I was new to Matlab,and this time I want to create a function for its image process.
Firstly, I download a picture from the Internet.Then I named it "map.jpg",and copy to my workspace.latter,I create a M_files and type the code into the files.
for example:
function y=mean_data(gray)
s=size(gray);
sum=0;
for i=1:s(1)
for j=1:s(2)
sum=sum+gray(i,j);
end
end
y=sum/(s(1)*s(2));
Finally,the difference happenend:
if I call the function in this way:
I=imread('map.jpg');
J=rgb2gray(I);
mean=mean_data(double(J))
the result will be OK.
However if I call in this way:
I=imread('map.jpg');
J=rgb2gray(I);
mean=mean_data(J)
the result will be zero.
So why does the result being so different?And thank you for helping me!!!
This is becuase the default output format of the data read by imread
is uint8 that is - 8 bit per R/G/B. With 8 bit you can't get any integer
higher than 255. Take a look:
>> uint8(250) + uint8(5)
ans =
255
>> uint8(250) + uint8(6)
ans =
255
So then, during division in your function this thing happens:
>> uint8(255) / 12345
ans =
0
However, when you use double() you change the representation of
your data to 64 bit floating point - a lot more room for representing big
numbers.
Instead of using your loop function you can use matlab's mean
function - it works well with uint8 format:
>> mean(uint8([255, 231]))
ans =
243
So you can use:
mean_dat = mean(mean(J));
% it is also not a good idea to name a variable 'mean'
% if you are going to use the mean function so I renamed
% your variable to mean_dat
Related
I'm relatively new to Z3 and experimenting with it in python. I've coded a program which returns the order in which different actions is performed, represented with a number. Z3 returns an integer representing the second the action starts.
Now I want to look at the model and see if there is an instance of time where nothing happens. To do this I made a list with only 0's and I want to change the index at the times where each action is being executed, to 1. For instance, if an action start at the 5th second and takes 8 seconds to be executed, the index 5 to 12 would be set to 1. Doing this with all the actions and then look for 0's in the list would hopefully give me the instances where nothing happens.
The problem is: I would like to write something like this for coding the problem
list_for_check = [0]*total_time
m = s.model()
for action in actions:
for index in range(m.evaluate(action.number) , m.evaluate(action.number) + action.time_it_takes):
list_for_check[index] = 1
But I get the error:
'IntNumRef' object cannot be interpreted as an integer
I've understood that Z3 isn't returning normal ints or bools in their models, but writing
if m.evaluate(action.boolean):
works, so I'm assuming the if is overwritten in a way, but this doesn't seem to be the case with range. So my question is: Is there a way to use range with Z3 ints? Or is there another way to do this?
The problem might also be that action.time_it_takes is an integer and adding a Z3int with a "normal" int doesn't work. (Done in the second part of the range).
I've also tried using int(m.evaluate(action.number)), but it doesn't work.
Thanks in advance :)
When you call evaluate it returns an IntNumRef, which is an internal z3 representation of an integer number inside z3. You need to call as_long() method of it to convert it to a Python number. Here's an example:
from z3 import *
s = Solver()
a = Int('a')
s.add(a > 4);
s.add(a < 7);
if s.check() == sat:
m = s.model()
print("a is %s" % m.evaluate(a))
print("Iterating from a to a+5:")
av = m.evaluate(a).as_long()
for index in range(av, av + 5):
print(index)
When I run this, I get:
a is 5
Iterating from a to a+5:
5
6
7
8
9
which is exactly what you're trying to achieve.
The method as_long() is defined here. Note that there are similar conversion functions from bit-vectors and rationals as well. You can search the z3py api using the interface at: https://z3prover.github.io/api/html/namespacez3py.html
I am trying to return a 4D array of image data from a function call in MATLAB. I'm not very advanced in MATLAB and I don't know what type of data I have to return from the function. Here is my function:
function classimg = loadImages(classdir,ext)
% create array of all images in directory
neg = dir([classdir ext]);
% get size of array (to loop through images)
numFileNeg = max(size(neg));
% create a 4D array to store our images
classimg = zeros(51,51,3,numFileNeg);
% loop through directory
for i=1:numFileNeg
classimg(:,:,:,i) = imread([myDir neg(i).name]);
end
end
Here is the function call:
negativeImgs = loadImages("C:\Users\example\Documents\TrainingImages\negatives\","*.jpg");
I cannot find any online documentation for the return type? Does anyone know what this would be? classimg is populated correctly so the code inner works.
You initialize classimg to be a 51x51x3xnumFileNeg matrix of zeros. You use the zeros function, so the datatype is double. To see this clearly, call your function from the command window, and then type "whos" to see both the size and datatype of classimg.
As Mike correctly points out, since you initialize classimg using zeros, and the default data type is double, your image data will be converted to double from whatever data type imread returns (often uint8).
If you would like classimg to be the same data type as your images (which I'm assuming all have the same type), you can load one image, get its class, and initialize classimg with that specific class. Here's how you could rewrite your function:
function classimg = loadImages(classdir, ext)
neg = dir(fullfile(classdir, ext));
numFileNeg = numel(neg);
tempImage = imread(fullfile(classdir, neg(1).name));
classimg = zeros(51, 51, 3, numFileNeg, class(tempImage));
classimg(:, :, :, 1) = tempImage;
for i = 2:numFileNeg
classimg(:, :, :, i) = imread(fullfile(classdir, neg(i).name));
end
end
Note that I made a couple of other changes. I used fullfile instead of concatenation of the directory and file names, since it handles any issues with file separators for you. I also used numel to get the number of files as Justin suggested in a comment.
Keep getting this error sometimes when mid is ZERO:
Invalid procedure call or argument: 'Mid'
How would I fix this?
Function CreateRandomString(iSize)
Const VALID_TEXT = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890"
Dim sNewSearchTag
Dim I
For I = 0 To iSize
Randomize
sNewSearchTag = sNewSearchTag & Mid(VALID_TEXT,Round(Rnd * Len(VALID_TEXT)),1)
Next
CreateRandomString = sNewSearchTag
End Function
For the random range to be correct you need to make sure the random value generated is between 1 and the length of the VALID_TEXT string value.
The simple formula to do this using Rnd() is
(Rnd() * Len(VALID_TEXT)) + 1
also move Randomize() outside the loop, as it is you'll just make it less random as you're resetting the seed with every iteration of the loop.
The reason for the error is Mid() expects a valid start and size, which a zero value is not. See this question for more information.
More information about random number ranges can be found in this answer to another question.
The second argument of Mid is 1 based. That means that if you did:
Mid(VALID_TEXT,1,1)
you will get "a", not "b" as you might be expecting.
An easy fix would be to add 1 to the second argument, but then you'll run into the same problem on the top end. Typically people will round a random number down after multiplying it instead of using Math.Round, either view Math.Floor or Integer truncation.
I am writing a simple code in matlab which has the purpose of creating the histogram of a grayscale image without using the function hist. I am stuck at the point in which mathlab displays the error "Subscript indices must either be real positive integers or logicals." Can you help me finding where is the wrong indices?
indirizzo='file.jpg';
immagine=imread(indirizzo);
immaginebn=rgb2gray(immagine);
n=zerps(0,255);
for x=0:255;
numeroennesimo=sum(sum(immaginebn==x));
n(x)=numeroennesimo;
end
plot(x,n)
you cant use 0 as index. Either make n(x+1) or for x = 1:256 and substract the 1 in your comparison. And there is a typo, I guess it means zeros instead of zerps, which also doesnt work with a 0. And one more, your plot will also not work as the x has only a size of 1 while n is an array of 266 and for a histogram I would use a barplot instead.
indirizzo='file.jpg';
immagine=imread(indirizzo);
immaginebn=rgb2gray(immagine);
n=zeros(1,256);
for x=0:255;
numeroennesimo=sum(sum(immaginebn==x-1));
n(x+1)=numeroennesimo;
end
bar(0:255,n)
or
indirizzo='file.jpg';
immagine=imread(indirizzo);
immaginebn=rgb2gray(immagine);
n=zeros(1,256);
xplot=zeros(1,256);
for x=1:256;
numeroennesimo=sum(sum(immaginebn==x-1));
n(x)=numeroennesimo;
xplot(x) = x-1;
end
plot(xplot,n)
I have a data file where decimal points aren't specified for a decimal number. The number is just described in the layout for the data file as first 2 digits as real and next 2 digits as decimal and it varies for different fields, the real and decimal part
So an actual number 12345.6789 is specified as 123456789. When I want this to be rounded off to 2 decimal points to match the value in application, I use the below logic
Public Function Rounding(NumberValue, DecimalPoints, RoundOff)
Rounder= Roundoff+1
Difference = DecimalPoints - Rounder
NumberValue = Mid(NumberValue, 1, Len(NumberValue)-Difference)
RealNumber=Mid(NumberValue,1,Len(NumberValue)-Rounder)
DecimalNumber=Right(NumberValue,Rounder)
NumberValue = RealNumber&"."&DecimalNumber
NumberValue = Cdbl(NumberValue)
NumberValue = Round(NumberValue, Roundoff)
Rounding = FormatNumber(NumberValue,Difference+1,,,0)
End Function
However the problem with this logic is that I am not able to round off decimals when the number has 0 as the decimal value
For an Example, lets take 12345.0000 which I want to round off to 2 decimal points
My function returns it as 12345 whereas I want this to be returned as 12345.00
Any ideas on how this logic could be tweaked to get the desired output or is that not possible at all?
To get the decimal places, use the Formatnumber function. See http://msdn.microsoft.com/en-us/library/ws343esk(v=vs.84).aspx - the default is normally 2 decimal places, but it is region settings specific when using the defaults.
Your script also has a small issue if the decimalpoints variable matches the roundoff variable - it will not populate Rounding with a result. I am also not sure why you are comparing DecimalPoints to Roundoff (-1) ?
I've revised the entire routine - it should do what you want (although I don't know what values you are feeding it) - So now it will work like this:
Doing 4 digits:
Rounding (123450001, 4, 2)
Result:
12345.00
Doing 2 digits:
Rounding (123450001, 2, 2)
Result:
1234500.01
Doing 4 digits (increments if > .5)
Rounding (876512345678, 8, 4)
Result:
8765.1235
Revised simplified function that should do everything you are asking:
Public Function Rounding(NumberValue, DecimalPoints, RoundOff )
RealNumber = Mid(NumberValue, 1, Len(NumberValue)-DecimalPoints)
DecimalNumber = Round("." & Right(NumberValue,DecimalPoints), RoundOff)
Rounding = FormatNumber(RealNumber + DecimalNumber,RoundOff,,,0)
End Function
Here's a working version of your Function:
Public Function Rounding(NumberValue, DecimalPoints, RoundOff)
RealNumber=left(NumberValue,Len(NumberValue)-DecimalPoints)
DecimalNumber="." & Right(NumberValue,DecimalPoints)
NumberValue = RealNumber + DecimalNumber
NumberValue = Round(NumberValue,RoundOff)
Rounding = FormatNumber(NumberValue, RoundOff,,,0)
End Function
I'm pretty sure you won't be able to use the Round() function for what you need. Take a look at the FormatNumber() or FormatCurrency() functions as they have the option to "IncludeLeadingZero".
Take a look at the answer from the following link for more information:
vbscript round to 2 decimal places using Ccur