How do I find the difference between two values without knowing which is larger?

I was wondering if there was a function built into Python that can determine the distance between two rational numbers but without me telling it which number is larger. e.g.

>>>distance(6,3)
3
>>>distance(3,6)
3

Obviously I could write a simple definition to calculate which is larger and then just do a simple subtraction:

def distance(x, y):
    if x >= y:
        result = x - y
    else:
        result = y - x
    return result
   

but I'd rather not have to call a custom function like this. From my limited experience I've often found Python has a built in function or a module that does exactly what you want and quicker than your code does it. Hopefully someone can tell me there is a built in function that can do this.

abs(x-y) will do exactly what you're looking for:

In [1]: abs(1-2)
Out[1]: 1

In [2]: abs(2-1)
Out[2]: 1

Although abs(x - y) and equivalently abs(y - x) work, the following one-liners also work:

• math.dist((x,), (y,)) (available in Python ≥3.8)

• math.fabs(x - y)

• max(x - y, y - x)

• -min(x - y, y - x)

• max(x, y) - min(x, y)

• (x - y) * math.copysign(1, x - y), or equivalently (d := x - y) * math.copysign(1, d) in Python ≥3.8

• functools.reduce(operator.sub, sorted([x, y], reverse=True))

All of these return the euclidean distance(x, y).

Just use abs(x - y). This'll return the net difference between the two as a positive value, regardless of which value is larger.

If you have an array, you can also use numpy.diff:

import numpy as np
a = [1,5,6,8]
np.diff(a)
Out: array([4, 1, 2])

This does not address the original question, but I thought I would expand on the answer zinturs gave. If you would like to determine the appropriately-signed distance between any two numbers, you could use a custom function like this:

import math
    
def distance(a, b):
    if (a == b):
        return 0
    elif (a < 0) and (b < 0) or (a > 0) and (b > 0):
        if (a < b):
            return (abs(abs(a) - abs(b)))
        else:
            return -(abs(abs(a) - abs(b)))
    else:
        return math.copysign((abs(a) + abs(b)),b)

print(distance(3,-5))  # -8

print(distance(-3,5))  #  8

print(distance(-3,-5)) #  2

print(distance(5,3))   # -2

print(distance(5,5))   #  0

print(distance(-5,3))  #  8

print(distance(5,-3))  # -8

Please share simpler or more pythonic approaches, if you have one.

So simple just use abs((a) - (b)).

will work seamless without any additional care in signs(positive , negative)

def get_distance(p1,p2):
     return abs((p1) - (p2))

get_distance(0,2)
2

get_distance(0,2)
2

get_distance(-2,0)
2

get_distance(2,-1)
3

get_distance(-2,-1)
1

use this function.

its the same convention you wanted. using the simple abs feature of python.

also - sometimes the answers are so simple we miss them, its okay :)

>>> def distance(x,y):
	return abs(x-y)

If you plan to use the signed distance calculation snippet posted by phi (like I did) and your b might have value 0, you probably want to fix the code as described below:

import math

def distance(a, b):
    if (a == b):
        return 0
    elif (a < 0) and (b < 0) or (a > 0) and (b >= 0): # fix: b >= 0 to cover case b == 0
        if (a < b):
            return (abs(abs(a) - abs(b)))
        else:
            return -(abs(abs(a) - abs(b)))
    else:
        return math.copysign((abs(a) + abs(b)),b)

The original snippet does not work correctly regarding sign when a > 0 and b == 0.

abs function is definitely not what you need as it is not calculating the distance. Try abs (-25+15) to see that it's not working. A distance between the numbers is 40 but the output will be 10. Because it's doing the math and then removing "minus" in front. I am using this custom function:

def distance(a, b):
if (a < 0) and (b < 0) or (a > 0) and (b > 0):
return abs( abs(a) - abs(b) )
if (a < 0) and (b > 0) or (a > 0) and (b < 0):
return abs( abs(a) + abs(b) )




print distance(-25, -15)
print distance(25, -15)
print distance(-25, 15)
print distance(25, 15)

You can try: a=[0,1,2,3,4,5,6,7,8,9];

[abs(x[1]-x[0]) for x in zip(a[1:],a[:-1])]