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.
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])]
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)
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])
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.
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
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)
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.
If you don't need a signed integer, this is an alternative that uses a slightly different approach, is easy to read and doesn't require an import:
def distance(a, b):
if a > 0 and b > 0:
return max(a, b) - min(a, b)
elif a < 0 and b < 0:
return abs(a - b)
elif a == b:
return 0
return abs(a - 0) + abs(b - 0)
Although abs(x - y) or equivalently abs(y - x) is preferred, if you are curious about a different answer, the following one-liners also work:
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))
Source: Stackoverflow.com