Quickly Find the Index in an Array Closest to Some Value

Question

I have an array of values, t, that is always in increasing order (but not always uniformly spaced). I have another single value, x. I need to find the index in t such that

Answer 1

Use searchsorted:

t = np.arange(10,100000)         # Not always uniform, but in increasing order
x = np.random.uniform(10,100000)

print t.searchsorted(x)

Edit:

Ah yes, I see that's what you do in f1. Maybe f3 below is easier to read than f1.

def f3(t, x):
    ind = t.searchsorted(x)
    if ind == len(t):
        return ind - 1 # x > max(t)
    elif ind == 0:
        return 0
    before = ind-1
    if x-t[before] < t[ind]-x:
        ind -= 1
    return ind

Answer 2

np.searchsorted is binary search (split the array in half each time). So you have to implement it in a way it return the last value smaller than x instead of returning zero.

Look at this algorithm (from here):

def binary_search(a, x):
    lo=0
    hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        midval = a[mid]
        if midval < x:
            lo = mid+1
        elif midval > x: 
            hi = mid
        else:
            return mid
    return lo-1 if lo > 0 else 0

just replaced the last line (was return -1). Also changed the arguments.

As the loops are written in Python, it may be slower than the first one... (Not benchmarked)

Answer 3

This seems much quicker (for me, Python 3.2-win32, numpy 1.6.0):

from bisect import bisect_left
def f3(t, x):
    i = bisect_left(t, x)
    if t[i] - x > 0.5:
        i-=1
    return i

---

Output:

[   10    11    12 ..., 99997 99998 99999]
37854.22200356027
37844
37844
37844
37854
37854
37854
f1 0.332725
f2 1.387974
f3 0.085864