NumPy - catching the index of first three consecutive negative numbers

Question

I need to find the index of the first occurrence of three consecutive negative numbers. In the normal Python way I would do it like this:

a = [1,-1,1,-1,1,-1,1,-1,-1,-1,1,-1,1]
b=0
for i,v in enumerate(a):
    if v<0:
        b+=1
    else:
        b=0
    if b==3:
        break
indx = i-2

Anyone has an idea how to do it in a smarter NumPy way?

You could try to take a look at this link. It seems like it would be able to help you. [Searching a sequence in a numpy array](https://stackoverflow.com/questions/36522220/searching-a-sequence-in-a-numpy-array?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa) — TMK, Jun 14 '18 at 09:12
*index of the first occurrence of **three consecutive negative numbers*** - so you expect it to be `7`? or `[7,8,9]` ? — RomanPerekhrest, Jun 14 '18 at 09:48
Yes, the first_consecutive_negative_island is doing exactly what I had to do..thanks — Ohm, Jun 15 '18 at 12:51

Divakar · Accepted Answer · 2018-06-14T10:27:54.373

Here's a vectorized solution with help from convolution -

def first_consecutive_negative_island(a, N=3):
    mask = np.convolve(np.less(a,0),np.ones(N,dtype=int))>=N
    if mask.any():
        return mask.argmax() - N + 1
    else:
        return None

Sample run -

In [168]: a = [1,-1,1,-1,1,-1,1,-1,-1,-1,1,-1,1]

In [169]: first_consecutive_negative_island(a, N=3)
Out[169]: 7

Works irrespective of where the group exists -

In [175]: a
Out[175]: [-1, -1, -1, 1, -1, 1, -1, -1, -1, 1, -1, 1]

In [176]: first_consecutive_negative_island(a, N=3)
Out[176]: 0

With no negative numbers, it gracefully returns None -

In [183]: a
Out[183]: [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]

In [184]: first_consecutive_negative_island(a, N=3)

For exactly three consecutive negative numbers search, we can use slicing, like so -

def first_consecutive_negative_island_v2(a):
    m =  np.less(a,0)
    mask = m[:-2] & m[1:-1] & m[2:]
    if mask.any():
        return mask.argmax()
    else:
        return None

Timings -

In [270]: a = np.random.randint(-1,2,(1000000)).tolist()

In [271]: %timeit first_consecutive_negative_island(a, N=3)
10 loops, best of 3: 44.5 ms per loop

In [272]: %timeit first_consecutive_negative_island_v2(a)
10 loops, best of 3: 38.7 ms per loop

You could also convolve `np.signbit(a)` instead of `np.less(a,0)` for a bit more speed (maybe). — Daniel F, Jun 14 '18 at 10:25

score 0 · Answer 2 · answered Jun 14 '18 at 09:21

0

import numpy as np
a = np.array([1,-1,1,-1,-1,-1,1,-1,-1,-1,1,-1,1])
item_index = np.where(a < 0)
for a,b in zip( item_index[0],item_index[0][2:]):
    if b-a == 2: 
        break
index_ = a

answered Jun 14 '18 at 09:21

ken

35
1
7

This is not a `numpy`-style answer. It uses `zip` AND a for loop. This is neither pythonic nor efficient numpy style. – JE_Muc Jun 14 '18 at 09:44

JE_Muc · Answer 3 · 2018-06-14T10:03:09.470

There is no looping needed at all. Also no zipping. Just do the following:

a = np.array([1, -1, 1, -1, -1, -1, 1, -1, -1, -1, 1, -1, 1])
idx = np.where(a < 0)[0][:3]

np.where returns a tuple with the indices of the chosen condition. [0] indexes this tuple's first dimension (for a 1D-input array this is the only dimension) and [:3] slices these indices to only use the first three indices.

If the occurrence of the first three consecutive negative numbers is needed, you can do the following:

idx = np.where(np.diff(np.where(a < 0), n=2) == 0)[1][0] + 2

This will give you the index of the first negative number of three consecutive negative numbers. If you want the indices of all three first negative consecutive numbers, do this:

idx_start = np.where(np.diff(np.where(a < 0), n=2) == 0)[1][0] + 2
idx = np.arange(idx_start, idx_start + 3)

This will work as long as the first two numbers in the array are not part of three consecutive negative numbers.

NumPy - catching the index of first three consecutive negative numbers

3 Answers3