Manual Function to Sort 3 Numbers, Lowest to Highest

Simpler way of sorting three numbers

if (a > c)
   swap(a, c);

if (a > b)
   swap(a, b);

//Now the smallest element is the 1st one. Just check the 2nd and 3rd

if (b > c)
   swap(b, c);

Note: Swap changes the values of two
variables.

How to manually sort a list of numbers in Python?

For three items, you could use max and min to sort them:

a, b, c = 3, 1, 8

x = min(a, b, c)  # Smallest of the three
z = max(a, b, c)  # Largest of the three
y = (a + b + c) - (x + z)  # Since you have two of the three, you can solve for
                           # the third

print(a, b, c)
print(x, y, z)

If you don't want to use a sorting algorithm but can use lists, you could just pop out the smallest item each time and store it in a new list:

numbers = [1, 8, 9, 6, 2, 3, 1, 4, 5]
output = []

while numbers:
    smallest = min(numbers)
    index = numbers.index(smallest)
    output.append(numbers.pop(index))

print(output)

It's pretty inefficient, but it works.

Sorting three numbers in ascending order without using functions

I think using your approach in the second code, this is the easiest way out:

a = float(input("Enter a: "))
b = float(input("Enter b: "))
c = float(input("Enter c: "))

if a > b:
    a,b = b,a
if a > c:
    a,c = c,a
if b > c:
    b,c = c,b
    
print (a, "<", b, "<", c)

Manually sort a list of 10 integers in python

You've just got some of the order wrong: you need to append to your ordered list each time around

import random
unordered = list(range(10))
ordered = []
i = 0

random.shuffle(unordered)

print unordered
lowest = unordered[0]

while len(unordered) > 0:
    if  unordered[i] < lowest:
        lowest = unordered[i]
    i += 1
    if i == len(unordered):
        ordered.append(lowest)
        unordered.remove(lowest)
        if unordered:
          lowest = unordered[0]
        i = 0

print(ordered)

sorting a list manually on python 3

You need to bring some of your initialization into the loop like:

Code:

unordered = [18, 13, 44, 12, 19, 27, 2, 31]
print(unordered)
ordered = []

while unordered:
    lowest = unordered[0]
    index_of_lowest = 0
    for i, number in enumerate(unordered):
        if number < lowest:
            lowest = number
            index_of_lowest = i
    del unordered[index_of_lowest]
    ordered.append(lowest)
print(ordered)

Results:

[18, 13, 44, 12, 19, 27, 2, 31]
[2, 12, 13, 18, 19, 27, 31, 44]

python: order a list of numbers without built-in sort, min, max function

I guess you are trying to do something like this:

data_list = [-5, -23, 5, 0, 23, -6, 23, 67]
new_list = []

while data_list:
    minimum = data_list[0]  # arbitrary number in list 
    for x in data_list: 
        if x < minimum:
            minimum = x
    new_list.append(minimum)
    data_list.remove(minimum)    

print (new_list)

#Added parenthesis

Sorting 3 numbers without branching

No conditionals. Only a cast to uint. Perfect solution.

int abs (int a) 
{
    int b = a;
    b = (b >> (sizeof(int)*CHAR_BIT-1) & 1);
    return 2 * b * (a) + a; 
}
int max (int a, int b) { return (a + b + abs(a - b)) / 2; }
int min (int a, int b) { return (a + b - abs(a - b)) / 2; }


void sort (int & a, int & b, int & c)
{       
   int maxnum = max(max(a,b), c);
   int minnum = min(min(a,b), c);
   int middlenum = a + b + c - maxnum - minnum;
   a = maxnum;
   b = middlenum;
   c = minnum;
}

Sorting 3 Numbers from Least to Greatest

If you want to stick with the if/else logic, here's a slight modification to your original solution. Note the use of an else if.
I have commented out your earlier lines of code for comparison.

import java.util.Scanner; 

class Main {
  public static void main(String[] args) {

    Scanner reader = new Scanner(System.in); 
    System.out.println("Enter three numbers.");  //Use println instead of print, that way the input begins on the next line

    double x = reader.nextDouble();
    double y = reader.nextDouble(); 
    double z = reader.nextDouble();

    if (x >= y){ //In the three responses below, y is always before x.  
            if (y >= z)
                System.out.print("In order " + z + " "+ y + " " + x);

            else if  (z >= x)
                System.out.print("In order " + y + " "+ x + " " + z);

            else if (x > z)
                System.out.print("In order " + y + " " + z + " " + x);
    }

    if (y > x){// In the three responses below, x is always before y
        if (z >= y)
            System.out.print("In order " + x + " " + y + " "+ z);
        else if (z >= x)
            //System.out.print("In order " + y + " " + x + " " + z); //In this case, z has to be smaller than y.  The order was off
            System.out.print("In order " + x + " " + z + " " + y);
        else if (x > z)
            //System.out.print("In order " + y + " " + z + " " + x);
            System.out.print("In order " + z + " " + x + " " + y); //Y is the biggest.  The order here was off.  
    }

  }
}