Fortran 二维数组插值

Fortran 二维数组插值可以通过使用内置的插值函数来实现。下面是一个简单的示例代码，演示了如何在 Fortran 中使用插值函数来插值一个二维数组：

program interp_2d_array
    implicit none
    
    integer, parameter :: n = 5
    integer :: i, j
    real :: x(n), y(n), f(n,n), xi, yi, fi
    
    ! Generate some data
    do i = 1, n
        x(i) = real(i)
        y(i) = real(i)
        do j = 1, n
            f(i,j) = real(i*j)
        end do
    end do
    
    ! Interpolate the data
    xi = 2.5
    yi = 3.5
    call interp2d(x, y, f, xi, yi, fi)
    
    ! Print the interpolated value
    write(*,*) fi
    
contains

    subroutine interp2d(x, y, f, xi, yi, fi)
        implicit none
        
        integer :: n, m, i, j
        real, dimension(:), intent(in) :: x, y, f
        real, intent(in) :: xi, yi
        real, intent(out) :: fi
        
        ! Get the dimensions of the input arrays
        n = size(x)
        m = size(y)
        
        ! Check that the input arrays have the correct dimensions
        if (size(f) /= n*m) then
            write(*,*) 'Error: f array has incorrect size'
            stop
        end if
        
        ! Call the interpolation function
        call interp2(x, y, f, n, m, xi, yi, fi)
        
    end subroutine interp2d
    
    subroutine interp2(x, y, f, n, m, xi, yi, fi)
        implicit none
        
        integer :: n, m, i, j, k
        real, dimension(n), intent(in) :: x
        real, dimension(m), intent(in) :: y
        real, dimension(n,m), intent(in) :: f
        real, intent(in) :: xi, yi
        real, intent(out) :: fi
        
        ! Declare variables
        real :: dx, dy, t, u, v
        
        ! Find the index of the x-coordinate closest to xi
        i = locate(x, xi)
        
        ! Find the index of the y-coordinate closest to yi
        j = locate(y, yi)
        
        ! Check that xi and yi are within the bounds of the input data
        if (i < 1 .or. i > n-1 .or. j < 1 .or. j > m-1) then
            write(*,*) 'Error: xi or yi is out of range'
            stop
        end if
        
        ! Calculate the distances between xi and x(i) and yi and y(j)
        dx = xi - x(i)
        dy = yi - y(j)
        
        ! Calculate the coefficients for the bilinear interpolation
        t = dx / (x(i+1) - x(i))
        u = dy / (y(j+1) - y(j))
        v = 1.0 - t
        fi = v * (f(i,j) + u * (f(i,j+1) - f(i,j))) + t * (f(i+1,j) + u * (f(i+1,j+1) - f(i+1,j)))
        
    end subroutine interp2
    
    function locate(a, x) result(i)
        implicit none
        
        integer :: n, i
        real, dimension(:), intent(in) :: a
        real, intent(in) :: x
        
        ! Find the index of the value closest to x
        n = size(a)
        i = minloc(abs(a - x))
        
    end function locate
    
end program interp_2d_array

在上面的代码中，我们首先生成了一个大小为 $5\times5$ 的二维数组，然后调用 `interp2d` 子程序来插值该数组。`interp2d` 子程序首先检查输入数据的维度是否正确，然后调用 `interp2` 函数来执行实际的插值操作。`interp2` 函数使用双线性插值来计算插值值。最后，我们使用 `locate` 函数来找到最接近目标 $x$ 和 $y$ 坐标的数组索引。