform random numbers in half-open interval [2.0, 5.0)
r2 = 5

a = 1  # Create tensor shape 1 x 7
b = 7

We can break down the expression (r1 - r2) * torch.rand(a, b) + r2 as follows:

1. torch.rand(a, b) produces an axb (1x7) tensor with numbers uniformly distributed in the range [0.0, 1.0).

x = torch.rand(a, b)
print(x)
# tensor([[0.5671, 0.9814, 0.8324, 0.0241, 0.2072, 0.6192, 0.4704]])

2. (r1 - r2) * torch.rand(a, b) produces numbers distributed in the uniform range [0.0, -3.0)

print((r1 - r2) * x)
tensor([[-1.7014, -2.9441, -2.4972, -0.0722, -0.6216, -1.8577, -1.4112]])

3. (r1 - r2) * torch.rand(a, b) + r2 produces numbers in the uniform range [5.0, 2.0)

print((r1 - r2) * x + r2)
tensor([[3.2986, 2.0559, 2.5028, 4.9278, 4.3784, 3.1423, 3.5888]])

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Now, let's break down the answer suggested by @Jonasson: (r2 - r1) * torch.rand(a, b) + r1

1. Again, torch.rand(a, b) produces (1x7) numbers uniformly distributed in the range [0.0, 1.0).

x = torch.rand(a, b)
print(x)
# tensor([[0.5671, 0.9814, 0.8324, 0.0241, 0.2072, 0.6192, 0.4704]])

2. (r2 - r1) * torch.rand(a, b) produces numbers uniformly distributed in the range [0.0, 3.0).

print((r2 - r1) * x)
# tensor([[1.7014, 2.9441, 2.4972, 0.0722, 0.6216, 1.8577, 1.4112]])

3. (r2 - r1) * torch.rand(a, b) + r1 produces numbers uniformly distributed in the range [2.0, 5.0)

print((r2 - r1) * x + r1)
tensor([[3.7014, 4.9441, 4.4972, 2.0722, 2.6216, 3.8577, 3.4112]])

In summary , (r1 - r2) * torch.rand(a, b) + r2 produces numbers in the range [r2, r1), while (r2 - r1) * torch.rand(a, b) + r1 produces numbers in the range [r1, r2).

Answer 2

torch.FloatTensor(a, b).uniform_(r1, r2)

Answer 3

Utilize the torch.distributions package to generate samples from different distributions.

For example to sample a 2d PyTorch tensor of size [a,b] from a uniform distribution of range(low, high) try the following sample code

import torch
a,b = 2,3   #dimension of the pytorch tensor to be generated
low,high = 0,1 #range of uniform distribution

x = torch.distributions.uniform.Uniform(low,high).sample([a,b]) 

Answer 4

To get a uniform random distribution, you can use

torch.distributions.uniform.Uniform()

example,

import torch
from torch.distributions import uniform

distribution = uniform.Uniform(torch.Tensor([0.0]),torch.Tensor([5.0]))
distribution.sample(torch.Size([2,3])

This will give the output, tensor of size [2, 3].

Answer 5

Please Can you try something like:

import torch as pt
pt.empty(2,3).uniform_(5,10).type(pt.FloatTensor)

Answer 6

This answer uses NumPy to first produce a random matrix and then converts the matrix to a PyTorch tensor. I find the NumPy API to be easier to understand.

import numpy as np

torch.from_numpy(np.random.uniform(low=r1, high=r2, size=(a, b)))

Answer 7

PyTorch has a number of distributions built in. You can build a tensor of the desired shape with elements drawn from a uniform distribution like so:

from torch.distributions.uniform import Uniform

shape = 3,4
r1, r2 = 0,1

x = Uniform(r1, r2).sample(shape) 

Answer 8

For those who are frustratingly bashing their keyboard yelling "why isn't this working." as I was... note the underscore behind the word uniform.

torch.FloatTensor(a, b).uniform_(r1, r2)
                               ^ here

Answer 9

See this for all distributions: https://pytorch.org/docs/stable/distributions.html#torch.distributions.uniform.Uniform

This is the way I found works:

# generating uniform variables

import numpy as np

num_samples = 3
Din = 1
lb, ub = -1, 1

xn = np.random.uniform(low=lb, high=ub, size=(num_samples,Din))
print(xn)

import torch

sampler = torch.distributions.Uniform(low=lb, high=ub)
r = sampler.sample((num_samples,Din))

print(r)

r2 = torch.torch.distributions.Uniform(low=lb, high=ub).sample((num_samples,Din))

print(r2)

# process input
f = nn.Sequential(OrderedDict([
    ('f1', nn.Linear(Din,Dout)),
    ('out', nn.SELU())
]))
Y = f(r2)
print(Y)

but I have to admit I don't know what the point of generating sampler is and why not just call it directly as I do in the one liner (last line of code).

Comments:

• sampler are good for it's so you can transform/compose/cache/etc distributions. see https://arxiv.org/abs/1711.10604 , and the top of the docs of https://pytorch.org/docs/stable/distributions.html# and https://arxiv.org/abs/1506.05254
• you can feed in tensors to uniform to let it know the high dimensional interval (hypercube) to generate the uniform samples (that's why it receives tensors as input rather than simply numbers)

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Reference:

• How to get a uniform distribution in a range [r1,r2] in PyTorch?
• https://discuss.pytorch.org/t/generating-random-tensors-according-to-the-uniform-distribution-pytorch/53030/8
• https://github.com/pytorch/pytorch/issues/24162

Answer 10

Pytorch (now?) has a random integer function that allows:

torch.randint(low=r1, high=r2, size=(1,), **kwargs)

and returns uniformly sampled random integers of shape size in range [r1, r2).

https://pytorch.org/docs/stable/generated/torch.randint.html