体温计编码 [英] Numpy thermometer encoding

问题描述

我正在尝试使用numpy优化的内置函数来生成温度计编码.如果给定长度为1，温度计编码基本上会生成 n 数量.例如，在8长度中，3将被编码为:

I am trying to use numpy optimized in-built functions to generate thermometer encoding. Thermometer encoding is basically generating n amount if 1's in a given length. For example in 8-length, 3 will be encoded as:

1 1 1 0 0 0 0 0

使用numpy根据整数输入生成矢量基本上是切片并设置1.

Using numpy to generate that vector based on a integer input is basically slicing and setting 1.

stream[:num_ones] = 1

所以给我的问题一个 vector 作为输入，什么是生成矩阵输出的最佳方法，例如:

So my question is given a vector as input what will be best way to generate a matrix output for instance:

[2 3 4 1]

输入应产生:

[[1 1 0 0 0 0 0 0],
 [1 1 1 0 0 0 0 0],
 [1 1 1 1 0 0 0 0],
 [1 0 0 0 0 0 0 0]]

我当前的解决方案是遍历所需大小的零矩阵，并使用我上面编写的切片方法将所需元素数设置为1.我有更快的方法吗?

My current solution is iterating over the a zero matrix of required size and setting the required number of elements to 1 using the slicing method I wrote above. Is there a faster way for me to do this?

推荐答案

我以前从未听说过温度计编码"，但是当您意识到它与一次性编码如此相似时，很显然您可以到达那里使用移位操作:

I'd never heard of "thermometer encoding" before, but when you realise how it's so similar to one-hot encoding, it becomes clear you can get there using bit shift ops:

>>> a = np.array([2, 3, 4, 1], dtype=np.uint8)
>>> print(np.fliplr(np.unpackbits((1 << a) - 1).reshape(-1,8)))
[[1 1 0 0 0 0 0 0]
 [1 1 1 0 0 0 0 0]
 [1 1 1 1 0 0 0 0]
 [1 0 0 0 0 0 0 0]]

您可以通过处理8个列块，将其概括为任意大小的整数:

You can generalise the idea to arbitrary size integers by working in 8 column chunks:

a = np.array([2, 13, 4, 0, 1, 17], dtype=np.uint8)
out = np.empty((len(a), 0), dtype=np.uint8)
while a.any():
    block = np.fliplr(np.unpackbits((1 << a) - 1).reshape(-1,8))
    out = np.concatenate([out, block], axis=1)
    a = np.where(a<8, 0, a-8)

print(out)
[[1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0]
 [1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0]]

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