Numpy splitting into array based on division

Question

I want to have something that could split an 1-D array :

np.array([600, 400, 300, 600, 100, 0, 2160])

into a 2-D array based on a value, e.g. 500 such that the resulting array should look like

500 | 100 | 0   | 0   | 0   
400 | 0   | 0   | 0   | 0   
300 | 0   | 0   | 0   | 0   
500 | 100 | 0   | 0   | 0   
100 | 0   | 0   | 0   | 0   
0   | 0   | 0   | 0   | 0   
500 | 500 | 500 | 500 | 160

where we fill from the left with how many 500's there could be and the last one as a reminder.

I am thinking of using np.divmod() but have no idea how to structure the array itself.

Can you expand on whats happening in your calculation of the result? — Tom McLean
– Tom McLean, Commented May 20, 2021 at 11:56
So for first number 600 = 500 * 1 + 100 so I want to have, for the first row, 500 100 for the second number 400 = 500 * 0 + 400 so I want to have, for the second row, only one number 400 and so on — hkbgner
– hkbgner, Commented May 20, 2021 at 11:59

Tls Chris · Accepted Answer · 2021-05-20 14:32:44Z

6

It's a min/max problem not division.

import numpy as np

arr = np.array([600, 400, 300, 600, 100, 0, 2160 ])    
res = np.zeros( (7, 6), dtype = np.int64)
res[:] = arr[:,None]

res -= np.arange( 0, 3000, 500 ) # Subtract successive 500s from arr.

res = np.clip( res, 0, 500 ) # Clip results to lie >= 0 and <= 500

res 

# array([[500, 100,   0,   0,   0,   0],
#        [400,   0,   0,   0,   0,   0],
#        [300,   0,   0,   0,   0,   0],
#        [500, 100,   0,   0,   0,   0],
#        [100,   0,   0,   0,   0,   0],
#        [  0,   0,   0,   0,   0,   0],
#        [500, 500, 500, 500, 160,   0]])

Or as a one liner

np.clip( arr[:,None] - np.arange(0,3000,500), 0, 500 )

Following the Mad Physicist's comment below a more general function

def steps_of( arr, step ):
    temp = arr[:, None] - np.arange( 0, step * ( arr.max() // step + 1), step)
    return np.clip( temp, 0, step )

edited May 20, 2021 at 14:32

answered May 20, 2021 at 12:48

Tls Chris

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1 Comment

Mad Physicist Over a year ago

Part of the challenge is computing 6 and 3000 based on arr rather than hard coding them. Also, OP has 5 columns, not 6. I'd suggest doing something like m = arr.shape[0], n = np.ceil(arr.max() / 500).astype(int), res = np.zeros_like(arr, shape=(m, n)) - 500 * np.arange(n)

obchardon · Accepted Answer · 2021-05-20 14:23:40Z

One other solution, with the same logic than @TlsChris's answer, using np.cumsum() and broadcasting:

# Divisor
n = 500
# Input data in 2D.
x = np.array([[600, 400, 300, 600, 100, 0, 2160]])
# Dimension of the output array (here : (7,5))
d = (len(x[0]),x.max()//n+1) 
# Compute the result
r = np.clip(x.T*np.ones((1,d[1]))-np.cumsum(np.ones(d)*n,1)+n,0,n)

The last line simply compute:

#array([[ 600.,  600.,  600.,  600.,  600.],        array([[1000., 1500., 2000., 2500., 3000.],
#       [ 400.,  400.,  400.,  400.,  400.],               [1000., 1500., 2000., 2500., 3000.],
#       [ 300.,  300.,  300.,  300.,  300.],               [1000., 1500., 2000., 2500., 3000.],
#       [ 600.,  600.,  600.,  600.,  600.],      -        [1000., 1500., 2000., 2500., 3000.],  
#       [ 100.,  100.,  100.,  100.,  100.],               [1000., 1500., 2000., 2500., 3000.],
#       [   0.,    0.,    0.,    0.,    0.],               [1000., 1500., 2000., 2500., 3000.],
#       [2160., 2160., 2160., 2160., 2160.]])              [1000., 1500., 2000., 2500., 3000.]])

And then apply the np.clip() function

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