I have a question in sort of connected component. I have a binary image ( onlye 0 and 1) I run the function from matlab:
f=
1 0 0 1 0 0 0 1 0 0
1 1 0 1 1 1 0 0 1 0
0 0 0 0 0 0 0 1 1 1
1 0 0 0 1 0 1 0 1 1
1 1 0 0 0 0 0 1 1 1
0 0 0 1 0 0 1 0 0 0
0 0 0 1 0 1 1 0 1 1
1 1 0 0 1 0 0 0 1 0
1 1 0 1 1 1 0 1 0 0
1 1 0 0 1 0 0 0 1 0
[L num]=bwlabel(f);
suppose that they give me the ma trix:
1 0 0 4 0 0 0 5 0 0
1 1 0 4 4 4 0 0 5 0
0 0 0 0 0 0 0 5 5 5
2 0 0 0 6 0 5 0 5 5
2 2 0 0 0 0 0 5 5 5
0 0 0 5 0 0 5 0 0 0
0 0 0 5 0 5 5 0 7 7
3 3 0 0 5 0 0 0 7 0
3 3 0 5 5 5 0 7 0 0
3 3 0 0 5 0 0 0 7 0
But you can see in this resul, the order of matrix is follow the column. Now I want to change this in to the oder rows, that mean number 4 is 2 , number 5 is 3... so on.
The oder is left-> right and top -> down. How can I do that ( the oder of reading )??
Thank you so much
f=f';
[L num]=bwlabel(f);
L=L';
does this solves your problem?
Related
I have to generate variables entry_1, entry_2 and entry_3 which will adopt the value 1 if id_i for that particular month had entry=1.
Example.
id month entry entry_1 entry_2 entry_3
1 1 1 1 0 0
1 2 0 0 0 0
1 3 0 0 1 1
1 4 0 0 0 0
2 1 0 1 0 0
2 2 0 0 0 0
2 3 1 0 1 1
2 4 0 0 0 0
3 1 0 1 0 0
3 2 0 0 0 0
3 3 1 0 1 1
3 4 0 0 0 0
Would anyone be so kind to propose an idea of how to implement a loop in order to do this?
I am thinking of something like this:
forvalues i=1(1)3 {
gen entry`i'=0
replace entry`i'=1 if on that particular month id=`i' had entry=1
}
You could do something like this (although your data don't quite look right for the question you're asking):
forvalues i = 1/3 {
gen entry_`i' = id == `i' & entry == 1
}
This generates a dummy variable entry_i for each i in the forvalues loop where entry_i = 1 if id is i and entry is 1, and 0 otherwise.
The code can be simplified down to at most one loop.
clear
input id month entry entry_1 entry_2 entry_3
1 1 1 1 0 0
1 2 0 0 0 0
1 3 0 0 1 1
1 4 0 0 0 0
2 1 0 1 0 0
2 2 0 0 0 0
2 3 1 0 1 1
2 4 0 0 0 0
3 1 0 1 0 0
3 2 0 0 0 0
3 3 1 0 1 1
3 4 0 0 0 0
end
forval j = 1/4 {
egen entry`j' = total(entry & id == `j'), by(month)
}
list id month entry entry? , sepby(id)
+--------------------------------------------------------+
| id month entry entry1 entry2 entry3 entry4 |
|--------------------------------------------------------|
1. | 1 1 1 1 0 0 0 |
2. | 1 2 0 0 0 0 0 |
3. | 1 3 0 0 1 1 0 |
4. | 1 4 0 0 0 0 0 |
|--------------------------------------------------------|
5. | 2 1 0 1 0 0 0 |
6. | 2 2 0 0 0 0 0 |
7. | 2 3 1 0 1 1 0 |
8. | 2 4 0 0 0 0 0 |
|--------------------------------------------------------|
9. | 3 1 0 1 0 0 0 |
10. | 3 2 0 0 0 0 0 |
11. | 3 3 1 0 1 1 0 |
12. | 3 4 0 0 0 0 0 |
+--------------------------------------------------------+
I don't know if this question is considered to be related to stackoverflow (I'm sorry if it's not but I have searched and did not find an answer anywhere).
I have coded a full adder
Output:
Truth Table :
a1 a2 b1 b2 S1 S2 C
______________________________
0 0 0 0 0 0 0
0 0 0 1 0 1 0
0 0 1 0 1 0 0
0 0 1 1 1 1 0
0 1 0 0 0 1 0
0 1 0 1 0 0 1
0 1 1 0 1 1 0
0 1 1 1 1 0 1
1 0 0 0 1 0 0
1 0 0 1 1 1 0
1 0 1 0 0 1 0
1 0 1 1 0 0 1
1 1 0 0 1 1 0
1 1 0 1 1 0 1
1 1 1 0 0 0 1
1 1 1 1 0 1 1
If somebody has ever calculated this, can they tell me if my output is correct
a1 a2 b1 b2 S1 S2 C a b s c
______________________________
0 0 0 0 0 0 0 0 0 0 0 nothing plus nothing is nothing
0 0 0 1 0 1 0 0 2 2 0 nothing plus two is two
0 0 1 0 1 0 0 0 1 1 0 nothing plus one is one
0 0 1 1 1 1 0 0 3 3 0 nothing plus three is three
0 1 0 0 0 1 0 2 0 2 0 two plus nothing is two
0 1 0 1 0 0 1 2 2 0 1 two plus two is four (four not in 0-3)
0 1 1 0 1 1 0 2 1 3 0 two plus 1 is three
0 1 1 1 1 0 1 2 3 1 1 two plus three is five (one and four)
1 0 0 0 1 0 0 1 0 1 0 one plus nothing is one
1 0 0 1 1 1 0 1 2 3 0 one plus two is three
1 0 1 0 0 1 0 1 1 2 0 one plus one is two
1 0 1 1 0 0 1 1 3 0 1 one plus three is four
1 1 0 0 1 1 0 3 0 3 0 three plus nothing is three
1 1 0 1 1 0 1 3 2 1 1 three plus two is five (one and four)
1 1 1 0 0 0 1 3 1 0 1 three plus one is four
1 1 1 1 0 1 1 3 3 2 1 three plus three is 6 (two and four)
Looks right. Ordering your 16 rows a little differently would make them flow in a more logical order.
It's an adder! Just check if it's adding. Let's take this row:
a2 a1 b2 b1 C S2 S2
1 0 1 1 1 0 1
Here I have reordered the columns in an easier to read manner: higher order bits first.
The a input is 10 = 2 (base 10). The b input is 11 = 3 (base 10). The output is 101, which
is 5 (base 10). So this one is right: 2 + 3 == 5.
I'll let you check the other rows.
Not sure if there is any name for this algorithm I'm currently developing - "growing neighbourhood algorithm" sounds like an appropriate name. So what is my problem about?
I would like to draw a stroke around an alpha transparent image to outline it. The size of the stroke should be user-definable.
I have an array which is filled by zeros and ones, consider each item of the array as a cell like in Game of Life. An item with 0 is empty (transparent pixel), an item with 1 is a first generation cell (non transparent pixel), the number of generations is defined by the size of the surrounding stroke.
This example depicts an rectangle surrounded by alpha values:
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 1 1 1 1 0 0 0
0 0 0 1 1 1 1 0 0 0
0 0 0 1 1 1 1 0 0 0
0 0 0 1 1 1 1 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
Then I would like to let the ones grow a new generation by surrounding every 0-generation Moore neighbour. It's the second generation (stroke with 1px) - thus the array looks after growing as follows:
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 2 2 2 2 2 2 0 0
0 0 2 1 1 1 1 2 0 0
0 0 2 1 1 1 1 2 0 0
0 0 2 1 1 1 1 2 0 0
0 0 2 1 1 1 1 2 0 0
0 0 2 2 2 2 2 2 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
3rd and 4th generation (stroke with 3px):
4 4 4 4 4 4 4 4 4 4
4 3 3 3 3 3 3 3 3 4
4 3 2 2 2 2 2 2 3 4
4 3 2 1 1 1 1 2 3 4
4 3 2 1 1 1 1 2 3 4
4 3 2 1 1 1 1 2 3 4
4 3 2 1 1 1 1 2 3 4
4 3 2 2 2 2 2 2 3 4
4 3 3 3 3 3 3 3 3 4
4 4 4 4 4 4 4 4 4 4
So far so good. I'm achieving this simple task by the following code snippet:
for (int gen = 1; gen <= 4; gen++)
{
for (int x = 1; x < arrayWidth - 1; x++)
{
for (int y = 1; y < arrayHeight - 1; y++)
{
// See if this cell is in the current generation.
if (_generation[x + arrayWidth * y] == gen)
{
// Generate next generation.
for (int i = x - 1; i <= x + 1; i++)
{
for (int j = y - 1; j <= y + 1; j++)
{
if (_generation[i + arrayWidth * j] == 0 || _generation[i + arrayWidth * j] > gen)
{
_generation[i + arrayWidth * j] = gen + 1;
}
}
}
}
}
}
}
This approach works perfectly for simple shapes like a rectangle for example. But how can I do this for an ellipse? As soon as we have kind of a stair pattern in the cells, I'm getting messy results:
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 1 1 1 1 0 0 0 0 0
0 0 0 0 1 1 1 1 1 1 0 0 0 0
0 0 0 1 1 1 1 1 1 1 1 0 0 0
0 0 1 1 1 1 1 1 1 1 1 1 0 0
0 0 1 1 1 1 1 1 1 1 1 1 0 0
0 0 1 1 1 1 1 1 1 1 1 1 0 0
0 0 1 1 1 1 1 1 1 1 1 1 0 0
0 0 1 1 1 1 1 1 1 1 1 1 0 0
0 0 0 1 1 1 1 1 1 1 1 0 0 0
0 0 0 0 1 1 1 1 1 1 0 0 0 0
0 0 0 0 0 1 1 1 1 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 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 2 2 2 2 2 2 0 0 0 0
0 0 0 2 2 1 1 1 1 2 2 0 0 0
0 0 2 2 1 1 1 1 1 1 2 2 0 0
0 2 2 1 1 1 1 1 1 1 1 2 2 0
0 2 1 1 1 1 1 1 1 1 1 1 2 0
0 2 1 1 1 1 1 1 1 1 1 1 2 0
0 2 1 1 1 1 1 1 1 1 1 1 2 0
0 2 1 1 1 1 1 1 1 1 1 1 2 0
0 2 1 1 1 1 1 1 1 1 1 1 2 0
0 2 2 1 1 1 1 1 1 1 1 2 0 0
0 0 2 2 1 1 1 1 1 1 2 2 0 0
0 0 0 2 2 1 1 1 1 2 2 0 0 0
0 0 0 0 2 2 2 2 2 2 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 3 3 3 3 3 3 3 3 0 0 0
0 0 3 3 2 2 2 2 2 2 3 3 0 0
0 3 3 2 2 1 1 1 1 2 2 3 3 0
3 3 2 2 1 1 1 1 1 1 2 2 3 3
3 2 2 1 1 1 1 1 1 1 1 2 2 3
3 2 1 1 1 1 1 1 1 1 1 1 2 3
3 2 1 1 1 1 1 1 1 1 1 1 2 3
3 2 1 1 1 1 1 1 1 1 1 1 2 3
3 2 1 1 1 1 1 1 1 1 1 1 2 3
3 2 1 1 1 1 1 1 1 1 1 1 2 3
3 2 2 1 1 1 1 1 1 1 1 2 2 3
3 3 2 2 1 1 1 1 1 1 2 2 3 3
0 3 3 2 2 1 1 1 1 2 2 3 3 0
0 0 3 3 2 2 2 2 2 2 3 3 0 0
0 0 0 3 3 3 3 3 3 3 3 0 0 0
When applying this algorithm to an ellipse, the outline looks kinda weird because of this problem (left: algorithm result, right: requested result):
The problem here is that I do not want have those 2 2 and 3 3 duplicate blocks which occur every time I have this "stair" pattern:
1 0 0 0 0 0 0 1
0 1 0 0 0 0 1 0
0 0 1 0 0 1 0 0
0 0 0 1 1 0 0 0
I want the above 2nd and 3rd generation calculations look like this:
0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 2 2 2 2 0 0 0 0 0
0 0 0 0 2 1 1 1 1 2 0 0 0 0
0 0 0 2 1 1 1 1 1 1 2 0 0 0
0 0 2 1 1 1 1 1 1 1 1 2 0 0
0 2 1 1 1 1 1 1 1 1 1 1 2 0
0 2 1 1 1 1 1 1 1 1 1 1 2 0
0 2 1 1 1 1 1 1 1 1 1 1 2 0
0 2 1 1 1 1 1 1 1 1 1 1 2 0
0 2 1 1 1 1 1 1 1 1 1 1 2 0
0 0 2 1 1 1 1 1 1 1 1 2 0 0
0 0 0 2 1 1 1 1 1 1 2 0 0 0
0 0 0 0 2 1 1 1 1 2 0 0 0 0
0 0 0 0 0 2 2 2 2 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 3 3 3 3 0 0 0 0 0
0 0 0 0 3 2 2 2 2 2 3 0 0 0
0 0 0 3 2 1 1 1 1 2 3 0 0 0
0 0 3 2 1 1 1 1 1 1 2 3 0 0
0 3 2 1 1 1 1 1 1 1 1 2 3 0
3 2 1 1 1 1 1 1 1 1 1 1 2 3
3 2 1 1 1 1 1 1 1 1 1 1 2 3
3 2 1 1 1 1 1 1 1 1 1 1 2 3
3 2 1 1 1 1 1 1 1 1 1 1 2 3
3 2 1 1 1 1 1 1 1 1 1 1 2 3
0 3 2 1 1 1 1 1 1 1 1 2 3 0
0 0 3 2 1 1 1 1 1 1 2 3 0 0
0 0 0 3 2 1 1 1 1 2 3 0 0 0
0 0 0 3 2 2 2 2 2 2 3 0 0 0
0 0 0 0 3 3 3 3 3 3 0 0 0 0
I've tried numerous methods to filter out those duplicate cell blocks, but I can't find an easy and generic solution for solving the problem.
Any ideas how to get stroke/outline like I get from Photoshop or Paint.NET?
Thanks!
Cheers
P
The proper name is dilation, check out morphological operations. You should try dilation with circle element, this will give you the requested result.
Here is a Matlab code that shows how it is done:
im = imcircle(70);
im = padarray(im,[20,20]);
figure;imshow(im);
im2 = imdilate(im,strel('disk',8));
figure;imshow(im2);
I have a binarized image like the folowing matrix:
1 1 0 0 1 1 0 0 1 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0 1 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0 1 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0 1 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1
The problem is that the image stars and end with 101, so how can i turn that into this.
1 0 1 0 1 0 0 1 1 0 1 0 1
1 0 1 0 1 0 0 1 1 0 1 0 1
1 0 1 0 1 0 0 1 1 0 1 0 1
1 0 1 0 1 0 0 1 1 0 1 0 1
I am trying to the decode the image binary code.
it seems like resizing original image with scale (0.5,1) using Nearest Neighbor method.
If you are using Matlab or any other language with similar array processing capabilities (APL, Fortran 90, Mathematica, C++ +Boost, ...) you could turn your input into your desired output with a statement similar to this:
arr(:,1:2:end)
if your array of pixels is called arr of course.
This does not return the median of close pixels, but then nor does the suggested output in the question.
how to convert these 7-segment decoder to boolean expression??
BCD 7-Segment decoder
A B C D a b c d e f g
0 0 0 0 0 0 0 0 0 0 1
0 0 0 1 1 0 0 1 1 1 1
0 0 1 0 0 0 1 0 0 1 0
0 0 1 1 0 0 0 0 1 1 0
0 1 0 0 1 0 0 1 1 0 0
0 1 0 1 0 1 0 0 1 0 0
0 1 1 0 0 1 0 0 0 0 0
0 1 1 1 0 0 0 1 1 1 1
1 0 0 0 0 0 0 0 0 0 0
1 0 0 1 0 0 0 0 1 0 0
I suggest you use a karnough map.
You'll need to use one for each result column, so 7 4x4 tables.
There are even a few karough map generators on the web that you can use.