Can someone explain what the percent parameter is for on the min_mas_grey() operator in Halcon?
min_max_gray(Regions, Image : : Percent : Min, Max, Range)
The documentation for this operator can be found here:
https://www.mvtec.com/doc/halcon/2005/en/min_max_gray.html
To elaborate a little on the explanation if you are having trouble following it:
calculates the number of pixels Percent corresponding to the area of
the input image. Then it goes inwards on both sides of the histogram
by this number of pixels and determines the smallest and the largest
gray value
Essentially, if percent is 0 you will obtain min/max as you'd expect however if you give a percentage it will subtract this percentage (as a pixel value) from either side of the histogram and give these values as min and max instead. If percent is 50 that then means min and max are the same and signify the median.
Let's look at a simplified example:
image in an image with 10 pixels who have the following values:
[0, 0, 1, 2, 3, 3, 3, 3, 4, 5]
The histogram would be like this:
0: 2
1: 1
2: 1
3: 4
4: 1
5: 1
If percent is 0 then min = 0 and max = 5.
Percent 10 would mean you take one pixel away at the edges of the histogram, thus min = 0 and max = 4...
percent 20, equals 2 pixels and thus min = 1 max = 3
percent 30, equals 3 pixels thus min = 2, max = 3
percent 50, min=max=3 which is the median
Related
I am trying to draw the flowchart for this question and stuck.
Given a list A of n numbers, count the number of times 0 occurs in the list.It might be an easy question but I am a beginner and don't have much idea about most questions. Please help me draw that!
You need to iterate through the list A and for each value, check if the value is equal to zero. If it is, add 1 to a running total.
In pseudo code:
total = 0
listA = [1, 5, 0, 2, 0, ...]
for i = 0 to len(listA):
if listA[i] == 0:
total = total + 1
next i
I am given this algorithmic problem, and need to find a way to return the count in a list S and another list L that is between some variable x and some variable y, inclusive, that runs in O(1) time:
I've issued a challenge against Jack. He will submit a list of his favorite years (from 0 to 2020). If Jack really likes a year,
he may list it multiple times. Since Jack comes up with this list on the fly, it is in no
particular order. Specifically, the list is not sorted, nor do years that appear in the list
multiple times appear next to each other in the list.
I will also submit such a list of years.
I then will ask Jack to pick a random year between 0 and 2020. Suppose Jack picks the year x.
At the same time, I will also then pick a random year between 0 and 2020. Suppose I
pick the year y. Without loss of generality, suppose that x ≤ y.
Once x and y are picked, Jack and I get a very short amount of time (perhaps 5
seconds) to decide if we want to re-do the process of selecting x and y.
If no one asks for a re-do, then we count the number of entries in Jack's list that are
between x and y inclusively and the number of entries in my list that are between x and
y inclusively.
More technically, here is the situation. You are given lists S and L of m and n integers,
respectively, in the range [0, k], representing the collections of years selected by Jack and
I. You may preprocess S and L in O(m+n+k) time. You must then give an algorithm
that runs in O(1) time – so that I can decide if I need to ask for a re-do – that solves the
following problem:
Input: Two integers, x as a member of [0,k] and y as a member of [0,k]
Output: the number of entries in S in the range [x, y], and the number of entries in L in [x, y].
For example, suppose S = {3, 1, 9, 2, 2, 3, 4}. Given x = 2 and y = 3, the returned count
would be 4.
I would prefer pseudocode; it helps me understand the problem a bit easier.
Implementing the approach of user3386109 taking care of edge case of x = 0.
user3386109 : Make a histogram, and then compute the accumulated sum for each entry in the histogram. Suppose S={3,1,9,2,2,3,4} and k is 9. The histogram is H={0,1,2,2,1,0,0,0,0,1}. After accumulating, H={0,1,3,5,6,6,6,6,6,7}. Given x=2 and y=3, the count is H[y] - H[x-1] = H[3] - H[1] = 5 - 1 = 4. Of course, x=0 is a corner case that has to be handled.
# INPUT
S = [3, 1, 9, 2, 2, 3, 4]
L = [2, 9, 4, 6, 8, 5, 3]
k = 9
x = 2
y = 3
# Histogram for S
S_hist = [0]*(k+1)
for element in S:
S_hist[element] = S_hist[element] + 1
# Storing prefix sum in S_hist
sum = S_hist[0]
for index in range(1,k+1):
sum = sum + S_hist[index]
S_hist[index] = sum
# Similar approach for L
# Histogram for L
L_hist = [0] * (k+1)
for element in L:
L_hist[element] = L_hist[element] + 1
# Stroing prefix sum in L_hist
sum = L_hist[0]
for index in range(1,k+1):
sum = sum + L_hist[index]
L_hist[index] = sum
# Finding number of elements between x and y (inclusive) in S
print("number of elements between x and y (inclusive) in S:")
if(x == 0):
print(S_hist[y])
else:
print(S_hist[y] - S_hist[x-1])
# Finding number of elements between x and y (inclusive) in S
print("number of elements between x and y (inclusive) in L:")
if(x == 0):
print(L_hist[y])
else:
print(L_hist[y] - L_hist[x-1])
Is there efficient way to downscale number of elements in array by decimal factor?
I want to downsize elements from one array by certain factor.
Example:
If I have 10 elements and need to scale down by factor 2.
1 2 3 4 5 6 7 8 9 10
scaled to
1.5 3.5 5.5 7.5 9.5
Grouping 2 by 2 and use arithmetic mean.
My problem is what if I need to downsize array with 10 elements to 6 elements? In theory I should group 1.6 elements and find their arithmetic mean, but how to do that?
Before suggesting a solution, let's define "downsize" in a more formal way. I would suggest this definition:
Downsizing starts with an array a[N] and produces an array b[M] such that the following is true:
M <= N - otherwise it would be upsizing, not downsizing
SUM(b) = (M/N) * SUM(a) - The sum is reduced proportionally to the number of elements
Elements of a participate in computation of b in the order of their occurrence in a
Let's consider your example of downsizing 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 to six elements. The total for your array is 55, so the total for the new array would be (6/10)*55 = 33. We can achieve this total in two steps:
Walk the array a totaling its elements until we've reached the integer part of N/M fraction (it must be an improper fraction by rule 1 above)
Let's say that a[i] was the last element of a that we could take as a whole in the current iteration. Take the fraction of a[i+1] equal to the fractional part of N/M
Continue to the next number starting with the remaining fraction of a[i+1]
Once you are done, your array b would contain M numbers totaling to SUM(a). Walk the array once more, and scale the result by N/M.
Here is how it works with your example:
b[0] = a[0] + (2/3)*a[1] = 2.33333
b[1] = (1/3)*a[1] + a[2] + (1/3)*a[3] = 5
b[2] = (2/3)*a[3] + a[4] = 7.66666
b[3] = a[5] + (2/3)*a[6] = 10.6666
b[4] = (1/3)*a[6] + a[7] + (1/3)*a[8] = 13.3333
b[5] = (2/3)*a[8] + a[9] = 16
--------
Total = 55
Scaling down by 6/10 produces the final result:
1.4 3 4.6 6.4 8 9.6 (Total = 33)
Here is a simple implementation in C++:
double need = ((double)a.size()) / b.size();
double have = 0;
size_t pos = 0;
for (size_t i = 0 ; i != a.size() ; i++) {
if (need >= have+1) {
b[pos] += a[i];
have++;
} else {
double frac = (need-have); // frac is less than 1 because of the "if" condition
b[pos++] += frac * a[i]; // frac of a[i] goes to current element of b
have = 1 - frac;
b[pos] += have * a[i]; // (1-frac) of a[i] goes to the next position of b
}
}
for (size_t i = 0 ; i != b.size() ; i++) {
b[i] /= need;
}
Demo.
You will need to resort to some form of interpolation, as the number of elements to average isn't integer.
You can consider computing the prefix sum of the array, i.e.
0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 10
yields by summation
0 1 2 3 4 5 6 7 8 9
1 3 6 10 15 21 28 36 45 55
Then perform linear interpolation to get the intermediate values that you are lacking, like at 0*, 10/6, 20/6, 30/5*, 40/6, 50/6, 60/6*. (Those with an asterisk are readily available).
0 1 10/6 2 3 20/6 4 5 6 40/6 7 8 50/6 9
1 3 15/3 6 10 35/3 15 21 28 100/3 36 45 145/3 55
Now you get fractional sums by subtracting values in pairs. The first average is
(15/3-1)/(10/6) = 12/5
I can't think of anything in the C++ library that will crank out something like this, all fully cooked and ready to go.
So you'll have to, pretty much, roll up your sleeves and go to work. At this point, the question of what's the "efficient" way of doing it boils down to its very basics. Which means:
1) Calculate how big the output array should be. Based on the description of the issue, you should be able to make that calculation even before looking at the values in the input array. You know the input array's size(), you can calculate the size() of the destination array.
2) So, you resize() the destination array up front. Now, you no longer need to worry about the time wasted in growing the size of the dynamic output array, incrementally, as you go through the input array, making your calculations.
3) So what's left is the actual work: iterating over the input array, and calculating the downsized values.
auto b=input_array.begin();
auto e=input_array.end();
auto p=output_array.begin();
Don't see many other options here, besides brute force iteration and calculations. Iterate from b to e, getting your samples, calculating each downsized value, and saving the resulting value into *p++.
This is what I have right now. It just finds the median with an odd amount of numbers.
def median(height):
height.sort()
x = len(height)
x -= 1
posn = x // 2
return height[posn]
"The median is the numeric value separating the higher half of a sample data set from the lower half. The median of a data set can be found by arranging all the values from lowest to highest value and picking the one in the middle. If there is an odd number of data values then the median will be the value in the middle. If there is an even number of data values the median is the mean of the two data values in the middle." - Source
For the data set 1, 1, 2, 5, 6, 6, 9 the median is 5.
For the data set 1, 1, 2, 6, 6, 9 the median is 4. It is the mean of 2 and 6 or, (2+6)/2 = 4.
i'd like to be able to calculate the 'mean brightest point' in a line of pixels. It's for a primitive 3D scanner.
for testing i simply stepped through the pixels and if the current pixel is brighter than the one before, the brightest point of that line will be set to the current pixel. This of course gives very jittery results throughout the image(s).
i'd like to get the 'average center of the brightness' instead, if that makes sense.
has to be a common thing, i'm simply lacking the right words for a google search.
Calculate the intensity-weighted average of the offset.
Given your example's intensities (guessed) and offsets:
0 0 0 0 1 3 2 3 1 0 0 0 0 0
1 2 3 4 5 6 7 8 9 10 11 12 13 14
this would give you (5+3*6+2*7+3*8+9)/(1+3+2+3+1) = 7
You're looking for 1D Convolution which takes a filter with which you "convolve" the image. For example, you can use a Median filter (borrowing example from Wikipedia)
x = [2 80 6 3]
y[1] = Median[2 2 80] = 2
y[2] = Median[2 80 6] = Median[2 6 80] = 6
y[3] = Median[80 6 3] = Median[3 6 80] = 6
y[4] = Median[6 3 3] = Median[3 3 6] = 3
so
y = [2 6 6 3]
So here, the window size is 3 since you're looking at 3 pixels at a time and replacing the pixel around this window with the median. A window of 3 means, we look at the first pixel before and first pixel after the pixel we're currently evaluating, 5 would mean 2 pixels before and after, etc.
For a mean filter, you do the same thing except replace the pixel around the window with the average of all the values, i.e.
x = [2 80 6 3]
y[1] = Mean[2 2 80] = 28
y[2] = Mean[2 80 6] = 29.33
y[3] = Mean[80 6 3] = 29.667
y[4] = Mean[6 3 3] = 4
so
y = [28 29.33 29.667 4]
So for your problem, y[3] is the "mean brightest point".
Note how the borders are handled for y[1] (no pixels before it) and y[4] (no pixels after it)- this example "replicates" the pixel near the border. Therefore, we generally "pad" an image with replicated or constant borders, convolve the image and then remove those borders.
This is a standard operation which you'll find in many computational packages.
your problem is like finding the longest sequence problem. once you are able to determine a sequence( the starting point and the length), the all that remains is finding the median, which is the central element.
for finding the sequence, definition of bright and dark has to be present, either relative -> previous value or couple of previous values. absolute: a fixed threshold.