CODEWITHSUNDEEP
pythonarraysnumpyreshapenumba
mathguy
mathguy

Reputation: 1518

Reshaping this array efficiently

Given this array X:

[1 2 3 2 3 1 4 5 7 1]

and the row length array R:

[3 2 5]

that represents the lengths of each row after the transformation.

I am looking for a computationally efficient function to reshape X into this array Y:

[[ 1.  2.  3. nan nan]
 [ 2.  3. nan nan nan]
 [ 1.  4.  5.  7.  1.]]

These are merely simplified version of the actual arrays I am working on. My actual arrays are more like this:

R = np.random.randint(5, size = 21000)+1
X = np.random.randint(10, size = np.sum(R))

I have already got down a function to generate the reshaped array, but the function runs way too slow. I have tried some Numba features to speed it up, but they generate to many error messages to handle. My super slow function:

def func1(given_array, row_length):

    corresponding_indices = np.cumsum(row_length)



    desired_result = np.full([len(row_length),np.amax(row_length)], np.nan)
    desired_result[0,:row_length[0]] = given_array[:corresponding_indices[0]]

    for i in range(1,len(row_length)):
        desired_result[i,:row_length[i]] = given_array[corresponding_indices[i-1]:corresponding_indices[i]]

    return desired_result

This function takes a daunting 34ms per loop when the sizes of the input_arrays haven't exceeded 100K yet. I am looking for a function that does the same thing with the same sizes but in under 10ms per loop

Thank you in advance

Upvotes: 1

Views: 55

Answers (1)

Divakar
Divakar

Reputation: 221524

Here's a vectorized one leveraging broadcasting -

def func2(given_array, row_length):
    given_array = np.asarray(given_array)
    row_length = np.asarray(row_length)
    mask = row_length[:,None] > np.arange(row_length.max())
    out = np.full(mask.shape, np.nan)
    out[mask] = given_array
    return out

Sample run -

In [305]: a = [1, 2, 3, 2, 3, 1, 4, 5, 7, 1]
     ...: b = [3, 2, 5]

In [306]: func2(a,b)
Out[306]: 
array([[ 1.,  2.,  3., nan, nan],
       [ 2.,  3., nan, nan, nan],
       [ 1.,  4.,  5.,  7.,  1.]])

Timings and verification on large dataset -

In [323]: np.random.seed(0)
     ...: R = np.random.randint(5, size = 21000)+1
     ...: X = np.random.randint(10, size = np.sum(R))

In [324]: %timeit func1(X,R)
100 loops, best of 3: 17.5 ms per loop

In [325]: %timeit func2(X,R)
1000 loops, best of 3: 657 µs per loop

In [332]: o1 = func1(X,R)

In [333]: o2 = func2(X,R)

In [334]: np.allclose(np.where(np.isnan(o1),0,o1),np.where(np.isnan(o2),0,o2))
Out[334]: True

Upvotes: 1

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