shift numpy array column and row?

Question

I have numpy array like this, where I have one column and one row of ZEROS :

 ([[0. , 2.8, 3.5, 0. , 2.5, 1. , 0.8],
   [0. , 0. , 0. , 0. , 0. , 0. , 0. ],
   [3.5, 2.5, 0. , 0. , 2.8, 1.3, 1.1],
   [3.6, 3.8, 3.3, 0. , 2.5, 0.6, 0.4],
   [2.5, 1.5, 2.8, 0. , 0. , 3.1, 1.9],
   [1. , 0.8, 1.3, 0. , 3.1, 0. , 2.8],
   [0.8, 1.6, 1.1, 0. , 1.9, 2.8, 0. ]])

I want to shift the zero-row to the top and the zero-column either to the left or remove it :

 ([[0. , 0. , 0. , 0. , 0. , 0. ]
   [0. , 2.8, 3.5, 2.5, 1. , 0.8],
   [3.5, 2.5, 0. , 2.8, 1.3, 1.1],
   [3.6, 3.8, 3.3, 2.5, 0.6, 0.4],
   [2.5, 1.5, 2.8, 0. , 3.1, 1.9],
   [1. , 0.8, 1.3, 3.1, 0. , 2.8],
   [0.8, 1.6, 1.1, 1.9, 2.8, 0. ]])

any quick and easy way to do it ? BTW I know the col&row-number, so i doesnt have to search for it.

Do you want the zero row/column shifted to the sides or swapped with whatever is currently on the sides? — miradulo
– miradulo, Commented Jun 2, 2018 at 3:23

Zev · Accepted Answer · 2018-06-02 04:50:20Z

2

Delete both the column and row and add back in a row of zeros.

This works for your example:

import numpy as np
a =  np.array([[0. , 2.8, 3.5, 0. , 2.5, 1. , 0.8],
   [0. , 0. , 0. , 0. , 0. , 0. , 0. ],
   [3.5, 2.5, 0. , 0. , 2.8, 1.3, 1.1],
   [3.6, 3.8, 3.3, 0. , 2.5, 0.6, 0.4],
   [2.5, 1.5, 2.8, 0. , 0. , 3.1, 1.9],
   [1. , 0.8, 1.3, 0. , 3.1, 0. , 2.8],
   [0.8, 1.6, 1.1, 0. , 1.9, 2.8, 0. ]])

def remove_column_of_zeros_and_shift_row(a, row, col):
    without_row = np.delete(a, row, axis=0)
    without_row_and_col = np.delete(without_row, col, axis=1)
    z = np.zeros((1, len(without_row_and_col[0])))
    without_col_shifted_row = np.append(z, without_row_and_col, axis=0)
    return without_col_shifted_row

my_result = remove_column_of_zeros_and_shift_row(a, 1, 3)

answered Jun 2, 2018 at 4:50

Zev

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Comments

Paul Panzer · Accepted Answer · 2018-06-02 10:18:01Z

2

Use advanced indexing together with np.ix_:

>>> import numpy as np
>>> 
>>> X = np.array( ([[0. , 2.8, 3.5, 0. , 2.5, 1. , 0.8],
...    [0. , 0. , 0. , 0. , 0. , 0. , 0. ],
...    [3.5, 2.5, 0. , 0. , 2.8, 1.3, 1.1],
...    [3.6, 3.8, 3.3, 0. , 2.5, 0.6, 0.4],
...    [2.5, 1.5, 2.8, 0. , 0. , 3.1, 1.9],
...    [1. , 0.8, 1.3, 0. , 3.1, 0. , 2.8],
...    [0.8, 1.6, 1.1, 0. , 1.9, 2.8, 0. ]]))
>>> 
>>> row = 1; col = 3
>>> h, w = X.shape
>>> i = np.r_[row, :row, row+1:h]
>>> j = np.r_[:col, col+1:w]
>>> X[np.ix_(i, j)]
array([[0. , 0. , 0. , 0. , 0. , 0. ],
       [0. , 2.8, 3.5, 2.5, 1. , 0.8],
       [3.5, 2.5, 0. , 2.8, 1.3, 1.1],
       [3.6, 3.8, 3.3, 2.5, 0.6, 0.4],
       [2.5, 1.5, 2.8, 0. , 3.1, 1.9],
       [1. , 0.8, 1.3, 3.1, 0. , 2.8],
       [0.8, 1.6, 1.1, 1.9, 2.8, 0. ]])

answered Jun 2, 2018 at 10:18

Paul Panzer

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2 Comments

Zev Over a year ago

np.ix_ docs say "This function takes N 1-D sequences and returns N outputs with N dimensions each, such that the shape is 1 in all but one dimension and the dimension with the non-unit shape value cycles through all N dimensions." Can you clarify that at all? This answer is probably more clever but I'd need some comments to follow it if I ran into it in the wild.

Paul Panzer Over a year ago

@Zev what this does is setting the two 1D arrays up for broadcasting. Think of it as a more economical version of meshgrid.

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