Effectively finding non-zeros spacing in scipy / numpy in Python?

Question

Suppose I have a python list or a python 1-d array (represented in numpy). Suppose there is continuous stretching of elements, how can I find the start and end coordinates (i.e., indices) of a non-zero portion in this list or array? eg,

a = [0, 0, 0, 0, 1, 2, 3, 4] 

nonzero_coords (a) should return [4, 7]. for:

 b = [1, 2, 3, 4, 0, 0] 

nonzero_coords (b) must return [0, 2].

thanks.

Answer 1

Assuming a single continuous segment of nonzero elements ...

 x = nonzero(a)[0] result = [x[0], x[-1]] 

Answer 2

This worked for a few holes for me

 from numpy import * def nonzero_intervals(value): lvalue = array(value) lvalue[0] = 0 lvalue[-1] = 0 a = diff((lvalue==0) * 1) intervals = zip( find(a == -1),find(a == 1)) return intervals 

Answer 3

In fact, nonzero_coords (b) should return [0, 3]. Can there be many holes at the entrance? If so, then what to do? Naive solution: scan to the first nonzero el. Then scan to the last nonzero el. The code below (sorry, did not check it):

 a = [0, 0, 0, 0, 1, 2, 3, 4, 5, 0, 0, 0] start = 0 size = len(a) # while (start < size and a[start] != 0): start += 1 end = start while (end < size and a[end] != 0): end += 1 return (start, end) 

Answer 4

If you downloaded numpy anyway, go to tom10 answer.

If for some reason you want something that works without loading numpy (I can’t imagine why, to be honest), I would suggest something like this:

 from itertools import groupby def nonzero_coords(iterable): start = 0 for iszero, sublist in groupby(iterable, lambda x:x==0): if iszero: start += len(list(sublist)) else: return start, start+len(list(sublist))-1 

Answer 5

It would be more consistent with python indexing for nonzero_coords([0, 0, 0, 0, 1, 2, 3, 4]) to return (4, 8) than (4, 7) , because [0, 0, 0, 0, 1, 2, 3, 4][4:8] returns [1, 2, 3, 4] .

Here is a function that calculates nonzero intervals. It processes several intervals:

 def nonzero_intervals(vec): ''' Find islands of non-zeros in the vector vec ''' if len(vec)==0: return [] elif not isinstance(vec, np.ndarray): vec = np.array(vec) edges, = np.nonzero(np.diff((vec==0)*1)) edge_vec = [edges+1] if vec[0] != 0: edge_vec.insert(0, [0]) if vec[-1] != 0: edge_vec.append([len(vec)]) edges = np.concatenate(edge_vec) return zip(edges[::2], edges[1::2]) 

If you really want the answer to have end indices included in the island, you can simply change the last line to: return zip(edges[::2], edges[1::2]-1)

Tests:

 a = [0, 0, 0, 0, 1, 2, 3, 4] intervals = nonzero_intervals(a) assert intervals == [(4, 8)] a = [1, 2, 3, 4, 0, 0] intervals = nonzero_intervals(a) assert intervals == [(0, 4)] a=[1, 2, 0, 0, 0, 3, 4, 0] intervals = nonzero_intervals(a) assert intervals == [(0, 2), (5, 7)] a = [0, 4, 0, 6, 0, 6, 7, 0, 9] intervals = nonzero_intervals(a) assert intervals == [(1, 2), (3, 4), (5, 7), (8, 9)] a = [1, 2, 3, 4] intervals = nonzero_intervals(a) assert intervals == [(0, 4)] a = [0, 0, 0] intervals = nonzero_intervals(a) assert intervals == [] a = [] intervals = nonzero_intervals(a) assert intervals == []