Question

给出数字和编码长度，如何将数字转换为张量的二进制表示形式？

例如，给定数字6和宽度8，我如何获得张量：

(0, 0, 0, 0, 0, 1, 1, 0)

Answer 1


def binary(x, bits):
    mask = 2**torch.arange(bits).to(x.device, x.dtype)
    return x.unsqueeze(-1).bitwise_and(mask).ne(0).byte()

如果您想反转位的顺序，请改用torch.arange(bits-1,-1,-1)。

Answer 2

Tiana's answer是不错的选择。顺便说一句，要将Tiana的2基结果转换回10基数，可以这样做：

import torch
import numpy as np


def dec2bin(x, bits):
    # mask = 2 ** torch.arange(bits).to(x.device, x.dtype)
    mask = 2 ** torch.arange(bits - 1, -1, -1).to(x.device, x.dtype)
    return x.unsqueeze(-1).bitwise_and(mask).ne(0).float()


def bin2dec(b, bits):
    mask = 2 ** torch.arange(bits - 1, -1, -1).to(b.device, b.dtype)
    return torch.sum(mask * b, -1)


if __name__ == '__main__':
    NUM_BITS = 7
    d = torch.randint(0, 16, (3, 6))
    b = dec2bin(d, NUM_BITS)
    # print(d)
    # print(b)
    # print(b.shape)
    # print("num of total bits: {}".format(np.prod(b.shape)))

    d_rec = bin2dec(b, NUM_BITS)

    # print(d_rec)
    print(abs(d - d_rec).max())  # should be 0.

Answer 3

如果输入为无符号字节且输出宽度为8位：

>>> binary = np.unpackbits(np.array([0xaa, 0xf0], dtype=np.uint8))
>>> print(torch.tensor(binary))
tensor([1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0], dtype=torch.uint8)

请注意，unpackbits()仅与np.uint8一起使用。

Answer 4

def decimal_to_binary_tensor(value, width=0):
    string = format(value, '0{}b'.format(width))
    binary = [0 if c == '0' else 1 for c in string]
    return torch.tensor(binary, dtype=torch.uint8)

示例：

>>> print(decimal_to_binary_tensor(6, width=8))
tensor([0, 0, 0, 0, 0, 1, 1, 0], dtype=torch.uint8)

>>> print(decimal_to_binary_tensor(6))
tensor([1, 1, 0], dtype=torch.uint8)

将整数转换为二进制位的pytorch张量

4 个答案: