Delete a series of elements every nth time in numpy array

Question

I know how to delete every 4th element in a numpy array:

frame = np.delete(frame,np.arange(4,frame.size,4))

Now I want to know if there is a simple command that can delete every nth (e.g 4) times 3 values.

A basic example:

input: [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20....]

Would lead to:

output: [1,2,3,7,8,9,13,14,15,19,20,....]

I was hoping for a simple numpy / python functionality, rather than writing a function that has to iterate over the vector (cause it is quite long in my case,...).

Thank you for your help

Daniel F · Accepted Answer · 2017-04-11 13:00:38Z

A method using boolean indexing:

def block_delete(a, n, m):  #keep n, remove m
    mask = np.tile(np.r_[np.ones(n), np.zeros(m)].astype(bool), a.size // (n + m) + 1)[:a.size]
    return a[mask]

Compare with @Divakar:

def mod_delete(a, n, m):
    return a[np.mod(np.arange(a.size), n + m) < n]

a = np.arange(19) + 1

%timeit block_delete(a, 3, 4)
10000 loops, best of 3: 50.6 µs per loop

%timeit mod_delete(a, 3, 4)
The slowest run took 9.37 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 5.69 µs per loop

Let's try a longer array:

a = np.arange(999) + 1

%timeit block_delete(a, 3, 4)
The slowest run took 4.61 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 54.8 µs per loop

%timeit mod_delete(a, 3, 4)
The slowest run took 5.13 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 14.5 µs per loop

Still longer:

a = np.arange(999999) + 1

%timeit block_delete(a, 3, 4)
100 loops, best of 3: 3.93 ms per loop

%timeit mod_delete(a, 3, 4)
100 loops, best of 3: 12.3 ms per loop

So which is faster will depend on the size of your array

Community · Accepted Answer · 2017-05-23 12:26:17Z

Approach #1 : Here's one approach with modulus and boolean-indexing -

a[np.mod(np.arange(a.size),6)<3]

As a function, it would translate to :

def select_in_groups(a, M, N): # Keep first M, delete next N and so on.
    return a[np.mod(np.arange(a.size),M+N)<M]

Sample step-by-step run -

# Input array
In [361]: a
Out[361]: 
array([ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17,
       18, 19, 20])

# Create a range array that spans along the length of array
In [362]: np.arange(a.size)
Out[362]: 
array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16,
       17, 18, 19])

# Use modulus to create "intervaled" version of it that shifts at
# the end of each group of 6 elements
In [363]: np.mod(np.arange(a.size),6)
Out[363]: array([0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1])

# We need to select the first three as valid ones, so compare against 3
# creating a boolean array or mask
In [364]: np.mod(np.arange(a.size),6) < 3
Out[364]: 
array([ True,  True,  True, False, False, False,  True,  True,  True,
       False, False, False,  True,  True,  True, False, False, False,
        True,  True], dtype=bool)

# Use the mask to select valid elements off array
In [365]: a[np.mod(np.arange(a.size),6)<3]
Out[365]: array([ 1,  2,  3,  7,  8,  9, 13, 14, 15, 19, 20])

Approach #2 : For performance, here's another method with NumPy array strides -

def select_in_groups_strided(a, M, N): # Keep first M, delete next N and so on.
    K = M+N
    na = a.size
    nrows = (1+((na-1)//K))
    n = a.strides[0]
    out = np.lib.index_tricks.as_strided(a, shape=(nrows,K), strides=(K*n,n))
    N = M*(na//K) + (na - (K*(na//K)))
    return out[:,:M].ravel()[:N]

Sample runs -

In [545]: a = np.arange(1,21)

In [546]: a
Out[546]: 
array([ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17,
       18, 19, 20])

In [547]: select_in_groups_strided(a,3,3)
Out[547]: array([ 1,  2,  3,  7,  8,  9, 13, 14, 15, 19, 20])

In [548]: a = np.arange(1,25)

In [549]: a
Out[549]: 
array([ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17,
       18, 19, 20, 21, 22, 23, 24])

In [550]: select_in_groups_strided(a,3,3)
Out[550]: array([ 1,  2,  3,  7,  8,  9, 13, 14, 15, 19, 20, 21])

Runtime test

Using the same setup as in @Daniel Forsman's timing tests -

In [637]: a = np.arange(1,21)

In [638]: %timeit block_delete(a,3,3)
10000 loops, best of 3: 21 µs per loop

In [639]: %timeit select_in_groups_strided(a,3,3)
100000 loops, best of 3: 6.44 µs per loop

In [640]: a = np.arange(1,2100)

In [641]: %timeit block_delete(a,3,3)
10000 loops, best of 3: 27 µs per loop

In [642]: %timeit select_in_groups_strided(a,3,3)
100000 loops, best of 3: 9.1 µs per loop

In [643]: a = np.arange(999999) + 1

In [644]: %timeit block_delete(a,3,3)
100 loops, best of 3: 2.24 ms per loop

In [645]: %timeit select_in_groups_strided(a,3,3)
1000 loops, best of 3: 1.12 ms per loop

Strided one scales pretty well across varying sizes, if you are considering performance.

Thank you this works, can you explain what the command does? I do not really know what is meant with the ",6"
yeah, figured there'd be an as_strided answer, just couldn't wrap my head around it.

Erdal Bozkurt · Accepted Answer · 2021-06-26 13:20:22Z

0

import numpy as np
a = np.array([10, 0, 0, 20, 0, 30, 40, 50, 0, 60, 70, 80, 90, 100,0])
print("Original array:")
print(a)
index=np.zeros(0)
for i in range(len(a)):
    if a[i]==0:
        index=np.append(index, i)
print("index=",index)
new_a=np.delete(a,index)
print("new_a=",new_a)

answered Jun 26, 2021 at 13:20

Erdal Bozkurt

1

Collectives™ on Stack Overflow

Delete a series of elements every nth time in numpy array

3 Answers 3

Comments

2 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related