function returning the smallest difference between elements of two lists - python

Question

Let's say a and b are recorded at a constant rate of 2s and 3s respectively:

>>> a
0, 2, 4, 6, 8, 10, 12
>>> b
0, 3, 6, 9, 12

I'd like to write a function in python that returns

the smallest positive difference (i.e., bigger than zero) of a-b, and
the number of instances that takes b to reach the same value of a.

So in the previous example,

the smallest difference of a-b is 1, that is, when a==4 and b==3 (or a==10 and b==9)
it takes 3 instances of b to reach the same value of a (i.e., 0, 3, 6).

Ideally I'd like to use the function in this manner:

a = 2
b = 3
>>> my_fun(a,b)
>>> [1, 3] #1-smallest difference, 3-number of instances

@tcaswell: I thought of looping through every element of a and b and calculate the difference, but I'm sure there must be a more efficient way of doing this... — HappyPy, Sep 12 '13 at 18:17
If the values are random, there really isn't a better way. Are they monotonically increasing? Random? — Max, Sep 12 '13 at 18:24
@Max: They are not random, they are increasing in steps defined by a and b. If a is 2.34 then [0, 2.34, 4.68, etc] — HappyPy, Sep 12 '13 at 18:26
if a/b is irrational, you will never have the same value in both. — tacaswell, Sep 12 '13 at 18:34
@tcaswell: That's a good point. You're right a/b must not be irrational. — HappyPy, Sep 12 '13 at 18:45

score 2 · Answer 1 · edited May 23 '17 at 12:28

2

Using the lcm() function from this answer:

def my_fun(a, b):
    n = lcm(a, b)
    arange = range(0, n+1, n // a if n > a else a)
    brange = range(0, n+1, n // b if n > b else b)
    m = min(x - y for x in arange for y in brange if x > y)
    return [m, n//b + 1]

This is assuming your values a and b are integers, if you need to do this for floats just multiply by some power of 10 to make them ints. So for example with a=2.34 and b=3.73 you would multiply by 100, run my_fun(234, 373), and then divide the first value of the result by 100 (the second value can be used as is).

Here is the lcm() function and gcd() function that it depends on:

def gcd(a, b):
    """Return greatest common divisor using Euclid's Algorithm."""
    while b:      
        a, b = b, a % b
    return a

def lcm(a, b):
    """Return lowest common multiple."""
    return a * b // gcd(a, b)

edited May 23 '17 at 12:28

Community

1
1

answered Sep 12 '13 at 18:25

Andrew Clark

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.: Thanks! However, if I run my_fun(4,4) I get [1,2]. In this case the smallest difference is actually 4, but the function is returning 1... – HappyPy Sep 12 '13 at 18:49
In fact, the function is always returning 1 as the smallest difference, no matter which values a and b have. – HappyPy Sep 12 '13 at 18:50
@HappyPy check my solution, if you have more test cases – Roman Pekar Sep 12 '13 at 19:18
See my edit, I think the issue with the differences should be resolved. – Andrew Clark Sep 12 '13 at 21:45

Roman Pekar · Accepted Answer · 2013-09-13T17:55:30.343

2

def worker(a, b):
    i, j, k = a, b, a
    while a != b:
        if a < b:
            a += i
        else:
            n = a - b
            k = n if not k else min(k, n)
            b += j

    return k, b / j + 1

>>> worker(4, 4)
(4, 2)
>>> worker(2, 3)
(1, 3)

edited Sep 13 '13 at 17:55

answered Sep 12 '13 at 18:28

Roman Pekar

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`<>` has been deprecated, use `!=`. That's what happens when you use lot of sql ;) – Ashwini Chaudhary Sep 13 '13 at 17:45

function returning the smallest difference between elements of two lists - python

2 Answers2