Fortran 读入向量列并进行拟合 我的数据文件（expansioncefs.dat）像一个向量列，有33个x值，后面是15789个f（x）块，由五个整数分隔，我想读取x和任何f（x）块，并使用三次样条方法进行拟合 当我编译代码时，我的输出文件是空的

代码

样本数据

示例输出文件

---

我已经读了好几遍了，不明白你在问什么。请你把课文修改一下，澄清一下好吗？

    implicit none
    real(kind=8),dimension(:), allocatable::x, y
    real(kind=8) ,allocatable::a(:), b(:), c(:), d(:)
    Real,Allocatable::inputdat(:,:),param(:,:)
    real(8) :: mat( 1:3, 1:33, 1:15789)!,tab(1:33,1:15790)
    integer::n
    
end module interpolation ! end of interpolation module.

program main ! start of main program.
    use interpolation
    implicit none
    character (len=14) :: nameformat, nom
    real(kind = 8)::sj, xin, f
    integer:: j, i
         
    open(100, file='expansioncoefs.dat',status='unknown') 

    Allocate (x(33))                
    Allocate (inputdat(33,15789))
    Allocate (param(15789,5))

    do i=1,33 
    read(100,*) x(i) 
!   write(*,*)x(i)
    enddo 
    do j=1,15789 
    read(100,*) param(j,1),param(j,2),param(j,3),param(j,4),param(j,5)
    do i=1, 33 
    read(100,*) inputdat(i, j)

!   print*, x(i), inputdat(i, j)
    enddo
    enddo
!   Do i=1,33                   

!   

    close(100)

    allocate(y(0:32))
    

    x(1:33) = x
    

        !print*, y
    allocate(a(0:32), b(0:32), c(0:32), d(0:32))
    
    b = 0.0; c = 0.0; d = 0.0
    do i = 1, 15789
    a(0:32) = inputdat(1:33, i+1)
!   a(0:32) = tab(1:33, i+1)

    enddo  
     
   
!   allocate a, b, c, and d arrays.
    
!
    a = y
!
!   call natural cubic spline subroutine.
    call natural_cubic_spline(n,x,a,b,c,d)
    mat(1,1:33,i) = b(0:32)
    mat(2,1:33,i) = c(0:32)
    mat(3,1:33,i) = d(0:32)
      

!   Interpolate data and write to file.
    
    
    do j = 1, 15789
    b(0:32) = mat(1,1:33,j)
    c(0:32) = mat(2,1:33,j)
    d(0:32) = mat(3,1:33,j)
    enddo
  
    do i=1,15789
    nameformat="(A5,I0.5,A4)"
    write(nom,nameformat) "srf_V",i,".asc" !extension
    print*,i,nom

    open (i, file=nom, status='unknown')
    enddo
       
    do i=1,128 
    xin=4.0+dble(i-1)*(12.0-4.0)/(128-1)! grid of 128 points

    write(i,"(e25.14)") f(xin) 
 
    enddo
      
      
     
!
    deallocate(x,y)
    deallocate(a, b, c, d)
!
    stop
end program main ! end of main program.

    real(kind = 8) function f(xin) ! start of interpolant function f.
    use interpolation
    implicit none
    real(kind = 8),intent(in)::xin
    integer:: j, i
!
    do i = 0,n-1
        if((x(i) .le. xin).and.(xin .le. x(i+1)))then
            j = i
            exit
        endif
    enddo
!
    f = a(j) + b(j)*(xin - x(j)) + c(j)*(xin - x(j))**2 + d(j)*(xin - x(j))**3 
!
    return
end function f ! end of interpolant function f.



subroutine natural_cubic_spline(n,x,a,b,c,d) ! start of natural_cubic_spline subroutine.
    implicit none
    integer,intent(in)::n
    real(kind = 8),intent(in)::x(0:n)
    real(kind = 8),intent(inout):: a(0:n)
    real(kind = 8),intent(out):: b(0:n), c(0:n), d(0:n)
    integer:: i, j
    real(kind = 8):: h(0:n), alpha(0:n), l(0:n), mu(0:n), z(0:n)
!
    h = 0.0; alpha = 0.0
!
!   step 1:
    do i = 0,n-1
        h(i) = x(i+1) - x(i)
    enddo
!
!   step 2:
    do i = 1,n-1
        alpha(i) = (3.0/h(i))*(a(i+1) - a(i)) - (3.0/h(i-1))*(a(i) - a(i-1))
    enddo
!
!   step 3:
    l(0) = 1.0; mu(0) = 0.0; z(0) = 0.0;
!
!   step 4:
    do i = 1,n-1
        l(i) = 2.0*(x(i+1) - x(i-1)) - h(i-1)*mu(i-1)
        mu(i) = h(i)/l(i)
        z(i) = (alpha(i) - h(i-1)*z(i-1))/l(i)
    enddo
!
!   step 5:
    l(n) = 1.0; z(n) = 0.0; c(n) = 0.0
!
!   step 6:
    do j = n-1,0,-1
        c(j) = z(j) - mu(j)*c(j+1)
        b(j) = (a(j+1) - a(j))/h(j) - h(j)*(c(j+1) + 2.0*c(j))/3.0
        d(j) = (c(j+1) - c(j))/(3.0*h(j))
    enddo
!
    return
end subroutine natural_cubic_spline ! end of natural_cubic_spline subroutine.

    4.000
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   10.000
   10.250
   10.500
   10.750
   11.000
   11.250
   11.500
   11.750
   12.000
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 0 0 2 0 2

  ```
       1 srf_V00001.asc
       2 srf_V00002.asc
       3 srf_V00003.asc
       4 srf_V00004.asc
       5 srf_V00005.asc
       6 srf_V00006.asc