Minimal distance between elements in a vector

Question

I need the minimal distance between elements of an array.

I did:

numpy.min(numpy.ediff1d(numpy.sort(x)))

Is there a better / more efficient / more elegant / faster way of doing this?

Do you want more elegant or more efficient? – Robert Harvey Apr 11 '13 at 16:24 — Robert Harvey, Apr 11 '13 at 16:24
;-) Well, I think in my case I need the fastest – Charles Brunet Apr 11 '13 at 16:25 — Charles Brunet, Apr 11 '13 at 16:25

score 4 · Accepted Answer · answered Apr 11 '13 at 17:18

If you are after sheer speed, here are some timings:

In [13]: a = np.random.rand(1000)

In [14]: %timeit np.sort(a)
10000 loops, best of 3: 31.9 us per loop

In [15]: %timeit np.ediff1d(a)
100000 loops, best of 3: 15.2 us per loop

In [16]: %timeit np.diff(a)
100000 loops, best of 3: 7.76 us per loop

In [17]: %timeit np.min(a)
100000 loops, best of 3: 3.19 us per loop

In [18]: %timeit np.unique(a)
10000 loops, best of 3: 53.8 us per loop

The timing of unique was in hopes that it would be comparably fast to sort, and you could break out early without the calls to diff and min if the length of the unique array was shorter than the array itself (as that would mean your answer was 0). But the overhead of unique is more than any gain to be made.

So it seems the only potential improvement I can offer is replacing ediff1d with diff:

In [19]: %timeit np.min(np.diff(np.sort(a)))
10000 loops, best of 3: 47.7 us per loop

In [20]: %timeit np.min(np.ediff1d(np.sort(a)))
10000 loops, best of 3: 57.1 us per loop

Interresting. I was expecting `ediff1d` to be faster than `diff`, since it is for 1d arrays, but apparently `diff` is faster. — Charles Brunet, Apr 11 '13 at 17:23

Daniel Li · Answer 2 · 2013-04-11T16:51:36.597

2

Your current approach is definitely optimal. By sorting first, you're reducing the space in between each element and ediff1d will return a difference array. Here's a suggestion:

Since we know that the difference must be positive since we have an ascending-order sort, we can implement ediff1d manually and include a break where the difference is zero. That way, if you have the sorted array x:

[1, 1, 2, 3, 4, 5, 6, 7, ... , n]

Rather than going through n elements, your ediff1d function breaks early and covers only the first two elements, returning [0]. This also reduces the size of the difference array, reducing the amount of iterations required by your min call.

Here is an example without the use of numpy:

x = [1, 12, 3, 8, 4, 1, 4, 9, 1, 29, 210, 313, 12]

def ediff1d_custom(x):
    darr = []

    for i in xrange(len(x)):
        if i != len(x) - 1:
            diff = x[i + 1] - x[i]
            darr.append(diff)

            if diff == 0:
                break

    return darr

print min(ediff1d_custom(sorted(x))) # prints 0

edited Apr 11 '13 at 16:51

answered Apr 11 '13 at 16:26

Daniel Li

14,976
6
43
60

Interesting, but overkill and not exactly what I need. My need is for values inside a single 1d array. – Charles Brunet Apr 11 '13 at 16:28
2

Unless the expected min actually is zero, this will be significantly slower than OP's existing solution. – askewchan Apr 11 '13 at 17:12
How is that? Is it because numpy is precompiled? – Daniel Li Apr 11 '13 at 17:15
@DanielLi Pretty much, and the arrays are (usually) in contiguous memory instead of dynamic lists. See http://stackoverflow.com/q/993984/1730674 – askewchan Apr 11 '13 at 17:40

score 0 · Answer 3 · answered Apr 11 '13 at 17:08

0

try:
    min(x[i+1]-x[i] for i in xrange(0, len(x)-1))
except ValueError:
    print 'Array contains less than two values.'

answered Apr 11 '13 at 17:08

Vladimir Chub

461
6
19

`x` must still be sorted before you can do this, since this is basically just non-numpy `min(ediff1d(x))` – askewchan Apr 11 '13 at 17:09

Minimal distance between elements in a vector

3 Answers3