读取未知形状的多维数组

时间:2014-06-12 08:27:49

标签: arrays fortran fortran90 dynamic-arrays

我想从文件中获取数据,该文件的数据内容可能有不同的大小。但是,结构非常简单。 3列和未定义的行数。我认为使用可分配的多维数组和显式的DO循环让我解决了我的问题。这是我到目前为止的代码

program arraycall
    implicit none

    integer, dimension(:,:), allocatable :: array
    integer :: max_rows, max_cols, row, col

    allocate(array(row,col))

    open(10, file='boundary.txt', access='sequential', status='old', FORM='FORMATTED')

     DO row = 1, max_rows
       DO col = 1, max_cols
            READ (10,*) array (row, col)
       END DO
     END DO

     print *, array (row,col)

     deallocate(array)

 end program arraycall 

现在我面临的问题是,我不知道如何定义这些max_rows和max_cols,这些因为它的大小未知而产生共鸣。

示例文件可能看起来像

11 12 13

21 22 23

31 32 33

41 42 43

所以我想出了动态估计文件记录长度的方法。更新以供将来引用他人

!---------------------------------------------------------------------
! Estimate the number of records in the inputfile
!---------------------------------------------------------------------
    open(lin,file=inputfile,status='old',action='read',position='rewind')

    loop1: do
      read(lin,*,iostat=eastat) inputline
      if (eastat < 0) then
        write(*,*) trim(inputfile),": number of records = ", numvalues
        exit loop1
      else if (eastat > 0 ) then
        stop "IO-Error!"
      end if

      numvalues=numvalues+1
    end do loop1
!-----------------------------------------------------------------------
! Read the records from the inputfile
!-----------------------------------------------------------------------
    rewind(lin)
    allocate (lon(numvalues),lat(numvalues),value(numvalues))

    do i=1,numvalues
      read(lin,*) lon(i),lat(i),value(i)
    end do

    close(lin)

2 个答案:

答案 0 :(得分:2)

我认为你有3个选项,其中两个已被描述:

  1. 两次读取文件。首先读取行数,然后分配和读取值。正如您所说,如果I / O时间相关,则效率低下。

  2. 正如@AlexanderVogt建议的那样,估计最大行数。在所有代码中,您不需要随身携带这个大矩阵。您可以定义第二个数组并执行类似的操作(基于@AlexanderVogt代码):

    allocate(array2(3,tot_rows))
    array2 = array(:, :tot_rows)
    deallocate(array)
    

    不幸的是,我担心您需要2个不同的阵列,因为您无法就地调整大小。这也意味着,如果arrayarrray2很大,您将会在很短的时间内使用大量内存。

  3. 使用链接列表。这是最优雅的解决方案,允许您只需一次读取文件,而无需预先分配数组。但它是最难编码的。这是一个适用于其中一个数组的简单示例。您需要三个链接列表或一个链接列表:

    integer, dimension(3) :: data
    

    如果您希望它与3列一起使用。

  4. 链接列表代码:

    program LinkedList
    implicit none
      integer :: i, eastat, value, numvalues
      type node
          integer :: data
          type( node ), pointer :: next
      end type node
      integer, dimension(:), allocatable :: lon
      type( node ), pointer :: head, current, previous
    
    
      nullify( head )   ! Initialize list to point to no target.
    
      open(10,file='data.dat',status='old',action='read', position='rewind')
      numvalues = 0 
      do 
          read(10,*,iostat=eastat) value
          if (eastat < 0) then
              write(*,*) "number of records = ", numvalues
              exit
          else if (eastat > 0 ) then
              stop "IO-Error!"
          end if
          allocate( current )
          current%data = value
          current%next => head
          head => current
          numvalues=numvalues+1
      end do 
      close(10)
    ! The list is read. You can now convert it into an array, if needed for
    ! numerical efficiency
    
      allocate(lon(numvalues))  
      current => head
      ! You could transverse the list this way if you hadn't kept numvalues
      !do  while ( associated( current ) )
      do i= numvalues, 1, -1
          lon(i) = current%data
          previous => current
          current => current%next
    !       head => current
          deallocate(previous)
       end do
    
    
    ! Output the list, deallocating them after use.
    print *,"lon = ", lon
    
    end program LinkedList
    

答案 1 :(得分:1)

您可以定义最大允许的行数,并使用iostat检查文件的结尾(或错误):

program arraycall 

  implicit none
  integer, dimension(:,:), allocatable :: array
  integer           :: row
  integer           :: stat ! Check return values
  ! Define max. values
  integer,parameter :: max_rows=1000
  integer,parameter :: max_cols=3    ! As stated in the question
  ! Total number of rows in the file
  integer           :: tot_rows

  allocate( array(max_cols,max_rows), stat=stat)
  ! Always a good idea to check the return value
  if ( stat /= 0 ) stop 'Cannot allocate memory!'

  open(10, file='boundary.txt', access='sequential', &
       status='old', FORM='FORMATTED')

  DO row = 1, max_rows
    ! You can directly read in arrays! 
    READ (10,*,iostat=stat) array(:,row)
    if ( stat > 0 ) then
      stop 'An error occured while reading the file'
    elseif ( stat < 0 ) then
      tot_rows = row-1
      print *, 'EOF reached. Found a total of ', tot_rows, 'rows.'
      exit
    endif
  END DO

  close(10)

  ! Do stuff, e.g. re-allocate the array
  print *,array(:,:tot_rows)

  deallocate(array)
end program arraycall 

iostat > 0是错误,iostat < 0是文件结束(或某些编译器的记录结束)。

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