I have an ndarray, and I want to replace every value in the array with the mean of its adjacent elements. The code below can do the job, but it is super slow when I have 700 arrays all with shape (7000, 7000) , so I wonder if there are better ways to do it. Thanks!
a = np.array(([1,2,3,4,5,6,7,8,9],[4,5,6,7,8,9,10,11,12],[3,4,5,6,7,8,9,10,11]))
row,col = a.shape
new_arr = np.ndarray(a.shape)
for x in xrange(row):
for y in xrange(col):
min_x = max(0, x-1)
min_y = max(0, y-1)
new_arr[x][y] = a[min_x:(x+2),min_y:(y+2)].mean()
print new_arr
all() in Python. The numpy. all() function tests whether all array elements along the mentioned axis evaluate to True.
The [:, :] stands for everything from the beginning to the end just like for lists. The difference is that the first : stands for first and the second : for the second dimension. a = numpy. zeros((3, 3)) In [132]: a Out[132]: array([[ 0., 0., 0.], [ 0., 0., 0.], [ 0., 0., 0.]])
If an array has shape (n,) , that means it's a 1-dimensional array with a length of n along its only dimension. It's not a row vector or a column vector; it doesn't have rows or columns.
Well, that's a smoothing operation in image processing
, which can be achieved with 2D
convolution. You are working a bit differently on the near-boundary elements. So, if the boundary elements are let off for precision, you can use scipy's convolve2d
like so -
from scipy.signal import convolve2d as conv2
out = (conv2(a,np.ones((3,3)),'same')/9.0
This specific operation is a built-in in OpenCV module as cv2.blur
and is very efficient at it. The name basically describes its operation of blurring the input arrays representing images. I believe the efficiency comes from the fact that internally its implemented entirely in C
for performance with a thin Python wrapper to handle NumPy arrays.
So, the output could be alternatively calculated with it, like so -
import cv2 # Import OpenCV module
out = cv2.blur(a.astype(float),(3,3))
Here's a quick show-down on timings on a decently big image/array -
In [93]: a = np.random.randint(0,255,(5000,5000)) # Input array
In [94]: %timeit conv2(a,np.ones((3,3)),'same')/9.0
1 loops, best of 3: 2.74 s per loop
In [95]: %timeit cv2.blur(a.astype(float),(3,3))
1 loops, best of 3: 627 ms per loop
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