Finding X values in numpy array and substituting for random value

Question

Consider an list of numpy arrays with values either -1’s or 1’s allocated in random positions.

a = np.array([1,-1,1,1,-1,1,-1,-1,1,-1])
b = np.array([-1,-1,1,-1,1,1,-1,1,-1,-1])

I need to perform operations on these arrays like sum and point wise multiplication.

For example, after summing 2 arrays i will have a new one with values -2,0 and 2.

c = a + b
c = [ 0 -2 2 0 0 2 -2 0 0 -2]

Now i would like to “normalize” it back to -1’s and 1’s.

For the 2’s and -2’s it is easy:

c[c < 0] = -1

c[c > 0] = 1

The problem is the 0. For them i would like to randomly choose either a -1 or a 1.

The desired output would be like:

c =  [ 1 -1 1 -1 -1 1 -1 1 -1 -1]

In generalized terms my question is how to find all N values equal to x, in an array, then substitute each for a random number.

My question is how to do this in the most “pythonic”, and fastest, way?

Thank’s

Please give an example of input arrays, the intermediate step and the expected output. — line-o
– line-o, Commented Jan 26, 2020 at 0:08

Sicks · Accepted Answer · 2020-01-26 05:16:03Z

Just Posting the final results from the answers i got so far. If anyone in the future has a better solution please share it!

I timed the 3 solutions i found and one i did.

def Norm1(HV):
    HV[HV > 0] = 1
    HV[HV < 0] = -1
    zind = np.where(HV == 0)[0]
    HV[zind] = np.array([np.random.choice([1, -1]) for _ in zind])
    return HV

def norm2(HV):
    if HV == 0: 
        return np.random.choice(np.array([-1,1])) 
    else: 
        return HV / HV * np.sign(HV)                                                                         

Norm2 = np.vectorize(norm2)

def Norm3(HV):
    HV[HV > 0] = 1
    HV[HV < 0] = -1
    mask = HV==0;
    HV[mask] = np.random.choice((-1,1),HV[mask].shape)
    return HV

def generate(size):
    return np.random.binomial(1, 0.5, size=size) * 2 - 1

def Norm4(arr):
    np.floor_divide(arr, 2, out=arr)
    positions = (arr == 0)
    size = sum(positions)
    np.add.at(arr, positions, generate(size)

The timings were:

%%timeit
d = Norm1(c)
203 µs ± 5.9 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

%%timeit
d = Norm2(c)
33.4 ms ± 1.03 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%%timeit
d = Norm3(c)
217 µs ± 11.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%%timeit
d = Norm4(c)
21 ms ± 1.23 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

So as it stands it looks like answer 1 and 3 are the best ones. The difference between them looks minimal, but after trying some more runs the number 1 always come slightly on top.

Thanks for the Helps guys! I will add some references to HD computing in the question as this is a core problem in this application so it will be easier for someone to find it if needed.

mechanical_meat · Accepted Answer · 2020-01-26 00:55:12Z

0

I'm not in any way claiming this is the fastest nor most efficient approach.

c = np.array([ 0, -2, 2, 0, 0, 2, -2, 0, 0, -2])                       

def norm(a): 
    if a == 0: 
        return np.random.choice(np.array([-1,1])) 
    else: 
        return a / a * np.sign(a)                                                                         

v_norm = np.vectorize(norm)
norm_arr = v_norm(c)

Result:

In [64]: norm_arr                                                             
Out[64]: array([ 1, -1,  1,  1, -1,  1, -1,  1, -1, -1])

edited Jan 26, 2020 at 0:55

answered Jan 26, 2020 at 0:39

mechanical_meat

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2 Comments

Sicks Over a year ago

Hello! Thanks for the idea! I will try this one an compare the performance of it against other solutions.

mechanical_meat Over a year ago

Ok, you're welcome. Please post an answer with your timing results when you're done!

M Newville · Accepted Answer · 2020-01-26 01:01:58Z

0

You might use:

>>> c = [0, -2, 2, 0, 0, 2, -2, 0, 0, -2]
>>> c = np.array([0, -2, 2, 0, 0, 2, -2, 0, 0, -2])
>>> zind = np.where(c==0)[0]
>>> c[zind] = np.array([np.random.choice([1, -1]) for _ in zind])
>>> c
array([ 1, -2,  2, -1, -1,  2, -2, -1,  1, -2])

answered Jan 26, 2020 at 1:01

M Newville

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2 Comments

wwii Over a year ago

mask = c==0; a[mask] = np.random.choice((-1,1),a[mask].shape)

Sicks Over a year ago

It looks like this one might be really fast! Thanks! Will try it against other ideas to see how each runs.

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