Sorting an array without changing position of negative numbers

import javax.swing.*;  
public class FirstSwingExample {  
public static void main(String[] args) {  
JFrame f=new JFrame();//creating instance of JFrame  
          
JButton b=new JButton("click");//creating instance of JButton  
b.setBounds(130,100,100, 40);//x axis, y axis, width, height  
          
f.add(b);//adding button in JFrame  
          
f.setSize(400,500);//400 width and 500 height  
f.setLayout(null);//using no layout managers  
f.setVisible(true);//making the frame visible  
}  
}  

from PIL import Image
import glob
image_list = []
for filename in glob.glob('yourpath/*.gif'): #assuming gif
    im=Image.open(filename)
    image_list.append(im)

percent_missing = df.isnull().sum() * 100 / len(df)
missing_value_df = pd.DataFrame({'column_name': df.columns,
                                 'percent_missing': percent_missing})

days = 0
week = [‘Monday’, ‘Tuesday’, ‘Wednesday’, ‘Thursday’, ‘Friday’, ‘Saturday’, 3.‘Sunday’]
while day < 7:
print(“Today is” + week[days])
days += 1
                                
                                

ip addr | grep eth0 | grep inet | awk '{print $2}' | awk -F '/' '{print $1}' | awk '{printf "%s:3000", $0}' | clip.exe

>>> ["foo", "bar", "baz"].index("bar")
1

import numpy as np

def pagerank(M, num_iterations=100, d=0.85):
    N = M.shape[1]
    v = np.random.rand(N, 1)
    v = v / np.linalg.norm(v, 1)
    iteration = 0
    while iteration < num_iterations:
        iteration += 1
        v = d * np.matmul(M, v) + (1 - d) / N
    return v

This method gets vowels (‘a’, ‘e’, ‘i’, ‘o’, ‘u’) found in a string.
   
#make a function:
def get_vowels(string):

#return is the keyword which means function have to return value: 
 return [each for each in string if each in 'aeiou']


#assign the words and function will return vowels words.
get_vowels('foobar') # ['o', 'o', 'a']


get_vowels('gym') # []

options = webdriver.ChromeOptions()
options.add_experimental_option("excludeSwitches", ["enable-automation"])
options.add_experimental_option('useAutomationExtension', False)
options.add_argument("--disable-blink-features=AutomationControlled")
driver = webdriver.Chrome(options=options)

int maxSubArraySum(int a[], int size) 
{ 
int max_so_far = 0, max_ending_here = 0; 
for (int i = 0; i < size; i++) 
{ 
	max_ending_here = max_ending_here + a[i]; 
	if (max_ending_here < 0) 
		max_ending_here = 0; 

	/* Do not compare for all elements. Compare only 
		when max_ending_here > 0 */
	else if (max_so_far < max_ending_here) 
		max_so_far = max_ending_here; 
} 
return max_so_far; 
} 

<?php 
function count_num_finger( $n ) 
{ 
	$r = $n % 8; 
	if ($r == 1) 
		return $r; 
	if ($r == 5) 
		return $r; 
	if ($r == 0 or $r == 2) 
		return 2; 
	if ($r == 3 or $r == 7) 
		return 3; 
	if ($r == 4 or $r == 6) 
		return 4; 
}	 

// Driver Code 
$n = 30; 
echo(count_num_finger($n)); 
 
?> 

import java.util.Stack;

/**
 * Java Program to implement a binary search tree. A binary search tree is a
 * sorted binary tree, where value of a node is greater than or equal to its
 * left the child and less than or equal to its right child.
 * 
 * @author WINDOWS 8
 *
 */
public class BST {

    private static class Node {
        private int data;
        private Node left, right;

        public Node(int value) {
            data = value;
            left = right = null;
        }
    }

    private Node root;

    public BST() {
        root = null;
    }

    public Node getRoot() {
        return root;
    }

    /**
     * Java function to check if binary tree is empty or not
     * Time Complexity of this solution is constant O(1) for
     * best, average and worst case. 
     * 
     * @return true if binary search tree is empty
     */
    public boolean isEmpty() {
        return null == root;
    }

    
    /**
     * Java function to return number of nodes in this binary search tree.
     * Time complexity of this method is O(n)
     * @return size of this binary search tree
     */
    public int size() {
        Node current = root;
        int size = 0;
        Stack<Node> stack = new Stack<Node>();
        while (!stack.isEmpty() || current != null) {
            if (current != null) {
                stack.push(current);
                current = current.left;
            } else {
                size++;
                current = stack.pop();
                current = current.right;
            }
        }
        return size;
    }


    /**
     * Java function to clear the binary search tree.
     * Time complexity of this method is O(1)
     */
    public void clear() {
        root = null;
    }

}

import pandas as pd

sheets_dict = pd.read_excel('Book1.xlsx', sheetname=None)

full_table = pd.DataFrame()
for name, sheet in sheets_dict.items():
    sheet['sheet'] = name
    sheet = sheet.rename(columns=lambda x: x.split('\n')[-1])
    full_table = full_table.append(sheet)

full_table.reset_index(inplace=True, drop=True)

print full_table

const arrSum = arr => arr.reduce((a,b) => a + b, 0)
// arrSum([20, 10, 5, 10]) -> 45