Using logic to compare absolute values

Question

Is there a way to make my code simpler/more Pythonic?

Given three int values (a, b, c), return True if one of b or c is "close" (differing from a by at most 1), while the other is "far", differing from both other values by 2 or more. Note: abs(num) computes the absolute value of a number.

close_far(1, 2, 10) → True
close_far(1, 2, 3) → False
close_far(4, 1, 3) → True

def close_far(a, b, c):
    if abs(a - b) <= 1 or abs(a - c) <= 1:
        if abs(c - a) >= 2 <= abs(c - b) or abs(b - a) >= 2 <= abs(b - c):
            return True       
    return False

Answer 1

Some tips:

• You could replace the innermost if with return
• Adding doctests would be nice
• Note that abs(a - b) is the same as abs(b - a). To avoid repeated calculation of abs values, it would be more optimal to cache in local variables

Like this:

def close_far(a, b, c):
    """
    >>> close_far(1, 2, 10)
    True
    >>> close_far(1, 2, 3)
    False
    >>> close_far(4, 1, 3)
    True
    """
    diff_ab = abs(a - b)
    diff_ac = abs(a - c)
    if diff_ab <= 1 or diff_ac <= 1:
        diff_bc = abs(c - b)
        return diff_ac >= 2 <= diff_bc or diff_ab >= 2 <= diff_bc
    return False