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I have a matrix as 2d-np.array and I would like to remove all rows that contain an element x in a specific column. My goal is to return a matrix without these rows, so it should be smaller. My function looks like this:

def delete_rows(matrix, x, col):
    for i in range(matrix.shape[0]-1):
        if(matrix[i,col] == x):
            np.delete(matrix, i, axis = 0)
    return matrix

Sadly in my test the shape of the matrix stayed the same after removing rows. I think the deleted rows were substituted by rows with 0s. Any advice on how I can achieve my goal?

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  • this is solvable using boolean indexing quite simply arr[~(arr == val).any(1),:] which is equal to checking for the value in the matrix, and then checking which rows have atleast one instance of the value using any, then inversing the Trues and False to finally filter. Check my answer for details. Commented May 27, 2022 at 12:27

4 Answers 4

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EDIT: added condition for the specific column to check

You don't need to use any apply methods for this. It can be solved with basic boolean indexing as follows -

arr[~(arr[:,col] == val),:]
  1. arr[:,col] selects the specific column from array
  2. arr[:,col] == val checks for the value and returns True where it exists, else False
  3. ~(arr[:,col] == val) inverses the True and False
  4. arr[~(arr[:,col] == val),:] keeps only the rows which have the boolean index as True and discards all False

Example solution

arr = np.array([[12, 10, 12,  0,  9,  4, 12, 11],
                [ 3, 10, 14,  5,  4,  3,  6,  6],
                [12, 10,  1,  0,  5,  7,  5, 10],
                [12,  8, 14, 14, 12,  3, 14, 10],
                [ 9, 14,  3,  8,  1, 10,  9,  6],
                [10,  3, 11,  3, 12, 13, 11, 10],
                [ 0,  6,  8,  8,  5,  5,  1, 10], #<- this to remove
                [13,  6,  1, 10,  7, 10, 10, 13],
                [ 3,  3,  8, 10, 13,  0,  0, 10], #<- this to remove
                [ 6,  2, 13,  5,  8,  2,  8, 10]])
#                         ^
#                 this column to check


#boolean indexing approach
val, col = 8,2 #value to check is 8 and column to check is 2

out = arr[~(arr[:,col] == val),:] #<-----
out

array([[12, 10, 12,  0,  9,  4, 12, 11],
       [ 3, 10, 14,  5,  4,  3,  6,  6],
       [12, 10,  1,  0,  5,  7,  5, 10],
       [12,  8, 14, 14, 12,  3, 14, 10],
       [ 9, 14,  3,  8,  1, 10,  9,  6],
       [10,  3, 11,  3, 12, 13, 11, 10],
       [13,  6,  1, 10,  7, 10, 10, 13],
       [ 6,  2, 13,  5,  8,  2,  8, 10]])

If you want to check for the value in all columns then try this -

arr[~(arr == val).any(1),:]

And if you want to keep ONLY rows with the value instead, just remove ~ from the condition.

arr[(arr[:,col] == val),:]

If you want to remove the column as well, using np.delete -

np.delete(arr[~(arr[:,col] == val),], col, axis=1)

Note: You cannot remove both rows and columns at once using np.delete so if you plan to use it, you will need to do np.delete two times once for axis = 0 (rows) and once for axis = 1 (columns)

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8 Comments

very nice. Just out of curiosity: How would I change this to return a matrix that ONLY consists of the rows which contain the element? So here e.g. all rows containing 8
Just remove the ~ from the condition. That inverses the condition to keep only rows WITH the column. Also, I have updated the code to check for specific column.
I have added few more variations that might interest you in my answer. Like checking all columns at once instead of single column and another one to keep rows that contain the value instead of removing them.
thank you so much. Is there any way I can remove a whole column using this logic? I am supposed to remove the column I was using for searching after deleting the rows. But am not getting this np.delete to work. It substitutes all values with 0 in the column instead. I think it has something to do with reshaping the matrix size. Do you have any idea?
thats simple too. i have added that in the last part. its basically np.delete(arr[~(arr[:,col] == val),], col, axis=1)
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Assuming you have an array like this:

array([[12,  5,  0,  3, 11,  3,  7,  9,  3,  5],
       [ 2,  4,  7,  6,  8,  8, 12, 10,  1,  6],
       [ 7,  7, 14,  8,  1,  5,  9, 13,  8,  9],
       [ 4,  3,  0,  3,  5, 14,  0,  2,  3,  8],
       [ 1,  3, 13,  3,  3, 14,  7,  0,  1,  9],
       [ 9,  0, 10,  4,  7,  3, 14, 11,  2,  7],
       [12,  2,  0,  0,  4,  5,  5,  6,  8,  4],
       [ 1,  4,  9, 10, 10,  8,  1,  1,  7,  9],
       [ 9,  3,  6,  7, 11, 14,  2, 11,  0, 14],
       [ 3,  5, 12,  9, 10,  4, 11,  4,  6,  4]])

You can remove all rows containing a 3 like this:

row_mask = np.apply_along_axis(np.any, 1, arr == 3)
arr = arr[~row_mask]

Your new array looks like this


array([[ 2,  4,  7,  6,  8,  8, 12, 10,  1,  6],
       [ 7,  7, 14,  8,  1,  5,  9, 13,  8,  9],
       [12,  2,  0,  0,  4,  5,  5,  6,  8,  4],
       [ 1,  4,  9, 10, 10,  8,  1,  1,  7,  9]])
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2 Comments

There is still a 3 in the last row :D
@NiklasKlein Hahaha there was, because I mixed up the axis. The 2nd parameter of apply_along_axis is the axis. If you plug in a 0 you do it with columns and 1 with rows. I corrected it.
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This can be simply done in one line:

import numpy as np

def delete_rows(matrix, x, col):
    return matrix[matrix[:,col]!=x,:]

For example, if we want to remove all the rows that contain 5 in the second column from matrix A:

>>> A = np.array([[1,2,3], [4,5,6], [7,8,9]])
>>> print(delete_rows(A, 5, 1))
[[1 2 3]
 [7 8 9]]
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1 Comment

I believe the op was interested in removing all the rows containing something and not just in a specific column.
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import pandas as pd
import numpy as np

matrix = np.array([[1, 2, 3],
                   [4, 5, 6],
                   [7, 8, 9]])
   
def f(matrix,num,col):
    rowsToDelete = matrix[matrix[:,col] != num, :]
    return rowsToDelete

# Remove rows where column 1 equals 5
res = f(matrix, 5, 1)  
print(res)  
'''
[[1 2 3]
 [7 8 9]]
'''
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