矢量方式的填充numpy阵列

Question

我有一些二进制字符串s一样001010.我想将它转换为numpy数组a,其中a[i] = np.array([[1], [0]])if s[i] == '0'和to np.array([[0], [1]])否则.

所以我写了这样的代码:

a = np.empty([len(s), 2, 1])
for i, char in enumerate(s):
    if char == '0':
        a[i] = np.array([[1], [0]])
    elif char == '1':
        a[i] = np.array([[0], [1]])

是否可以在没有for循环的情况下将其重写为矢量化形式？

我的预期输出看起来像:

array([[[1.],
        [0.]],

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

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

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

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

       [[1.],
        [0.]]])

Answer 1

方法#1:这里有一个NumPy char数组 -

sa = np.frombuffer(s,dtype='S1')
out = np.where(sa[:,None,None]=='0',[[1],[0]],[[0],[1]])

方法#2:再一次作为一个班轮 -

((np.frombuffer(s,dtype=np.uint8)[:,None]==[48,49])[...,None]).astype(float)

方法#3:最后一个完全专注于表现 -

a = np.zeros([len(s), 2, 1])
idx = np.frombuffer(s,dtype=np.uint8)-48
a[np.arange(len(idx)),idx] = 1

在一串100000字符上的时间-

In [2]: np.random.seed(0)

In [3]: s = ''.join(map(str,np.random.randint(0,2,(100000)).tolist()))

# @yatu's soln
In [4]: %%timeit
     ...: a = np.array(list(s), dtype=int)
     ...: np.where(a==0, np.array([[1], [0]]), np.array([[0], [1]])).T[:,:,None]
10 loops, best of 3: 36.3 ms per loop

# App#1 from this post    
In [5]: %%timeit
     ...: sa = np.frombuffer(s,dtype='S1')
     ...: out = np.where(sa[:,None,None]=='0',[[1],[0]],[[0],[1]])
100 loops, best of 3: 3.56 ms per loop

# App#2 from this post    
In [6]: %timeit ((np.frombuffer(s,dtype=np.uint8)[:,None]==[48,49])[...,None]).astype(float)
1000 loops, best of 3: 1.81 ms per loop

# App#3 from this post    
In [7]: %%timeit
     ...: a = np.zeros([len(s), 2, 1])
     ...: idx = np.frombuffer(s,dtype=np.uint8)-48
     ...: a[np.arange(len(idx)),idx] = 1
1000 loops, best of 3: 1.81 ms per loop