使用适当的填充将整数转换为二进制数组

问题描述

我的整数范围为0..2**m - 1，我想将它们转换为长度为m的二进制numpy数组.例如，说m = 4.现在15 = 1111为二进制，因此输出应为(1,1,1,1). 2 = 10为二进制，因此输出应为(0,0,1,0).如果m是3，则2应该转换为(0,1,0).

I have integers in the range 0..2**m - 1 and I would like to convert them to binary numpy arrays of length m. For example, say m = 4. Now 15 = 1111 in binary and so the output should be (1,1,1,1). 2 = 10 in binary and so the output should be (0,0,1,0). If m were 3 then 2 should be converted to (0,1,0).

我尝试了np.unpackbits(np.uint8(num))，但是没有给出正确长度的数组.例如

I tried np.unpackbits(np.uint8(num)) but that doesn't give an array of the right length. For example,

np.unpackbits(np.uint8(15))
Out[5]: array([0, 0, 0, 0, 1, 1, 1, 1], dtype=uint8)

我想要一种适用于代码中所有m的方法.

I would like a method that worked for whatever m I have in the code.

推荐答案

您应该能够对此向量化，例如

You should be able to vectorize this, something like

>>> d = np.array([1,2,3,4,5])
>>> m = 8
>>> (((d[:,None] & (1 << np.arange(m)))) > 0).astype(int)
array([[1, 0, 0, 0, 0, 0, 0, 0],
       [0, 1, 0, 0, 0, 0, 0, 0],
       [1, 1, 0, 0, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 0, 0, 0],
       [1, 0, 1, 0, 0, 0, 0, 0]])

仅获取适当的比特权重，然后按位进行取和:

which just gets the appropriate bit weights and then takes the bitwise and:

>>> (1 << np.arange(m))
array([  1,   2,   4,   8,  16,  32,  64, 128])
>>> d[:,None] & (1 << np.arange(m))
array([[1, 0, 0, 0, 0, 0, 0, 0],
       [0, 2, 0, 0, 0, 0, 0, 0],
       [1, 2, 0, 0, 0, 0, 0, 0],
       [0, 0, 4, 0, 0, 0, 0, 0],
       [1, 0, 4, 0, 0, 0, 0, 0]])

有很多方法可以将其转换为非零的(> 0)*1，.astype(bool).astype(int)等.我基本上是随机选择一个.

There are lots of ways to convert this to 1s wherever it's non-zero (> 0)*1, .astype(bool).astype(int), etc. I chose one basically at random.

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