I wrote a program to calculate a square finite difference matrix, where you can enter the number of rows (equals the number of columns) -> this is stored in the variable matrix. The program works fine:
program fin_diff_matrix
implicit none
integer, dimension(:,:), allocatable :: A
integer :: matrix,i,j
print *,'Enter elements:'
read *, matrix
allocate(A(matrix,matrix))
A = 0
A(1,1) = 2
A(1,2) = -1
A(matrix,matrix) = 2
A(matrix,matrix-1) = -1
do j=2,matrix-1
A(j,j-1) = -1
A(j,j) = 2
A(j,j+1) = -1
end do
print *, 'Matrix A: '
write(*,1) A
1 format(6i10)
end program fin_diff_matrix
For the output I want that matrix is formatted for the output, e.g. if the user enters 6 rows the output should also look like:
2 -1 0 0 0 0
-1 2 -1 0 0 0
0 -1 2 -1 0 0
0 0 -1 2 -1 0
0 0 0 -1 2 -1
0 0 0 0 -1 2
The output of the format should also be variable, e.g. if the user enters 10, the output should also be formatted in 10 columns. Research on the Internet gave the following solution for the format statement with angle brackets:
1 format(<matrix>i<10)
If I compile with gfortran in Linux I always get the following error in the terminal:
fin_diff_matrix.f95:37.12:
1 format(<matrix>i10)
1
Error: Unexpected element '<' in format string at (1)
fin_diff_matrix.f95:35.11:
write(*,1) A
1
Error: FORMAT label 1 at (1) not defined
What doesn't that work and what is my mistake?
The syntax you are trying to use is non-standard, it works only in some compilers and I discourage using it.
Also, forget the FORMAT() statements for good, they are obsolete.
You can get your own number inside the format string when you construct it yourself from several parts
character(80) :: form
form = '( (i10,1x))'
write(form(2:11),'(i10)') matrix
write(*,form) A
You can also write your matrix in a loop per row and then you can use an arbitrarily large count number or a * in Fortran 2008.
do i = 1, matrix
write(*,'(999(i10,1x))') A(:,i)
end do
do i = 1, matrix
write(*,'(*(i10,1x))') A
end do
Just check if I did not transpose the matrix inadvertently.
Related
i seen a post on the site about it and i didn't understand the answer, can i get explanation please:
question:
Write code to determine if a number is divisible by 3. The input to the function is a single bit, 0 or 1, and the output should be 1 if the number received so far is the binary representation of a number divisible by 3, otherwise zero.
Examples:
input "0": (0) output 1
inputs "1,0,0": (4) output 0
inputs "1,1,0,0": (6) output 1
This is based on an interview question. I ask for a drawing of logic gates but since this is stackoverflow I'll accept any coding language. Bonus points for a hardware implementation (verilog etc).
Part a (easy): First input is the MSB.
Part b (a little harder): First input is the LSB.
Part c (difficult): Which one is faster and smaller, (a) or (b)? (Not theoretically in the Big-O sense, but practically faster/smaller.) Now take the slower/bigger one and make it as fast/small as the faster/smaller one.
answer:
State table for LSB:
S I S' O
0 0 0 1
0 1 1 0
1 0 2 0
1 1 0 1
2 0 1 0
2 1 2 0
Explanation: 0 is divisible by three. 0 << 1 + 0 = 0. Repeat using S = (S << 1 + I) % 3 and O = 1 if S == 0.
State table for MSB:
S I S' O
0 0 0 1
0 1 2 0
1 0 1 0
1 1 0 1
2 0 2 0
2 1 1 0
Explanation: 0 is divisible by three. 0 >> 1 + 0 = 0. Repeat using S = (S >> 1 + I) % 3 and O = 1 if S == 0.
S' is different from above, but O works the same, since S' is 0 for the same cases (00 and 11). Since O is the same in both cases, O_LSB = O_MSB, so to make MSB as short as LSB, or vice-versa, just use the shortest of both.
thanks for the answers in advanced.
Well, I suppose the question isn't entirely off-topic, since you asked about logic design, but you'll have to do the coding yourself.
You have 3 states in the S column. These track the value of the current full input mod 3. So, S0 means the current input mod 3 is 0, and so is divisible by 0 (remember also that 0 is divisible by 3). S1 means the remainder is 1, S2 means that the remainder is 2.
The I column gives the current input (0 or 1), and S' gives the next state (in other words, the new number mod 3).
For 'LSB', the new number is the old number << 1, plus either 0 or 1. Write out the table. For starters, if the old modulo was 0, then the new modulo will be 0 if the input bit was 0, and will be 1 if the new input was 1. This gives you the first 2 rows in the first table. Filling in the rest is left as an exercise for you.
Note that the O column is just 1 if the next state is 0, as expected.
I am trying to alter some voxel values in Matlab.
I am using the following code:
for p=1:100
Vol(Vol(:,:,p) > 0) = 65535; %altering voxel values in the volume to 65535 if value > 0.
end
Unfortunately, I find all the values being altered, as if the condition is not working, although if i write Vol(Vol(:,:,1)>0)= 65535 immediately in the command line it works perfectly.
Any clue where the error is?
The reason why is because you are not indexing each slice properly in your volume. When you are doing this for loop, what will happen is that the Boolean condition that is provided in Vol is modifying only the first channel.
Consider this small example. Let's create a 3 x 3 x 3 matrix of all 1s.
A = ones(3,3,3)
A(:,:,1) =
1 1 1
1 1 1
1 1 1
A(:,:,2) =
1 1 1
1 1 1
1 1 1
A(:,:,3) =
1 1 1
1 1 1
1 1 1
Let's set the first slice all to 65535 according to your condition:
A(A(:,:,1) > 0) = 65535
A(:,:,1) =
65535 65535 65535
65535 65535 65535
65535 65535 65535
A(:,:,2) =
1 1 1
1 1 1
1 1 1
A(:,:,3) =
1 1 1
1 1 1
1 1 1
This certainly works as we expect. Now let's try going to the second channel:
A(A(:,:,2) > 0) = 65535
A(:,:,1) =
65535 65535 65535
65535 65535 65535
65535 65535 65535
A(:,:,2) =
1 1 1
1 1 1
1 1 1
A(:,:,3) =
1 1 1
1 1 1
1 1 1
Oh no! It didn't work! It only worked for the first channel.... why? The reason why is because A(:,:,1) or any other channel provides a 2D matrix. If you provide a single 2D matrix, it only modifies the first slice of the volume. As such, as your loop keeps progressing, only the first channel gets modified (if at all). If you wanted to modify the second channel, you would have to create a 3D matrix, where the first slice would have all logical false, while the second slice contains the Boolean mask from Vol(:,:,2) > 0.
The 3D slicing stuff is probably complicated, especially for someone new to MATLAB. As such, I would recommend you do this to make things simpler. If you want to modify each slice, consider placing each binary mask as a temporary variable, modifying that temporary variable, then manually assigning this back to each slice. In other words:
for p=1:100
temp = Vol(:,:,p); %//Extract p'th channel
temp(temp > 0) = 65535; %// Find non-zero pixels and set to 65535
Vol(:,:,p) = temp; %// Set back to p'th channel.
end
Another recommended suggestion
Instead of using for loops, I would like to recommend this simple one-liner:
Vol(Vol > 0) = 65535;
This will automatically create a 3D Boolean matrix that will index Vol, and it will find those locations that are greater than 0, and set all of those locations to 65535. This avoids the need of any unnecessary for loops. This one-line essentially performs what the above for loop is doing, but is much more quicker... and I daresay much easier to read.
For your problem, I would just do :
Vol(Vol(:,:,1:100) > 0) = 65535;
No need for loop.
I need an algorithm in Matlab which counts how many adjacent and non-overlapping (1,1) I have in each row of a matrix A mx(n*2) without using loops. E.g.
A=[1 1 1 0 1 1 0 0 0 1; 1 0 1 1 1 1 0 0 1 1] %m=2, n=5
Then I want
B=[2;3] %mx1
Specific case
Assuming A to have ones and zeros only, this could be one way -
B = sum(reshape(sum(reshape(A',2,[]))==2,size(A,2)/2,[]))
General case
If you are looking for a general approach that must work for all integers and a case where you can specify the pattern of numbers, you may use this -
patt = [0 1] %%// pattern to be found out
B = sum(reshape(ismember(reshape(A',2,[])',patt,'rows'),[],2))
Output
With patt = [1 1], B = [2 3]
With patt = [0 1], B = [1 0]
you can use transpose then reshape so each consecutive values will now be in a row, then compare the top and bottom row (boolean compare or compare the sum of each row to 2), then sum the result of the comparison and reshape the result to your liking.
in code, it would look like:
A=[1 1 1 0 1 1 0 0 0 1; 1 0 1 1 1 1 0 0 1 1] ;
m = size(A,1) ;
n = size(A,2)/2 ;
Atemp = reshape(A.' , 2 , [] , m ) ;
B = squeeze(sum(sum(Atemp)==2))
You could pack everything in one line of code if you want, but several lines is usually easier for comprehension. For clarity, the Atemp matrix looks like that:
Atemp(:,:,1) =
1 1 1 0 0
1 0 1 0 1
Atemp(:,:,2) =
1 1 1 0 1
0 1 1 0 1
You'll notice that each row of the original A matrix has been broken down in 2 rows element-wise. The second line will simply compare the sum of each row with 2, then sum the valid result of the comparisons.
The squeeze command is only to remove the singleton dimensions not necessary anymore.
you can use imresize , for example
imresize(A,[size(A,1),size(A,2)/2])>0.8
ans =
1 0 1 0 0
0 1 1 0 1
this places 1 where you have [1 1] pairs... then you can just use sum
For any pair type [x y] you can :
x=0; y=1;
R(size(A,1),size(A,2)/2)=0; % prealocarting memory
for n=1:size(A,1)
b=[A(n,1:2:end)' A(n,2:2:end)']
try
R(n,find(b(:,1)==x & b(:,2)==y))=1;
end
end
R =
0 0 0 0 1
0 0 0 0 0
With diff (to detect start and end of each run of ones) and accumarray (to group runs of the same row; each run contributes half its length rounded down):
B = diff([zeros(1,size(A,1)); A.'; zeros(1,size(A,1))]); %'// columnwise is easier
[is js] = find(B==1); %// rows and columns of starts of runs of ones
[ie je] = find(B==-1); %// rows and columns of ends of runs of ones
result = accumarray(js, floor((ie-is)/2)); %// sum values for each row of A
Suppose you have this program
Subroutine readDIM which reads the dimensions (rows, columns) of a matrix from a txt file. (In order to simplify, let it be an INTEGER). ReadDIM works using tokens and it works fine by assumption.
A text file containing for example:
1 2 3 4
1 2 20 5
3 0 333 3
Returns nrow = 3, ncol = 4
Since readDIM has given the true dimensions of the matrix, I want to allocate space to:
REAL, DIMENSION (:,:), ALLOCATABLE :: vMatrix
To read the matrix from a txt file and to store it into the 2d-array. So I've written the following
SUBROUTINE buildVMatrix
OPEN(UNIT=1, FILE = filename, STATUS ='OLD',IOSTAT=ios);
ALLOCATE(vMatrix(nrow,ncol));
WRITE(*,*) "Register matrix from file:", filename;
WRITE(*,*) "-------------------------------------------------------";
DO i = 1, UBOUND(vMatrix,1)
READ(1,*, IOSTAT = ios) (vMatrix(i,j),j=1,UBOUND(vMatrix,2));
!IF(ios /= 0 ) EXIT
END DO
CLOSE(1)
END SUBROUTINE
When I print vMatrix the output is:
matrix.txt : 1 2 3 4 buildVMatrix output (once printed) 1 2 3 4
1 2 20 5 1 2 20 5
3 0 333 3 3 0 333 0
It doesn't read the last number. I know it's caused by the DO loop inside buildVMatrix, but can't explain myself this and have no idea how to fix it writing a different code.
It's because there's no line ending at the last line in your txt file, try to type a return after the last number.
Hi I am struggling with reading data from a file quickly enough. ( Currently left for 4hrs, then crashed) must be a simpler way.
The text file looks similar like this:
From To
1 5
3 2
2 1
4 3
From this I want to form a matrix so that there is a 1 in the according [m,n]
The current code is:
function [z] = reed (A)
[m,n]=size(A);
i=1;
while (i <= n)
z(A(1,i),A(2,i))=1;
i=i+1;
end
Which output the following matrix, z:
z =
0 0 0 0 1
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
My actual file has 280,000,000 links to and from, this code is too slow for this size file. Does anybody know a much faster was to do this in matlab?
thanks
You can do something along the lines of the following:
>> A = zeros(4,5);
>> B = importdata('testcase.txt');
>> A(sub2ind(size(A),B.data(:,1),B.data(:,2))) = 1;
My test case, 'testcase.txt' contains your sample data:
From To
1 5
3 2
2 1
4 3
The result would be:
>> A
A =
0 0 0 0 1
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
EDIT - 1
After taking a look at your data, it seems that even if you modify this code appropriately, you may not have enough memory to execute it as the matrix A would become too large.
As such, you can use sparse matrices to achieve the same as given below:
>> B = importdata('web-Stanford.txt');
>> A = sparse(B.data(:,1),B.data(:,2),1,max(max(B.data)),max(max(B.data)));
This would be the approach I'd recommend as your A matrix will have a size of [281903,281903] which would usually be too large to handle due to memory constraints. A sparse matrix on the other hand, maintains only those matrix entries which are non-zero, thus saving on a lot of space. In most cases, you can use sparse matrices more-or-less as you use normal matrices.
More information about the sparse command is given here.
EDIT - 2
I'm not sure why it isn't working for you. Here's a screenshot of how I did it in case that helps:
EDIT - 3
It seems that you're getting a double matrix in B while I'm getting a struct. I'm not sure why this is happening; I can only speculate that you deleted the header lines from the input file before you used importdata.
Basically it's just that my B.data is the same as your B. As such, you should be able to use the following instead:
>> A = sparse(B(:,1),B(:,2),1,max(max(B)),max(max(B)));