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How to get index of NumPy multidimensional array in reverse order?
• Last Updated : 05 Apr, 2021

In this article, we will see how to get the index of the NumPy multidimensional array in reverse order.

Approach

• First, we import the NumPy library and initialize the necessary parameters which include the matrix which we need to work upon and other required parameters.
• Now we are going to loop over each row, flip it and find the desired element’s index in each row in reverse order and store it in our list(result) which we had created.
• Now we got all our indexes from the reverse of the rows in the array. We now need to convert the indexing suitable to be read from left to right so, we subtract the length of the row with each index and further reduce it by one to get our array suitable for zero indexing.

Final Index= Total Elements In Rows – Current Index-1

Example:

Input:

[[1,2,3,4,2],

[2,3,4,1,5],

[2,2,4,3,2],

[1,3,4,2,4]]

Output:

[4 0 4 3]

Explanation: In the above example,  we are trying to find the first occurrence indexing of ‘2’ in reverse order and we had 5 elements in each row and after backward indexing we get an array like [0, 4, 0, 1] now using our formula above.

final_list => Total Elements In Rows – Current Index- 1 => 5-0-1 =>4

final_list => Total Elements In Rows – Current Index- 1 => 5-4-1 =>0

final_list => Total Elements In Rows – Current Index- 1 => 5-0-1 =>4

final_list => Total Elements In Rows – Current Index- 1 => 5-1-1 =>3

Example 1:

## Python3

 `#import Modules``import` `numpy as np`` ` `# initialize parameters``x ``=` `np.array([[``1``, ``2``, ``3``, ``4``, ``2``],``              ``[``2``, ``3``, ``4``, ``1``, ``5``],``              ``[``2``, ``2``, ``4``, ``3``, ``2``],``              ``[``1``, ``3``, ``4``, ``2``, ``4``]])  ``num_cols ``=` `len``(x[``0``])  ``result ``=` `[]  `` ` `# loop over each row``for` `row ``in` `x:  ``    ``row ``=` `np.flip(row)  ``    ``index ``=` `np.where(row ``=``=` `2``)  ``    ``result.append(index[``0``][``0``])  `` ` `# get the final indexes``# Store the result as of the initial arrays``final_list ``=` `num_cols``-``np.array(result)``-``1`` ` `# print``print``(final_list)  `

Output:

[4 0 4 3]

Example 2:

The above method could work for string as well. In the example below we are trying to find the first occurrence indexing of ‘Sam’ in reverse order.

## Python3

 `#import Modules``import` `numpy as np`` ` `# initialize parameters``x ``=` `np.array([[``"Sam"``, ``"John"``, ``"Lilly"``],``              ``[``"Sam"``, ``"Sam"``, ``"Kate"``],``              ``[``"Jack"``, ``"John"``, ``"Sam"``],``              ``[``"Sam"``, ``"Jack"``, ``"Rose"``]]) ``num_cols ``=` `len``(x[``0``]) ``result ``=` `[] `` ` `# loop over each row``for` `row ``in` `x: ``    ``row ``=` `np.flip(row)  ``    ``index ``=` `np.where(row ``=``=` `"Sam"``)  ``    ``result.append(index[``0``][``0``]) `` ` `# get the final indexes``# Store the result as of the initial arrays``final_list ``=` `num_cols``-``np.array(result)``-``1`` ` `# print``print``(final_list)  `

Output:

[0 1 2 ]

Example 3:

For Boolean data, we have the same approach, but since it has only 0 or 1 as the value we can use argmax() which will find the index of the highest value (for each row with axis=1). Since True is equivalent to 1 and False to 0, it’ll record the index of the first True value.

## Python3

 `# import Modules``import` `numpy as np`` ` `# initialize parameters``a ``=` `np.array([[``True``, ``False``, ``True``, ``True``],``              ``[``False``, ``False``, ``True``, ``False``],``              ``[``False``, ``True``, ``True``, ``True``],``              ``[``True``, ``False``, ``False``, ``True``]])`` ` `reversed_array ``=` `a[:, ::``-``1``]``max_val ``=` `np.argmax(reversed_array, axis``=``1``)``num_rows ``=` `a.shape[``1``]  ``final_list ``=` `num_rows``-``1``-``max_val  `` ` `print``(final_list)  `

Output:

[3 2 3 3]

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