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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Concept explainers

Hash Table

A hash table also called a hash map, is a data structure that maps key-value pairs. The hash table can be used in complex systems to provide fast access and compact storage. A hash table is a structure that stores data in an abstract data type kind of associative array. It keeps data in an array format, with a distinct index value for every data value. If the index of the required data is known, then the data can be accessed very easily.

Question

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: \(s[2] = 3\).)

```python
1 pass
2 def loss(x, q):
3
4 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.

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