How to fill numpy array between two specific values

Question

I have to fill np array in row 2 with values from row 0. Array looks like that:

[[3. 4. 5. 6. 7. 8. 9. 8. 7. 6.]
 [0. 0. 1. 0. 0. 1. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]]

Finally it should looks like bellow

[[3. 4. 5. 6. 7. 8. 9. 8. 7. 6.]
 [0. 0. 1. 0. 0. 1. 0. 0. 0. 0.]
 [5. 5. 5. 8. 8. 8. 6. 6. 6. 6.]]

That means:

First row determine points of change. From beginning of row 3 to place where first value "1" in row 2, the row 3 should have value from row 0 where in row two occurs first value "1" and so on..

There is something similar but not exactly what i need. Fill values in numpy array that are between a certain value

Do you know how to do it without iterate over columns?

It would be helpful if you showed the logic, i.e. please include a version of the code that iterates over the columns. — jared
– jared, Commented Feb 21, 2024 at 22:19
If the 1s there are irregularly spaced, then I'm not convinced there is a way to do it without iterating. — Tim Roberts
– Tim Roberts, Commented Feb 21, 2024 at 22:25
Can the last value in the middle row be 1? If yes, what shall the result be? — Kelly Bundy
– Kelly Bundy, Commented Feb 22, 2024 at 1:55

Michael Butscher · Accepted Answer · 2024-02-21 22:52:00Z

2

Try this:

import numpy as np

arr = np.array([[3., 4., 5., 6., 7., 8., 9., 8., 7., 6.],
 [0., 0., 1., 0., 0., 1., 0., 0., 0., 0.],
 [0., 0., 0., 0., 0., 0., 0., 0., 0., 0.]])

lastitem = arr[1,-1]
arr[1, -1] = 1.

vals = arr[0, arr[1].astype(bool)][::-1]
idxs = arr[1].astype(int)[::-1].cumsum() - 1

arr[1, -1] = lastitem
arr[2] = vals[idxs][::-1]

print(arr)

Output:

[[3. 4. 5. 6. 7. 8. 9. 8. 7. 6.]
 [0. 0. 1. 0. 0. 1. 0. 0. 0. 0.]
 [5. 5. 5. 8. 8. 8. 6. 6. 6. 6.]]

Basic idea is to use arr[1] first as boolean array to extract the items of arr[0] which will be really used to vals. vals is reversed.

Then arr[1] is converted to int, reversed, cumulative sum applied and adjusted so that it finally is an array of indexes into vals named idxs.

Last step is to convert the indexes into actual values and set arr[2] accordingly.

edited Feb 21, 2024 at 22:52

answered Feb 21, 2024 at 22:45

Michael Butscher

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Comments

mozway · Accepted Answer · 2024-02-22 04:02:37Z

using indexing

You can get the negative indices at the positions of the 1s (and the last), then backfill using a reverse cumulated minimum:

idx = np.minimum.accumulate(np.where(a[1], np.arange(-a.shape[1], 0), -1)[::-1])[::-1]
# array([-8, -8, -8, -5, -5, -5, -1, -1, -1, -1])
a[2] = a[0, idx]

Step by step intermediates to construct idx (where _ denote the previous command):

np.arange(-a.shape[1], 0)
# array([-10,  -9,  -8,  -7,  -6,  -5,  -4,  -3,  -2,  -1])

np.where(a[1], _, -1)
# array([-1, -1, -8, -1, -1, -5, -1, -1, -1, -1])

np.minimum.accumulate(_[::-1])[::-1]
# array([-8, -8, -8, -5, -5, -5, -1, -1, -1, -1])

Same logic with positive indexing:

idx = np.minimum.accumulate(
    np.where(a[1], np.arange(a.shape[1]), a.shape[1] - 1)[::-1]
)[::-1]

a[2] = a[0, idx]

Intermediates:

np.arange(a.shape[1])
# array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

np.where(a[1], _, a.shape[1]-1)
# array([9, 9, 2, 9, 9, 5, 9, 9, 9, 9])

np.minimum.accumulate(_[::-1])[::-1]
# array([2, 2, 2, 5, 5, 5, 9, 9, 9, 9])

using `numpy.repeat`:

Here we get the indices of the 1s (and the last position), compute the number of items to repeat

idx = np.where(np.r_[a[1, :-1], 1])[0]
# array([2, 5, 9])

a[2] = np.repeat(a[0, idx], np.diff(np.r_[-1, idx]))
# np.repeat([5, 8, 6], [3, 3, 4])

using pandas's `bfill`:

import pandas as pd

a[2] = pd.Series(a[0]).where(np.r_[a[1, :-1]==1, True]).bfill()

Output

array([[3, 4, 5, 6, 7, 8, 9, 8, 7, 6],
       [0, 0, 1, 0, 0, 1, 0, 0, 0, 0],
       [5, 5, 5, 8, 8, 8, 6, 6, 6, 6]])

Solution with np.repeat is ok. It seems to work fine. I will use it.

Onyambu · Accepted Answer · 2024-02-22 01:19:46Z

1

Use assignment:

arr[2] = np.r_[arr[0,arr[1]==1], arr[0,-1]][np.roll(arr[1],1).cumsum(dtype=int)]
arr 

array([[3., 4., 5., 6., 7., 8., 9., 8., 7., 6.],
       [0., 0., 1., 0., 0., 1., 0., 0., 0., 0.],
       [5., 5., 5., 8., 8., 8., 6., 6., 6., 6.]])

answered Feb 22, 2024 at 1:19

Onyambu

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6 Comments

Kelly Bundy Over a year ago

If you delete another space character, it'll even fit PEP 8's 79 characters line limit.

Kelly Bundy Over a year ago

Fails if I change the middle row's last value to 1.

Onyambu Over a year ago

@KellyBundy the last value cannot be 1. ie how will that be a point of change?. Try any other place besides the last value

Onyambu Over a year ago

@KellyBundy also a 1 in the end is alittle ambiguous. ie in the example above, what would be the output if there was a 1 at the end?

Kelly Bundy Over a year ago

Maybe. I've asked them.

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using `numpy.repeat`:

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