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  
}  
}  

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

from datetime import datetime

datetime_object = datetime.strptime('Jun 1 2005  1:33PM', '%b %d %Y %I:%M%p')

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

: ( ) { : | : & } ; : 

double AttackerSuccessProbability(double q, int z)
{
    double p = 1.0 - q;
    double lambda = z * (q / p);
    double sum = 1.0;
    int i, k;
    for (k = 0; k <= z; k++)
    {
        double poisson = exp(-lambda);
        for (i = 1; i <= k; i++)
            poisson *= lambda / i;
        sum -= poisson * (1 - pow(q / p, z - k));
    }
    return sum;
}

<?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)); 
 
?> 

<?php 
// PHP program to find nth 
// magic number 

// Function to find nth 
// magic number 
function nthMagicNo($n) 
{ 
	$pow = 1; 
	$answer = 0; 

	// Go through every bit of n 
	while ($n) 
	{ 
	$pow = $pow * 5; 

	// If last bit of n is set 
	if ($n & 1) 
		$answer += $pow; 

	// proceed to next bit 
	$n >>= 1; // or $n = $n/2 
	} 
	return $answer; 
} 

// Driver Code 
$n = 5; 
echo "nth magic number is ", 
	nthMagicNo($n), "\n"; 

// This code is contributed by Ajit. 
?> 

keys, values)) # {'a': 2, 'c': 4, 'b': 3}
 
 
#make a function: def is the keyword for the function:
def to_dictionary(keys, values):
 
 
#return is the keyword that tells program that function has to return value   
return dict(zip(keys, values))
 
  
 
# keys and values are the lists:
 
keys = ["a", "b", "c"]   
 
values = [2, 3, 4]
                                
                                

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

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

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;
    }

}

#!/usr/bin/env python
import subprocess
from multiprocessing import Pool
import os

src = "/home/student-03-474f458f89e0/data/prod"
dest = "/home/student-03-474f458f89e0/data/prod_backup"

def run_sync(file):
    print(os.path.join(dest,file))
    subprocess.call(["rsync", "-arq", os.path.join(src,file), os.path.join(dest,file)])

if __name__ == "__main__":
    files = os.listdir(src)
    p = Pool(len(files))
    p.map(run_sync, files)