Position element in ordered list in logarithmic time Python

Question

I have a sorted list and i need to position an element within that list such that the previous element is <= and the next element in the list is > (the list is a list of floating point numbers)

I shall need to return the position of the element that is <= i.e. the previous element

how can i implement this in logarithmic time. i thought of using a method similar to binary seacrh but couldn't get it to work

Any help would be appreciated

P.S. an example is:if the list is

testlist=[0.0, 0.25, 0.5, 0.75, 1.0]

and i run the function for 0.27 the function will return 1 (the location of 0.25) and if i run it for 0.5 it'll return 2

If this is homework, you should tag it as such to avoid answers like the one that suggest using the `bisect` module. — K Mehta, Apr 01 '12 at 22:26

score 3 · Answer 1 · answered Apr 01 '12 at 15:52

3

There is a dedicated module for binary search: bisect

import bisect

testlist=[0.0, 0.25, 0.5, 0.75, 1.0]
print bisect.bisect(testlist, .27) - 1
## 1

answered Apr 01 '12 at 15:52

georg

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Thanks, the answer worked, now I can move on with the rest of my code – Marvin Android Apr 02 '12 at 17:46

Position element in ordered list in logarithmic time Python

1 Answers1