pandas how to find continuous values in a series whose differences are within a certain distance

Question

I have a pandas Series that is composed of ints

a = np.array([1,2,3,5,7,10,13,16,20])
pd.Series(a)

0  1
1  2
2  3
3  5
4  7
5  10
6  13
7  16
8  20

now I want to cluster the series into groups that in each group, the differences between two neighbour values are <= distance. For example, if the distance is defined as 1, we have

[1,2,3], [5], [7], [10], [13], [16], [20]

if the distance is 2, we have

[1,2,3,5,7], [10], [13], [16], [20]

if the distance is 3, we have

[1,2,3,5,7,10,13,16], [20]

how to do this using pandas/numpy?

Answer 1

This is the pandas way, using groupby.

n = 1

s

0     1
1     2
2     3
3     5
4     7
5    10
6    13
7    16
8    20
dtype: int64

m = ~s.diff().fillna(0).le(n)   
v = s.groupby(m.cumsum()).apply(lambda x: x.tolist()).tolist()

v
[[1, 2, 3], [5], [7], [10], [13], [16], [20]]

Answer 2

Here's one approach -

np.split(a,np.flatnonzero(np.diff(a)>d)+1)

As a function to output list of lists -

def splitme(a,d) : 
    return list(map(list,np.split(a,np.flatnonzero(np.diff(a)>d)+1)))

For performance, I would suggest using zip to get the start, stop indices and then slicing, thus avoiding np.split which might prove to be the bottleneck -

def splitme_zip(a,d) : 
    m = np.concatenate(([True],a[1:] > a[:-1] + d,[True]))
    idx = np.flatnonzero(m)
    l = a.tolist()
    return [l[i:j] for i,j in zip(idx[:-1],idx[1:])]

If you need the output as a list of arrays, skip the list conversion with .tolist/map(list,).

Sample runs -

In [122]: a = np.array([1,2,3,5,7,10,13,16,20])

In [123]: splitme(a,1)
Out[123]: [[1, 2, 3], [5], [7], [10], [13], [16], [20]]

In [124]: splitme(a,2)
Out[124]: [[1, 2, 3, 5, 7], [10], [13], [16], [20]]

In [125]: splitme(a,3)
Out[125]: [[1, 2, 3, 5, 7, 10, 13, 16], [20]]

Runtime test -

In [180]: a = np.sort(np.random.randint(1,10000*2,(10000)))

In [181]: s = pd.Series(a)

In [182]: d = 3

In [183]: %timeit pandas_way(s,d) #@cᴏʟᴅsᴘᴇᴇᴅ's soln
10 loops, best of 3: 55.1 ms per loop

In [184]: %timeit np.split(a,np.flatnonzero(np.diff(a)>d)+1)
     ...: %timeit splitme(a,d)
     ...: %timeit splitme_zip(a,d)
1000 loops, best of 3: 1.47 ms per loop
100 loops, best of 3: 2.87 ms per loop
1000 loops, best of 3: 516 µs per loop

In [185]: a
Out[185]: array([    2,     2,     2, ..., 19992, 19996, 19999])