About the numpy.where statement - numpy

Question

I would like to use the numpy.where to check the value of a previous row but don't know how to code
for n1 in range(len(image1)):
print('input image ',input_folder+'\\' + image1[n1])
print('\n')
print('image1[n1] ',image1[n1])
print('\n')
im = Image.open(input_folder+'\\'+image1[n1])
a = np.array(im, dtype='uint8')
width, height = im.size
print('width ',width)
print('height ',height)
a = np.where(a==[0,0,0],[255,255,255],a)
!-- Change the looping statement to np.where --!
for h in range(height):
for w in range(width):
if h <= (height - 2) and w <= (width - 2):
if a[h,w,0] != 255 and a[h,w,1] != 255 and a[h,w,2] != 255:
if (a[h-1,w,0] == 255 and a[h-1,w,1] == 255 and a[h-1,w,2] == 255 and a[h+1,w,0] == 255 and a[h+1,w,1] == 255 and a[h+1,w,2] == 255) or (a[h,w-1,0] == 255 and a[h,w-1,1] == 255 and a[h,w-1,2] == 255 and a[h,w+1,0] == 255 and a[h,w+1,1] == 255 and a[h,w+1,2] == 255):***
Change the above looping statement to np.where(a[-??] = [255,255,255] or a[+??] = [255,255,255]) so it can run more faster than the for loop statement. -->
a[h,w,0] = 255
a[h,w,1] = 255
a[h,w,2] = 255

Answer 1

I'm afraid, you can not use np.where here.
The reason is that:
the condition passed to np.where should indicate each element of the
source array,
whereas the criterion in your code actually relates only to first 2
dimensions of the source array.
So I came up with another, quite elegant and concise solution.
Part 1: How to get first two indices of elements, where all elements
in the third dimension are != 255:
To to it, on the whole array, you could run:
np.not_equal(a, 255).all(axis=2)
Part 2: How to limit the "range of operation" to elements having both
previous and next row and column.
You can do it passing to the above code a "subrange" of the original array:
np.not_equal(a[1:-1, 1:-1], 255).all(axis=2))
You should eliminate both the first and the last column and row (in
your code you failed to eliminate the first row / column).
But note that this time the resulting indices are by one less than before,
so at the later step you will have to add 1 to them.
Part 3: A function to check whether all elements along the third dimension
== 255, for some row (r) and column (c):
def all_eq(arr, r, c):
return np.equal(arr[r, c], 255).all()
(will be used soon).
Part 4: How to get the result:
res = a.copy()
for r, c in zip(*np.where(np.not_equal(a[1:-1, 1:-1], 255).all(axis=2))):
h = r + 1
w = c + 1
if all_eq(a, h-1, w) and all_eq(a, h+1, w) or\
all_eq(a, h, w-1) and all_eq(a, h, w+1):
res[h, w] = 255
Note that this code starts from making a copy of the original array
(it will hold the result).
Then, for r, c in zip(…) iterates over the indices found.
First 2 lines in the loop add 1 to the indices found, in the subrange
of the original array, so now h and w indicate row / column in the whole
original array.
Then if checks whether the respective adjacent pixels have 255 in all elements.
If they do, then put 255 in all elements of the "current" pixel, in the result.
You can't operate on the original array, since changed values in some pixels
would "falsify" the evaluation of conditions for subseqent pixels.
Edit
After some research I found, that it is possible to use np.where,
although the solution is a bit complicated and involving quite a big
number of Numpy methods:
# Mask 1: Pixels with all elements != 255
m1 = np.zeros((height, width), dtype='int8')
idx = np.where(np.not_equal(a, 255).all(axis=2))
m1[idx] = 1
# Pixels with all elements == 255
m2 = np.apply_along_axis(lambda px: np.equal(px, 255).all(), 2, a).astype('int8')
# Both adjacent pixels (left / right) == 255
m2a = np.logical_and(np.insert(m2, 0, 0, axis=1)[:,:-1], np.insert(m2,
width, 0, axis=1)[:,1:])
# Both adjacent pixels (up / down) == 255
m2b = np.logical_and(np.insert(m2, 0, 0, axis=0)[:-1,:], np.insert(m2,
height, 0, axis=0)[1:,:])
# Mask 2: Both adjacent pixels (either vertically or horizontally) == 255
m2 = np.logical_or(m2a, m2b)
# The "final" mask
msk = np.logical_and(m1, m2)
# Generate the result
result = np.where(np.expand_dims(msk, 2), 255, a)
This solution should be substantially faster than my first concept.