Iterate over an array and save to a Dict

Question

Given that array:

array([[0, 0, 0, ..., 0, 0, 0],
       [0, 0, 0, ..., 0, 0, 0],
       [0, 0, 0, ..., 0, 0, 0],
       ...,
       [0, 0, 0, ..., 0, 0, 0],
       [0, 0, 0, ..., 0, 0, 0],
       [0, 0, 0, ..., 0, 0, 0]], dtype=uint8)

I’m willing to stick it to count values on each line and then save them by line. The code below is doing the general sum. I would like to have the count per line.

def count(image):
    array = np.array(image)
    array[[ array == 0 ]] = 1
    array[[ array == 255 ]] = 0
    for row in array:
        unique, counts = np.unique(array[row,] , return_counts=True)
        d = dict(zip(unique, counts))
    return new

The result:

{0: 234710, 1: 515}

Answer 1

Good morning, Your answer is almost correct, for your script to work you should put the unique only in row not in array[row,], also there is a row identifier, you can create a dictionary for each row, or a list at each position a dictionary for the corresponding row:

a list with a dictionary for each line:

def count(image):
    array = np.array(image)
    array[[ array == 0 ]] = 1
    array[[ array == 255 ]] = 0
    colecao = list()
    for row in array:
        unique, counts = np.unique(row , return_counts=True)
        d = dict(zip(unique, counts))
        colecao.append(d)

a dictionary for each line:

def count(image):
    array = np.array(image)
    array[[ array == 0 ]] = 1
    array[[ array == 255 ]] = 0
    count = 0
    colecao = dict()
    for row in array:
        unique, counts = np.unique(row , return_counts=True)
        d = dict(zip(unique, counts))
        colecao[count] = d
        count += 1

Check if any of them are useful to you.