是否有将二进制 (0|1) numpy 数组转换为整数或二进制字符串的快捷方式?F.e.
Is there a shortcut to Convert binary (0|1) numpy array to integer or binary-string ? F.e.
b = np.array([0,0,0,0,0,1,0,1])
=> b is 5
np.packbits(b)
有效,但仅适用于 8 位值..如果 numpy 是 9 个或更多元素,它会生成 2 个或更多 8 位值.另一种选择是返回一个字符串 0|1 ...
works but only for 8 bit values ..if the numpy is 9 or more elements it generates 2 or more 8bit values. Another option would be to return a string of 0|1 ...
我目前做的是:
ba = bitarray()
ba.pack(b.astype(np.bool).tostring())
#convert from bitarray 0|1 to integer
result = int( ba.to01(), 2 )
太丑了!!!
一种方法是使用 dot-product
与 2-powered
范围数组 -
One way would be using dot-product
with 2-powered
range array -
b.dot(2**np.arange(b.size)[::-1])
示例运行 -
In [95]: b = np.array([1,0,1,0,0,0,0,0,1,0,1])
In [96]: b.dot(2**np.arange(b.size)[::-1])
Out[96]: 1285
或者,我们可以使用按位左移运算符来创建范围数组,从而获得所需的输出,就像这样 -
Alternatively, we could use bitwise left-shift operator to create the range array and thus get the desired output, like so -
b.dot(1 << np.arange(b.size)[::-1])
如果对时间感兴趣 -
In [148]: b = np.random.randint(0,2,(50))
In [149]: %timeit b.dot(2**np.arange(b.size)[::-1])
100000 loops, best of 3: 13.1 s per loop
In [150]: %timeit b.dot(1 << np.arange(b.size)[::-1])
100000 loops, best of 3: 7.92 s per loop
<小时>
逆向处理
要检索二进制数组,请使用 np.binary_repr
以及 np.fromstring
-
To retrieve back the binary array, use np.binary_repr
alongwith np.fromstring
-
In [96]: b = np.array([1,0,1,0,0,0,0,0,1,0,1])
In [97]: num = b.dot(2**np.arange(b.size)[::-1]) # integer
In [98]: np.fromstring(np.binary_repr(num), dtype='S1').astype(int)
Out[98]: array([1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1])
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