CODEWITHSUNDEEP
pythonlist
Greggy D
Greggy D

Reputation: 121

Rounding integers

I have 15 numbers, 

[1, 5, 10, 20, 30, 50, 70, 100, 150, 200, 500, 1000, 2000, 5000, 10000]

I have people entering the quantity and what I want is it to round to the lowest number. hence if someone enters 36 it will round to 30.

Upvotes: 0

Views: 197

Answers (3)

Akavall
Akavall

Reputation: 86188

Here is a recursive solution. It should be O(log n); it relies on the fact that the list is sorted.

L = [1, 5, 10, 20, 30, 50, 70, 100, 150, 200, 500, 1000, 2000, 5000, 10000]

def roundit(x,n):
    if len(x) == 1:
        return x[0]
    elif x[len(x)/2] > n:
        return roundit(x[0:len(x)/2],n)
    else:
        return roundit(x[len(x)/2 :],n)

Result:

>>> roundit(L,36)
30
>>> roundit(L,77)
70
>>> roundit(L,150)
150

Upvotes: 2

eumiro
eumiro

Reputation: 212885

bisect will do it in O(log N):

>>> import bisect
>>> L = [1, 5, 10, 20, 30, 50, 70, 100, 150, 200, 500, 1000, 2000, 5000, 10000]
>>> L[bisect.bisect(L, 36) - 1]
30

or with pure python and O(N):

>>> L = [1, 5, 10, 20, 30, 50, 70, 100, 150, 200, 500, 1000, 2000, 5000, 10000]
>>> next(elem for elem in reversed(L) if elem <= 36)
30

assumed the L list is sorted. Otherwise L.sort() it before.

Upvotes: 7

wim
wim

Reputation: 362756

With pure python:

>>> numbers = [1, 5, 10, 20, 30, 50, 70, 100, 150, 200, 500, 1000, 2000, 5000, 10000]
>>> x = 36
>>> max(n for n in numbers if n <= x)
30

note: Does not rely on numbers list being sorted.

Upvotes: 7

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