In Python, let's say I have a 1366x768 numpy array. And I want to delete each second row from it (0th row remains, 1st removed, 2nd remains, 3rd removed.. and so on), and replace the empty space with a duplicate from the row which was before it (the undeleted row) at the same time.
Is it possible in numpy?
One approach -
a[::2].repeat(2,axis=0)
To make the changes in the array, assign it back.
Sample run -
In [105]: a
Out[105]:
array([[2, 5, 1, 1],
[2, 0, 2, 5],
[1, 1, 5, 7],
[0, 7, 1, 8],
[8, 5, 2, 3],
[2, 1, 0, 6],
[5, 6, 1, 6],
[7, 1, 4, 7],
[3, 8, 1, 4],
[5, 8, 8, 8]])
In [106]: a[::2].repeat(2,axis=0)
Out[106]:
array([[2, 5, 1, 1],
[2, 5, 1, 1],
[1, 1, 5, 7],
[1, 1, 5, 7],
[8, 5, 2, 3],
[8, 5, 2, 3],
[5, 6, 1, 6],
[5, 6, 1, 6],
[3, 8, 1, 4],
[3, 8, 1, 4]])
If we care about performance, here's another approach using NumPy strides -
def strided_app(a):
m0,n0 = a.strides
m,n = a.shape
strided = np.lib.stride_tricks.as_strided
return strided(a,shape=(m//2,2,n),strides=(2*m0,0,n0)).reshape(-1,n)
Sample run -
In [154]: a
Out[154]:
array([[4, 8, 7, 7],
[5, 5, 1, 7],
[1, 8, 1, 3],
[6, 6, 5, 6],
[0, 2, 6, 3],
[6, 6, 8, 7],
[7, 6, 8, 1],
[7, 8, 8, 2],
[4, 0, 2, 8],
[5, 8, 1, 4]])
In [155]: strided_app(a)
Out[155]:
array([[4, 8, 7, 7],
[4, 8, 7, 7],
[1, 8, 1, 3],
[1, 8, 1, 3],
[0, 2, 6, 3],
[0, 2, 6, 3],
[7, 6, 8, 1],
[7, 6, 8, 1],
[4, 0, 2, 8],
[4, 0, 2, 8]])
Timings -
In [156]: arr = np.arange(1000000).reshape(1000, 1000)
# Proposed soln-1
In [157]: %timeit arr[::2].repeat(2,axis=0)
1000 loops, best of 3: 1.26 ms per loop
# @Psidom 's soln
In [158]: %timeit arr[1::2] = arr[::2]
1000 loops, best of 3: 928 µs per loop
In [159]: arr = np.arange(1000000).reshape(1000, 1000)
# Proposed soln-2
In [160]: %timeit strided_app(arr)
1000 loops, best of 3: 830 µs per loop
Looks like you have an even number of rows, in which case, you can use assignment (assign the odd rows values to corresponding even rows):
arr = np.array([[1,4],[3,1],[2,3],[2,2]])
arr[1::2] = arr[::2]
arr
#array([[1, 4],
# [1, 4],
# [2, 3],
# [2, 3]])
This avoids copying the entire array, but doesn't work if the array has odd number of rows.
Timing: Here is a comparison of the timing, the assignment does seem faster.
arr = np.arange(1000000).reshape(1000, 1000)
%timeit arr[::2].repeat(2,axis=0)
1000 loops, best of 3: 913 µs per loop
%timeit arr[1::2] = arr[::2]
1000 loops, best of 3: 655 µs per loop
strides. It's just crazy that strides is beating a simple assignment step.strides is still a quite new concept to me. -This works for both even and an odd number of rows.
for i in range(1,len(a),2):
a[i] = a[i-1]
