I have problem during low-pass interpolation, I have to fill 1D array with zeros before processing it.
I have something like this: [1 2 3 4 5 6]
I want to have array like this [1 0 2 0 3 0 4 0 5 0 6] so it is L-1 zeros in array where L is the number of all values inside array before zero stuffing.
How to do it in Python?
You can assign an unpadded list of values into a slice of another list of zeros:
original_list = range(1,7) # [1,2,3,4,5,6]
padded_list = [0]*(2*len(original_list)-1) # [0,0,0,0,0,0,0,0,0,0,0]
padded_list[::2] = original_list # [1,0,2,0,3,0,4,0,5,0,6]
This can translate to numpy too, though as Jaime pointed out in a comment, it's even easier to use numpy.insert:
import numpy as np
arr = np.arange(1, 7) # array([1, 2, 3, 4, 5, 6])
np.insert(arr, slice(1, None), 0) # array([1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6])
I had to apply that for bigger amount of data real time, so I checked the performance and np.insert seems to be the slowest one on Python 2.7:
np.insert: 1.393 ms
list: 0.406 ms
np.repeat then zero out: 0.409 ms
np.zeros then assign value: 0.073 ms
Here's the test code:
import time
import numpy as np
start = 0
limit = 30000000
def tic():
global start
start = time.clock()
def toc(text):
print("%-30s %.3f ms" % (text, time.clock()-start))
original_list = range(1,limit)
original_array = np.arange(1, limit)
tic()
insert_array = np.insert(original_array, slice(1, None), 0)
toc("np.insert:")
tic()
padded_list = [0]*(2*len(original_list)-1)
padded_list[::2] = original_list
toc("list:")
tic()
repeat_array = np.repeat(original_array, 2)
repeat_array[1::2] = 0
toc("np.repeat then zero out:")
tic()
padded_array = np.zeros(2*len(original_list)-1, dtype=original_array.dtype)
padded_array[::2] = original_array
toc("np.zeros then assign value:")
l = [1, 2, 3, 4, 5, 6]
def zero_pad(l):
r = []
for x in l:
r.append(x)
r.append(0)
return r[:-1]
print zero_pad(l)
Will print:
[1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6]
One simplistic approach:
>>> i = [1,2,3,4,5,6]
>>> [i.insert(i.index(x)+1,0) for x in i[:-1]]
[None, None, None, None, None]
>>> i
[1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6]
The same as above, expanded:
>>> for x in i[:-1]:
... i.insert(i.index(x)+1,0)
...
>>> i
[1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6]
Nones.i.index is not necessary, and it will give the wrong results if there are duplicate values in the list. Since you know that you want to insert a zero at every position in reverse order you should just use for x in range(len(i)-1,0,-1): i.insert(x, 0)def zero(array):
for i in range(1, len(array)):
array.insert(i+i-1,0)
return array
array = [1, 2, 3, 4, 5, 6]
newArray = zero(array)
Mandatory itertools approach:
>>> from itertools import izip, repeat, chain
>>> list(chain.from_iterable(izip(el, repeat(0))))[:-1]
[1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6]
Much better itertools approach...
from itertools import islice, izip, repeat, chain
blah = list (islice (chain.from_iterable (izip(repeat (0), your_seq)), 1, None))