How to remove several columns from a matrix - matrix

Integer :: NBE,ierr,RN,i,j
Real(kind=8), allocatable :: AA1(:,:),AA2(:,:)
NBE=40
RN=3*NBE-2
Allocate(AA1(3*NBE,3*NBE),AA2(3*NBE,RN),stat=ierr)
If (ierr .ne. 0) Then
print *, 'allocate steps failed 1'
pause
End If
Do i=1,3*NBE
Do j=1,3*NBE
AA1(i,j)=1
End Do
End Do
I want to remove columns 97 and 113 from the matrix AA1 and then this matrix becomes AA2. I just want to know if any command of Fortran can realize this operation?

Alexander Vogt's answer gives the concept of using a vector subscript to select the elements of the array for inclusion. That answer constructs the vector subscript array using
[(i,i=1,96),(i=98,112),(i=114,3*NBE)]
Some may consider
AA2 = AA1(:,[(i,i=1,96),(i=98,112),(i=114,3*NBE)])
to be less than clear in reading. One could use a "temporary" index vector
integer selected_columns(RN)
selected_columns = [(i,i=1,96),(i=98,112),(i=114,3*NBE)]
AA2 = AA1(:,selected_columns)
but that doesn't address the array constructor being not nice, especially in more complicated cases. Instead, we can create a mask and use our common techniques:
logical column_wanted(3*NBE)
integer, allocatable :: selected_columns(:)
! Create a mask of whether a column is wanted
column_wanted = .TRUE.
column_wanted([97,113]) = .FALSE.
! Create a list of indexes of wanted columns
selected_columns = PACK([(i,i=1,3*NBE)],column_wanted)
AA2 = AA1(:,selected_columns)

Here's a simple one liner:
AA2 = AA1(:,[(i,i=1,96),(i=98,112),(i=114,3*NBE)])
Explanation:
(Inner part) Construct a temporary array for the indices [1,...,96,98,...,112,114,...,3*NBE]
(Outer part) Copy the matrix and only consider the columns in the index array
OK, I yield to #IanBush... Even simpler would be to do three dedicated assignments:
AA2(:,1:96) = AA1(:,1:96)
AA2(:,97:111) = AA1(:,98:112)
AA2(:,112:) = AA1(:,114:)

I don't have a fortran compiler here at home, so I cant test it. But I'd do something line this:
i = 0
DO j = 1, 3*NBE
IF (j == 97 .OR. j == 113) CYCLE
i = i + 1
AA2(:, i) = AA1(:, j)
END DO
The CYCLE command means that the rest of the loop should not be executed any more and the next iteration should start. Thereby, i will not get incremented, so when j=96, then i=96, when j=98 then i=97, and when j=114 then i=112.
A few more words: Due to Fortran's memory layout, you want to cycle over the first index the fastest, and so forth. So your code would run faster if you changed it to:
Do j=1,3*NBE ! Outer loop over second index
Do i=1,3*NBE ! Inner loop over first index
AA1(i,j)=1
End Do
End Do
(Of course, such an easy initialisation can be done even easier with just AA1(:,:) = 1 of just AA1 = 1.

Related

How to find the dimension of a matrix?

I have a matrix that contains both character and reals and I want a program that reads this matrix (finds the dimensions by itself). Here is my code:
! A fortran95 program for G95
Program Project2nd
implicit none
character(len=40), allocatable :: a(:,:)
integer i,j,k,n,m,l,st
character(len=40) d
n=0; m=1; j=1;
open(10,file=&
'/Users/dariakowsari/Documents/Physics/Programming/Fortran95-Projects/Project2nd/input.txt', &
IOstat=st)
do while (st == 0)
read(10,*,IOstat=st) d
n=n+1
end do
st=0
do j=1,m
do while (st == 0)
allocate(a(1,m))
read(10,*,IOstat=st) (a(1,j),j=1,m)
m=m+1
deallocate(a)
end do
print*, n,m
end
Here is my Matrix:
a b 13 15.5 13.2
c d 16 16.75 19
e f 19.2 12.2 18.2
With this code I got (3,2) for the dimensions of my matrix.
There are a few errors in your example code which means it doesn't compile for me but after a few changes I managed to get a similar result to you.
*Update: As noted by #francescalus in the comments to my other (now deleted) answer, that approach involved undefined behaviour and as such is not an appropriate solution. This arose from trying to read more elements from the file than were present.)
Here's an alternative approach, which should avoid this undefined behaviour, but is probably pretty inefficient.
Program Project2nd
implicit none
character(len=40), allocatable :: a(:)
integer, allocatable :: ind(:)
integer, parameter :: maxElements = 100
integer i,j,n,m,st
character(len=40) d
n=0;
open(10,file='mat.txt',IOstat=st)
!Find number of lines
do while (st == 0)
read(10,*,IOstat=st) d
if(st ==0) n=n+1
end do
!Move back to the start of the file
rewind(10)
!Read all of the data
do m=n,maxElements,n
allocate(a(m))
read(10,*,IOstat=st) a
deallocate(a)
rewind(10)
if(st.ne.0) exit
enddo
m = m -n !Need to roll back m by one iteration to get the last which worked.
if(mod(m,n).ne.0) then
print*,"Error: Number of elements not divisible by number of rows."
stop
endif
!Number of columns = n_elements/nrow
m=m/n
print*, n,m
end Program Project2nd
Essentially this uses the same code as you had for counting the number of lines, however note that you only want to increment n when the read was successful (i.e. st==0). Note we do not exit the whilst block as soon as st becomes non-zero, it is only once we reach the end of the whilst block. After that we need to rewind the file so that the next read starts at the start of the file.
In a previous comment you mentioned that you'd rather not have to specify maxElement if you really want to avoid this then replace the second do loop with something like
st = 0 ; m = n
do while (st==0)
allocate(a(m))
read(10,*,IOstat=st) a
deallocate(a)
rewind(10)
if(st.ne.0) then
m = m - n !Go back to value of m that worked
exit
endif
m=m+n
enddo
here is how to do w/o rewinding.
implicit none
character(len=100) wholeline
character(len=20), allocatable :: c(:)
integer iline,io,ni,nums
open(20,file='testin.dat')
iline=0
do while(.true.)
read(20,'(a)',iostat=io)wholeline
if(io.ne.0)exit
iline=iline+1
ni=lineitems(wholeline)
allocate(c(ni))
read(wholeline,*)c
nums=ctnums(c)
write(*,*)'line',iline,' contains ',ni,'items',nums,
$ 'are numbers'
deallocate(c)
enddo
write(*,*)'total lines is ',iline
contains
integer function ctnums(c)
! count the number of items in a character array that are numbers
! this is a template,
! obviously you could assign the numbers to a real array here
character(len=*), allocatable :: c(:)
real f
integer i,io
ctnums=0
do i = 1,size(c)
read(c(i),*,iostat=io)f
if(io.eq.0)ctnums=ctnums+1
enddo
end function
integer function lineitems(line)
! count the number of items in a space delimited string
integer,parameter ::maxitems=100
character(len=*) line
character(len=80) :: c(maxitems)
integer iline,io
lineitems=0
do iline=1,maxitems
read(line,*,iostat=io)c(:iline)
if(io.ne.0)return
lineitems=iline
enddo
if(lineitems.eq.maxitems)write(*,*)'warning maxitems reached'
end function
end
output
line 1 contains 5 items 3 are numbers
line 2 contains 5 items 3 are numbers
total lines is 2

Read array of unknown size from keyboard

I want to insert an unknown number of values in an array (no matter the order).
I could first read how many values are to be inserted, then allocate the allocatable array, and finally read its values, as in the following code
PROGRAM try
IMPLICIT NONE
INTEGER :: N
REAL, DIMENSION(:), ALLOCATABLE :: x
WRITE (*,*) "how many values?"
READ (*,*) N
ALLOCATE(x(N))
WRITE (*,*) "insert the values"
READ (*,*) x
END PROGRAM
What if I want to insert the values without declaring how many before allocating the array?
I think I should use a DO WHILE cycle to insert the values in ascending order, till a descending value is insert, thus indicating the sequence is ended. I think a part of the code would be the following,
index = 1
WRITE(*,*) x
READ(*,*) x(index)
exit = .FALSE.
DO WHILE (exit.EQV..FALSE.)
index = index + 1
READ(*,*) x(index)
IF (x(index)>x(index-1)) THEN
exit = .TRUE.
index = index - 1
END IF
END DO
How to declare the array x?
I tried with the following solution, building on the concept "a lot of memory allocation and reallocation" expressed by #High Performance Mark.
PROGRAM COEFFS
USE COMPACT
IMPLICIT NONE
REAL, DIMENSION(:), ALLOCATABLE :: x,x2
INTEGER :: nL,nR,nT,index,oL,oR
LOGICAL :: exit
WRITE(*,*) "Input an increasing sequence of reals (end the sequence &
& with the first decreasing element, which will be discarded):"
index = 1
ALLOCATE(x(index))
READ(*,*) x(index)
ALLOCATE(x2(index))
x2 = x
DEALLOCATE(x)
exit = .FALSE.
DO WHILE (exit.EQV..FALSE.)
index = index + 1
ALLOCATE(x(index))
x(1:index-1) = x2
READ(*,*) x(index)
DEALLOCATE(x2)
ALLOCATE(x2(index))
x2 = x
DEALLOCATE(x)
IF (x2(index)<x2(index-1)) THEN
exit = .TRUE.
index = index - 1
ALLOCATE(x(index))
x = x2(1:index)
END IF
END DO
DEALLOCATE(x2)
WRITE(*,*) "x = ", x
END PROGRAM
With the array being input by keyboard, I don't think allocation/reallocation is a problem, since it happens at a much higher speed than that of my fingers typing the values, doesn't it?
Still I think the code could be made better. For instance, using two arrays is the only way to take advantage of allocation/reallocation?

Use of IF statement with matrices in fortran

I want to go through a matrix and check if any block of it is the same as a predefined unit. Here is my code. 'sd5' is the 2 by 2 predefined unit.
ALLOCATE (fList((n-1)**2,3))
fList = 0
p = 1
DO i = 1, n-1, 1
DO j = 1, n-1, 1
IF (TEST(i:i+1, j:j+1) == sd5) THEN
fList(p,1:3) = (i, j+1, 101) ! 101 should be replaced by submatrix number
END IF
p = p+1
END DO
END DO
The problem seems to be in the IF statement as four logical statements are returned in TEST(i:i+1, j:j+1) == sd5. I get this error:
Error: IF clause at (1) requires a scalar LOGICAL expression
I get another error:
fList(p,1:3) = (i, j+1, 101) ! 101 should be replaced by sub matrix number
1
Error: Expected PARAMETER symbol in complex constant at (1)
I do not understand this error, as all variables are integer which I defined.
First, if statements require scalar clauses.
(TEST(i:i+1, j:j+1) == sd5)
results in a 2x2 matrix containing .true. or .false.. Since you want to check all entries, the statement should read
IF ( all( TEST(i:i+1, j:j+1) == sd5) ) THEN
[ You could also use any if only one matching entry is sufficient. ]
The second statement is a little tricky, since you do not state what you want to achieve. As it is, it is not what you would expect. My guess is that you are trying to store a vector of length three, and the assignment should read
fList(p,1:3) = (/ i, j+1, 101 /)
or
fList(p,1:3) = [ i, j+1, 101 ]
The syntax you provided is in fact used to specify complex constants:
( Real, Imag )
In this form, Real and Imag need to be constants or literals themselves, cf. the Fortran 2008 Standard, R417.

How can I count occurrence of numbers in matrix in fortran?

I'm using fortran 90, and I hope to count the number of occurence, when two numbers appears in an array.
flag=0
q=0
do k=1,ncolumns
if (conn(m,k)==i .and. conn(m,k)==l) flag=1
enddo
if (flag==1) q=q+1
write (*,*) q
Here, conn(m,k) is the matrix, made up of m lines and k columns. I want to read the conn(m,k), and count the number of occurrence when both number i and l are included in conn(m,k). I know above code will not work because it prints out only 0, since that if loop have a problem. But I cannot use '.or.' because I want the count when i and l both are included in the conn(m,k). How can I check both number i and l are included in conn?
I modified the code above like
ncolumns=2
flag=0
q=0
do k=1,ncolumns
!!!if (conn(m,k)==i .and. conn(m,k)==l) flag=1
if (((conn(m,1)==i).and.(conn(m,2)==l)).or.((conn(m,1)==l).and.(conn(m,2)==i))) flag=1
enddo
if (flag==1) q=q+1
write (*,*) q
This works fine, but as you can see, this code is ridiculous since I need to manually define k, specially when 'ncolumns' is huge number. How can I do this with index?
Likewise, how can I check 2 or more specific numbers are included in the matrix like conn(m,k) in fortran? Thanks.
Something like this ought to do what you want:
nums = [2,12,-4,99] ! an array of the numbers you're looking for
q = 0 ! the count of rows containing all the numbers in nums
DO ix = 1, SIZE(conn,1) ! the number of rows in conn
nmatches = 0 ! the number of elements of nums found in conn(ix,:)
DO jx = 1, SIZE(nums)
IF(ANY(conn(ix,:)==nums(jx))) nmatches = nmatches+1 ! figure this out yourself
END DO
! if there are as many matches in this row as there are elements in nums, add 1 to q
IF(nmatches==SIZE(nums)) q = q+1
END DO
You can also use a dummy matrix (dummy_mat) to populate with ones where the values are located in the matrix you are searching (value_mat) and then sum the dummy matrix to get the count (num_entries):
nums = [2,12,-4,99]
do i=1,size(nums) ! loop over the values you are looking for
dummy_mat = 0 ! zero out dummy matrix that is the same size as your value matrix
where (value_mat(:,:) == nums(i))
dummy_mat(:,:) = 1
end where
num_entries(i) = SUM(dummy_mat)
end do
From the comment "if there are 3 lines in conn which have two elements (such as 3 and 12) together, the printed q should be 3".
You can do this with a single loop if you have Fortran95 (I forget if it is in the 90 spec) or later.
Here is an example:
Program Test_Count
Implicit None
Real(Kind=8), Dimension(3) :: nums = (/-2.0_8 , -3.0_8 , -4.0_8/)
Real(Kind=8), Dimension(4,4) :: test
Logical, Dimension(4) :: Mask
Integer :: i,j,NumberOfLines
! Fill test
Do i = 1,4
Do j = 1,4
test(i,j) = -4.0_8 + (j -1)
End Do
End Do
! Count the row that have all the nums in them
Mask = any(test == nums(1),2)
Do i = 2,3
Mask = Mask .and. any(test == num2(i),2)
End Do
NumberOfLines = count(Mask)
Write(*,*) NumberOfLines ! Prints out 4.
End Program Test_Count

MATLAB loop optimization

I have a matrix, matrix_logical(50000,100000), that is a sparse logical matrix (a lot of falses, some true). I have to produce a matrix, intersect(50000,50000), that, for each pair, i,j, of rows of matrix_logical(50000,100000), stores the number of columns for which rows i and j have both "true" as the value.
Here is the code I wrote:
% store in advance the nonzeros cols
for i=1:50000
nonzeros{i} = num2cell(find(matrix_logical(i,:)));
end
intersect = zeros(50000,50000);
for i=1:49999
a = cell2mat(nonzeros{i});
for j=(i+1):50000
b = cell2mat(nonzeros{j});
intersect(i,j) = numel(intersect(a,b));
end
end
Is it possible to further increase the performance? It takes too long to compute the matrix. I would like to avoid the double loop in the second part of the code.
matrix_logical is sparse, but it is not saved as sparse in MATLAB because otherwise the performance become the worst possible.
Since the [i,j] entry counts the number of non zero elements in the element-wise multiplication of rows i and j, you can do it by multiplying matrix_logical with its transpose (you should convert to numeric data type first, e.g matrix_logical = single(matrix_logical)):
inter = matrix_logical * matrix_logical';
And it works both for sparse or full representation.
EDIT
In order to calculate numel(intersect(a,b))/numel(union(a,b)); (as asked in your comment), you can use the fact that for two sets a and b, you have
length(union(a,b)) = length(a) + length(b) - length(intersect(a,b))
so, you can do the following:
unLen = sum(matrix_logical,2);
tmp = repmat(unLen, 1, length(unLen)) + repmat(unLen', length(unLen), 1);
inter = matrix_logical * matrix_logical';
inter = inter ./ (tmp-inter);
If I understood you correctly, you want a logical AND of the rows:
intersct = zeros(50000, 50000)
for ii = 1:49999
for jj = ii:50000
intersct(ii, jj) = sum(matrix_logical(ii, :) & matrix_logical(jj, :));
intersct(jj, ii) = intersct(ii, jj);
end
end
Doesn't avoid the double loop, but at least works without the first loop and the slow find command.
Elaborating on my comment, here is a distance function suitable for pdist()
function out = distfun(xi,xj)
out = zeros(size(xj,1),1);
for i=1:size(xj,1)
out(i) = sum(sum( xi & xj(i,:) )) / sum(sum( xi | xj(i,:) ));
end
In my experience, sum(sum()) is faster for logicals than nnz(), thus its appearance above.
You would also need to use squareform() to reshape the output of pdist() appropriately:
squareform(pdist(martrix_logical,#distfun));
Note that pdist() includes a 'jaccard' distance measure, but it is actually the Jaccard distance and not the Jaccard index or coefficient, which is the value you are apparently after.

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