Finding the leftmost and the rightmost cell that is True in a 2d-list of booleans (Python)

Question

I have a 2d-array (28 x 28) that has boolean values.

array([[False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False,  True,  True,
         True, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False,  True,  True,  True,
         True,  True, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False, False, False, False, False,  True,  True,  True,  True,
         True,  True, False, False, False, False, False, False, False,
        False],
       ...

This represents an image of a digit. In order to stretch this image so that the horizontal and vertical range of ink pixels runs the full horizontal/vertical range of the box.

Below are the steps that I am trying to take:

1. Find the leftmost (x_min) and the rightmost (x_max) pixels that have ink (after thresholding). 
2. Find the topmost (y_min) and bottom-most (y_max) pixels similarly. 
3. Find the centerpoint (x,y) by taking the mean of x_min and x_max and y_min and y_max respectively. 
4. Now, the 20x20 bounding is defined as img[y_center-10 : y_center + 10, x_center-10:x_center+10]

I tried to do Step 1 in a naive way, just iterating all pixels, but I know that is not the best way. What is the best way to find the left and rightmost pixel that is True?

That's the same as this https://stackoverflow.com/a/54286937/2836621 — Mark Setchell, Jan 25 '19 at 22:20

Michel Keijzers · Answer 1 · 2019-01-25T21:00:04.887

0

You can use the function index, which returns the first index of the value you search for:

e.g.

array = ([False, False, True, True, False])
print(array.index(True))

Returns

You can reverse the array to find the last one, by adding:

array.reverse();
print(array.index(True))

Which returns

However, reverse is very expensive operation for what you need.

edited Jan 25 '19 at 21:00

answered Jan 25 '19 at 20:57

Michel Keijzers

15,025
28
93
119

Thanks. Do I have to iterate through all rows in order to find the leftmost `True` pixel? – Dawn17 Jan 25 '19 at 20:59
What can I do to find the top-most and bottom-most pixel with `True`? – Dawn17 Jan 25 '19 at 20:59
@Dawn17 I added a new answer which results in all your requested values. – Michel Keijzers Jan 25 '19 at 21:52

Aiden Zhao · Answer 2 · 2019-01-25T21:30:18.763

Will this work for you? not most efficient, but works. the idea is to remove rows and columns that's all False, so you get your 'cropped' digit.

import pandas as pd

img = np.array([[False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False, False,  True,  True,
         True, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False, False, False, False, False, False,  True,  True,  True,
         True,  True, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False, False, False, False, False,  True,  True,  True,  True,
         True,  True, False, False, False, False, False, False, False,
        False]])

img = pd.DataFrame(img)
img.replace(False,np.nan, inplace = True)
newimage = img.dropna(axis = 0, how = 'all')
newimage = newimage.dropna(axis = 1, how = 'all')
newimage.T.values

Yes, this works, but for some reason, the black and white are flipped in the cropped image. — Dawn17, Jan 25 '19 at 21:27

Michel Keijzers · Answer 3 · 2019-01-25T22:30:46.903

array = [[False, False, False, False],
         [False, False, False, False],
         [True , False, False, False],
         [False, False, False, False]]

print('left: ')
print(min([x.index(True) if True in x else 100 for x in array]))

print('right: ')
print(min([x[::-1].index(True) if True in x else 100 for x in array]))

print('top')
print([any(x) for x in array].index(True))

print('bottom')    
print([any(x) for x in array[::-1]].index(True))

prints

left: 
0
right: 
3
top
2
bottom
1

Side notes:

I used 100 for left/right in case there is no True at all in a row.
If there is a chance for no True value at all in the entire matrix, conditions for top and bottom have to be changed slightly.
::-1 iterates the list in reverse order
any checks for the occurrence of a non 0/empty/False value.
min checks for the minimum value of a list
[...] is called list comprehension in case you want to find more about it

With your values I get:

left:
14
right:
8
top:
4
bottom:
0

Finding the leftmost and the rightmost cell that is True in a 2d-list of booleans (Python)

3 Answers3