Find Start / Stop Index Range For Values in NumPy Array Greater Than N

Here's another solution (which I believe can be improved):

import numpy as np
from numpy.lib.stride_tricks import as_strided

x = np.array([2, 3, 4, 0, 0, 1, 1, 4, 6, 5, 8, 9, 9, 4, 2, 0, 3])

# array of unique values of x bigger than 1
a = np.unique(x[x>=2])

step = len(a)  # if you encounter memory problems, try a smaller step
result = []
for i in range(0, len(a), step):
    ai = a[i:i + step]
    c = np.argwhere(x >= ai[:, None])
    c[:,0] = ai[c[:,0]]
    c =  np.pad(c, ((1,1), (0,0)), 'symmetric')

    d = np.where(np.diff(c[:,1]) !=1)[0]

    e = as_strided(d, shape=(len(d)-1, 2), strides=d.strides*2).copy()
    # e = e[(np.diff(e, axis=1) > 1).flatten()]
    e[:,0] = e[:,0] + 1 

    result.append(np.hstack([c[:,0][e[:,0, None]], c[:,1][e]]))

result = np.concatenate(result)

# array([[ 2,  0,  2],
#        [ 2,  7, 14],
#        [ 2, 16, 16],
#        [ 3,  1,  2],
#        [ 3,  7, 13],
#        [ 3, 16, 16],
#        [ 4,  2,  2],
#        [ 4,  7, 13],
#        [ 5,  8, 12],
#        [ 6,  8,  8],
#        [ 6, 10, 12],
#        [ 8, 10, 12],
#        [ 9, 11, 12]])

Sorry for not commenting what each step does -- if later I will find time I will fix it.

This is quite interesting problem, indeed. I attempted to solve it splitting it into three parts.

Grouping:

import numpy as np
import pandas as pd
x = np.array([2, 3, 4, 0, 0, 1, 1, 4, 6, 5, 8, 9, 9, 4, 2, 0, 3])
groups = pd.DataFrame(x).groupby([0]).indices

So groups is dictionary {0: [3, 4, 15], 1: [5, 6], 2: [0, 14], 3: [1, 16], 4: [2, 7, 13], 5: [9], 6: [8], 8: [10], 9: [11, 12]} and its values are numpy arrays of dtype=int64.

Masking:

In this part I iterate through several masks arrays x>=i for each unique value i in descending order:

mask_array = np.zeros(x.size).astype(int)
for group in list(groups)[::-1]:
    mask = mask_array[groups[group]] = 1
    # print(group, ':', mask_array)
    # output = find_slices(mask)

And these masks look like this:

9 : [0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0]
8 : [0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0]
6 : [0 0 0 0 0 0 0 0 1 0 1 1 1 0 0 0 0]
5 : [0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0]
4 : [0 0 1 0 0 0 0 1 1 1 1 1 1 1 0 0 0]
3 : [0 1 1 0 0 0 0 1 1 1 1 1 1 1 0 0 1]
2 : [1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 0 1]
1 : [1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 1]
0 : [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]

Extracting slices from masks:

I expect to construct some function called find_slices that extracts slice positions from mask arrays (if you uncomment it). This is what I did:

def find_slices(m):
    m1 = np.r_[0, m]
    m2 = np.r_[m, 0]
    starts, = np.where(~m1 & m2)
    ends, = np.where(m1 & ~m2)
    return np.c_[starts, ends - 1]

For example, slice positions of array [0 1 1 0 0 0 0 1 1 1 1 1 1 1 0 0 1] would be [[1, 2], [7, 13], [16, 16]]. Note that this is not a standard way of returning slices, end position is usually incremented by 1.

Final script

After all, one needs some manoeuvres to make the expected output, here is like it looks at the end:

import numpy as np
import pandas as pd
x = np.array([2, 3, 4, 0, 0, 1, 1, 4, 6, 5, 8, 9, 9, 4, 2, 0, 3])
groups = pd.DataFrame(x).groupby([0]).indices
mask_array = np.zeros(x.size).astype(bool)

m = []
for group in list(groups)[::-1]:
    mask_array[groups[group]] = True
    s = find_slices(mask_array)
    group_output = np.c_[np.repeat(group, s.shape[0]), s] #insert first column
    m.append(group_output) 
output = np.concatenate(m[::-1])
output = output[output[:,1]!= output[:,2]] #elimate slices with unit length

Output:

 [[ 0  0 16]
 [ 1  0  2]
 [ 1  5 14]
 [ 2  0  2]
 [ 2  7 14]
 [ 3  1  2]
 [ 3  7 13]
 [ 4  7 13]
 [ 5  8 12]
 [ 6 10 12]
 [ 8 10 12]
 [ 9 11 12]]