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...program pck implicit none real,dimension(10,2) :: a integer,dimension(10) :: b integer :: i,check open(10,file='pic.txt') do i=1,10 read(10,*)a(i,:) end do forall (i=1:10) where (a(i,:) == a(i,:)) check=check+1 check=b(i) end where end forall end program But: pck.f90:16.1...

My file: 3515 2.154 3516 2.126 3517 2.098 3518 2.063 3519 2.084 3520 2.114 3521 2.128 . . 13093 2.302 13094 2.294 13095 2.29 13096 2.314 13097 2.314 13098 2.294 13099 2.294 13100 2.279 9586 lines in the file. My code: program pr2 implicit none integer :: i,j real,dimension(9586) :: c...

< 1| 255.33 | This will be repeated 90000 times in file. So first number will change length.Also | | creates problem,should I read it as character?

Ok,I will try now.

character*72 ifname,ofname write(*, fmt="(/'Enter input file name')") read(5,85) ifname 85 format(a72) write(*, fmt="(/'Enter output file name')") read(5,85) ofname open(unit=11, file=ifname, form='unformatted', status='old') open(unit=12, file=ofname)...

I am translating some code form matlab to fortran90,I need to generate random numbers for matrix. This is how MATLAB code looks like,and gives me the same values all the time. rand('state',0) v = 2*(ones(c,1)*aa).*(rand(c,n)-0.5) + ones(c,1)*mm; Is there any idea how to solve this in FORTRAN?

SOLVED: forall(x=1:n, i=1:33,a(x)==b(i)) c(x+i-i)=i end forall

No, take a look: do x=1,n read(30,*)a(x) end do forall(x=1:n, i=1:33,b(i)==a(x)) c(i+x-x)=i end forall do i=1,33 write(31,88)c(i) 88 format(i8) I got exactly the same 33 instead of 27 values.

...:: a integer,dimension(33) :: c character(6) :: tfile=" .nmf" character(6) :: pfile=" .pos" allocate(a(n)) id1=iter/10 id2=iter-10*id1 tfile(1:2)=CHAR(id1+48)//CHAR(id2+48) pfile(1:2)=CHAR(id1+48)//CHAR(id2+48) open(30, file=tfile) open(31, file=pfile) do x=1,n...

...comp implicit none integer :: i,n real,dimension(17) :: a real,dimension(29) :: b,c open(10,file='01.txt',status='old') open(11,file='frekvencije.dat',status='old') read(10,*)a read(11,*)b do 20 n=1,17 do i=1,29 if (b(i)==a(n)) then c(i)=i end if end do 20 continue write(*,*)c end program

ok.For example a=(/2,5,6,7,9/) and b=(/1,2,3,4,5,6,7,8,9) What I want is to get c=(/0,2,0,0,5,6,7,0,9,/). I had tried with nested do looped but it didn't work well.

...:: i real,dimension(17) :: a real,dimension(29) :: b,c open(10,file='01.txt',status='old') open(11,file='frekvencije.dat',status='old') read(10,*)a read(11,*)b do i=1,29 if (b(i)==a(1)) then c(i)=i end if end do write(*,*)c end program How to solve this,I want to compare b with a,but how...

...open(11,file='time01.dat',status='unknown') do i=1,240 read (10,50) x(i),y(i) 50 format(2x,i3,2x,f9.6) end do write(11,*)y end forrtl: severe (64): input conversion error, unit 10, file /home/milenko/futog/s1.dat I have to put i3 because file has 240 rows. < 232| 940.52 | < 233| 982.62 | <...

< 1| 255.33 | < 2| 32.233353 | < 3| 27.115475 | < 4| 24.594143 | < 5| 32.685696 | < 6| 21.042452 | < 7| 47.647877 | < 8| 92.553001 | < 9| 96.665306 | How to format < and |.I have tried with a2 like character but it does not work.

I want to skip first 19 lines(description of file),the in the 20th start numerical values what I really need.How to write this?

Well,just to explain the code is not mine,I am using it for my Phd thesis.So to rewrite it FORTRAN90 does not make sense.The problem is that some parameters are within the code,so the code is compiled with them,the other ones are read from numerous input files.it is a little bit tricky to find...

...in this subroutine: subroutine atupv(m,n,u,v) include 'ray.par' parameter(nnzmax=100000, ndmax=1000) parameter(nsconst=3*nximax*nzimax, + nszero=7*nximax*nzimax, + nmmax=nximax*nzimax) integer row(nnzmax+nszero),column(nnzmax+nszero)...

Yes,I want to convert velocities into integer*2.Please,suggest some simple solutions.

...milenko@milenkons:~/fg77/f05k/bin$ od -h vel.mod 0000000 007a 0000 1388 1388 1388 1388 1388 1388 0000020 1388 1388 1388 1388 1388 1388 1388 1388 * 0000160 1388 1388 1388 1388 1388 1388 1388 007a 0000200 0000 007a 0000 141e 141e 141e 141e 141e 0000220 141e 141e 141e 141e 141e 141e 141e 141e *...

Ok, I have problem with later results.I have so far used only od -h to check vel.mod file. My question is for example if my programme can be written somehow in different way using do loops but to read these 7 data files?