do NOT use any LOOPS! Do not use FOR, WHILE, ETC. Use Numpy. **Problem 2**: Write a function `loss(x, q)` to return \(\sum_{{j=0, j \neq q}}^{{K-1}} \max(0, x_j - x_q + 1)\), where \(x\) is a 1-dimensional array of \(K\) floats, \(q\) is an integer with \(0 \leq q \leq K-1\), and \(x_j\) is the \((j + 1)\)-th element of array \(x\). **Sample**: If \(s = np.array([5, 1, 3, 6])\), then `loss(s, 2)` returns 7. (Note: \(s[2] = 3\).)```python1 pass2 def loss(x, q):34 s = np.array([5, 1, 3, 6])5 loss(s, 2)```**Explanation**:- The problem requires the implementation of a function `loss(x, q)` which computes a specific summation involving elements of the array \(x\), excluding the element at index \(q\).- The summation involves calculating \(\max(0, x_j - x_q + 1)\) for each element \(x_j\) in the array except when \(j = q\).- The example given sets \(x = [5, 1, 3, 6]\) and \(q = 2\) (which corresponds to the element 3 in array \(x\)). The calculation returns a result of 7.

ALGORITHM:- 1. Initialise the array. 2. Call the loss function with the corresponding array. 3.…

5K-1 Problem 2. Write a function loss(x, q) to return Eita max(0, x¡ – x, + 1), where x is a 1darray of K floats, q is an integer with 0 < q < K – 1, and x; is the (j + 1)-th element of array x. Sample: if s = np.array( [5, 1, 3, 6]), then loss(s, 2) returns 7. (Not s[2]=3 .) 1 pass 3 def loss(x,q): 4 5 6 s = np.array([5, 1, 3, 6]) 7 loss(s, 2) 8.

5K-1 Problem 2. Write a function loss(x, q) to return Eita max(0, x¡ – x, + 1), where x is a 1darray of K floats, q is an integer with 0 < q < K – 1, and x; is the (j + 1)-th element of array x. Sample: if s = np.array( [5, 1, 3, 6]), then loss(s, 2) returns 7. (Not s[2]=3 .) 1 pass 3 def loss(x,q): 4 5 6 s = np.array([5, 1, 3, 6]) 7 loss(s, 2) 8.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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