Python: how to find value in list smaller than target

Question

For example I have a non-ordered list of values [10, 20, 50, 200, 100, 300, 250, 150]

I have this code which returns the next greater value:

def GetN         


        

Answer 1

nextHighest = lambda seq,x: min([(i-x,i) for i in seq if x<=i] or [(0,None)])[1]
nextLowest  = lambda seq,x: min([(x-i,i) for i in seq if x>=i] or [(0,None)])[1]

Here's how this works: Looking at nextHighest, the argument to min is a list comprehension, that calculates the differences between each value in the list and the input x, but only for those values >= x. Since you want the actual value, then we need the list elements to include both the difference to the value, and the actual value. Tuples are compared value by value, left-to-right, so the tuple for each value i in the sequence becomes (i-x,i) - the min tuple will have the actual value in the [1]'th element.

If the input x value is outside the range of values in seq (or if seq is just empty), then the list comprehension will give us an empty list, which will raise a ValueError in min. In case this happens, we add the or [(0,None)] term inside the argument to min. If the list comprehension is empty, it will evaluate to False, in which case min will instead look at the sequence containing the single tuple (0,None). In the case, the [1]'th element is None, indicating that there were no elements in seq higher than x.

Here are some test cases:

>>> t = [10, 20, 50, 200, 100, 300, 250, 150]
>>> print nextHighest(t,55)
100
>>> print nextLowest(t,55)
50
>>> print nextHighest([],55)
None
>>> print nextLowest([],55)
None
>>> print nextHighest(t,550)
None