CODEWITHSUNDEEP
pythonnumpybooleancumsum
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Reputation: 5327

get index of the first block of at least n consecutive False values in boolean array

I have a numpy boolean array 

w=np.array([True,False,True,True,False,False,False])

I would like to get the index of the first time there are at n_at_least false values. For instance here 

`n_at_least`=1 -> desired_index=1

`n_at_least`=3 -> desired_index=4

I have tried

np.cumsum(~w)

which does increase every time a False value is encountered. However, when True is encountered the counter is not starting from 0 again so I only get the total count of False elements rather than the count of the last consecutive ones.

Upvotes: 8

Views: 1348

Answers (6)

Lukisn
Lukisn

Reputation: 190

I think for this linear search operation a python implementation is ok. My suggestion looks like this:

def find_block(arr, n_at_least=1):
    current_index = 0
    current_count = 0
    for index, item in enumerate(arr):
         if item:
             current_count = 0
             current_index = index + 1
         else:
             current_count += 1
         if current_count == n_at_least:
             return current_index
    return None # Make sure this is outside for loop

Running this function yields the following outputs:

>>> import numpy
>>> w = numpy.array([True, False, True, True, False, False, False])
>>> find_block(w, n_at_least=1)
1
>>> find_block(w, n_at_least=3)
4
>>> find_block(w, n_at_least=4)
>>> # None

Upvotes: 2

Divakar
Divakar

Reputation: 221524

Here's a vectorized solution that finds the start, stop indices and hence lengths of islands of zeros and finally uses argmax to get the starting index of the first island satisfying the criteria of zeros count being >= n -

def first_occ_index(w, n):
    idx = np.flatnonzero(np.r_[True, w, True])
    lens = np.diff(idx) - 1
    return idx[(lens >= n).argmax()]

Sample run -

In [107]: w
Out[107]: array([ True, False,  True,  True, False, False, False])

In [108]: first_occ_index(w, n=1)
Out[108]: 1

In [109]: first_occ_index(w, n=3)
Out[109]: 4

Upvotes: 7

jpp
jpp

Reputation: 164623

This is one way using a generator expression with slicing:

w = np.array([True,False,True,True,False,False,False])

n = 2
val = False

res = next((i for i, j in enumerate(w[k:k+n] for k in range(len(w)-n+1)) \
            if np.all(j==val)), None)

# 4

Upvotes: 1

Paul Panzer
Paul Panzer

Reputation: 53029

Here is an O(n) numpy solution:

>>> def first_consec(A, n):
...     A = np.r_[True, A, True]
...     switch, = np.where(A[:-1]!=A[1:])
...     runs = switch[1::2] - switch[::2]
...     idx = np.argmax(runs >= n)
...     if runs[idx] < n:
...         return None
...     return switch[2*idx]
... 
>>> first_consec(w, 4)
>>> first_consec(w, 3)
4
>>> first_consec(w, 2)
4
>>> first_consec(w, 1)
1

Upvotes: 4

wohe1
wohe1

Reputation: 775

It should work this way:

def n_at_least(n):
    for i in range(len(w)):
         if not any(w[i:i+n]):
             return i

However I don't know if there is a better way...

Upvotes: 1

FHTMitchell
FHTMitchell

Reputation: 12147

I think you're falling into the numpy trap of only wanting to use numpy functions. What's wrong with python? This solution is O(n)

def f(array, n_at_least):
    curr_found_false = 0
    curr_index = 0
    for index, elem in enumerate(array):
        if not elem:
            if curr_found_false == 0:
                curr_index = index
            curr_found_false += 1
            if curr_found_false == n_at_least:
                return curr_index
        else:
            curr_found_false = 0

Outputs

w=np.array([True,False,True,True,False,False,False])
f(w, 1)
# --> 1
f(w, 3)
# --> 4

Upvotes: 4

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