Vector way of populating numpy array

Question

I have some binary string s like 001010. I want to convert it to numpy array a where a[i] = np.array([[1], [0]]) if s[i] == '0' and to np.array([[0], [1]]) otherwise.

So I wrote such code:

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]])

Can it be rewritten to a vectorized form without for-loop in a more numpy way?

My expected output looks like:

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

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

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

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

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

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

Divakar · Accepted Answer · 2019-03-01 11:20:52Z

Approach #1 : Here's one with NumPy char array -

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

Approach #2 : One more as one-liner -

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

Approach #3 : Final one focused entirely on performance -

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

Timings on a string of 100000 chars -

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

yatu · Accepted Answer · 2019-03-01 11:01:39Z

3

A simple way to do so is by creating a list from the string, and then turn this list to a np.array of integers by specifying dtype=int:

s = '001010'

a = np.array(list(s), dtype=int)
# array([0, 0, 1, 0, 1, 0])

And then use np.where in order to select among np.array([[1], [0]]) or np.array([[0], [1]]) according to the values in a:

np.where(a==0, np.array([[1], [0]]), np.array([[0], [1]])).T[:,:,None]
array([[[1],
        [0]],

       [[1],
        [0]],

       [[0],
        [1]],

       [[1],
        [0]],

       [[0],
        [1]],

       [[1],
        [0]]])

edited Mar 1, 2019 at 11:01

answered Mar 1, 2019 at 10:47

yatu

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5 Comments

user8408080 Over a year ago

Actually OP wanted an array of arrays, but still +1 because I think, this is what he actually meant

yatu Over a year ago

Hmm not anymore after seing OPs update. Willl have to change

Kenenbek Arzymatov Over a year ago

sorry, could you look to updated version of the question?

yatu Over a year ago

Updated the answer @RomaKarageorgievich. Let me know if this is what you want. Otherwise please share expected output

yatu Over a year ago

@RomaKarageorgievich updated to match expected output

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