ind_resolve

Resolves a single list of indices to their multi-dimensional representation.

Prototype

	function ind_resolve (
		indices [*] : integer,  
		dsizes  [*] : integer   
	)

	return_val  :  integer

Arguments

A list of indices into a one-dimensional array that was converted to a one-dimensional array from a multi-dimensional array.

A list of dimension sizes of the multi-dimensioned array mentioned above.

Description

ind_resolve takes a list of indices into a one-dimensional array (that was converted from a multi-dimensional array) and returns the list of indices that they represent in the multi-dimensional array. The array returned will be dimensioned N x M, where N is the number of indices and M is the length of dsizes (the number of dimensions of the multi-dimensional array).

See Also

ind, where

Examples

Example 1

The following example demonstrates the usefulness of the ind_resolve function in conjunction with the ind function:

;
; Create sample multi-dimensioned array.
; The following creates a 3D array of size 2 x 2 x 4.
;
  a = (/(/(/1,2,3,4/), (/5,6,7,8/)/), (/(/9,10,9,8/),(/7,6,5,4/)/)/)
;
; Returns the index locations where "a" is greater than 5.
; NCL's ndtooned is used to temporarily
; reshape "a" as is required by the ind function.
;
;
  a1D      = ndtooned(a)
  dsizes_a = dimsizes(a)
  indices  = ind_resolve(ind(a1D.gt.5),dsizes_a)
  print(indices)

; Set all elements of "a" which are greater than 5 to -999.

  dim_ida  = dimsizes(indices)     
  npts     = dim_ida(0)       ; number of elements > 5 (here 9)
  ndim     = dim_ida(1)       ; rank of "a"  (here 3)
  
  do n=0,npts-1
     a(indices(n,0),indices(n,1),indices(n,2)) = -999
  end do
  print(a)

There are nine index locations where a is greater than 5: (0,1,1), (0,1,2), (0,1,3), (1,0,0), (1,0,1), (1,0,2), (1,0,3), (1,1,0), and (1,1,1). The above script, then, will return an array dimensioned 9 x 3 with the following values:

Variable: indices
Type: integer
Total Size: 108 bytes
            27 values
Number of Dimensions: 2
Dimensions and sizes:   [9] x [3]
Coordinates: 
Number Of Attributes: 1
  _FillValue :  -999
(0,0)   0
(0,1)   1
(0,2)   1
(1,0)   0
(1,1)   1
(1,2)   2
(2,0)   0
(2,1)   1
(2,2)   3
(3,0)   1
(3,1)   0
(3,2)   0
(4,0)   1
(4,1)   0
(4,2)   1
(5,0)   1
(5,1)   0
(5,2)   2
(6,0)   1
(6,1)   0
(6,2)   3
(7,0)   1
(7,1)   1
(7,2)   0
(8,0)   1
(8,1)   1
(8,2)   1

After the do loop, the array a contains the following:

Variable: a
Type: integer
Total Size: 64 bytes
           16 values
Number of Dimensions: 3
Dimensions and sizes:   [2] x [2] x [4]
Coordinates: 
(0,0,0) 1
(0,0,1) 2
(0,0,2) 3
(0,0,3) 4
(0,1,0) 5
(0,1,1) -999
(0,1,2) -999
(0,1,3) -999
(1,0,0) -999
(1,0,1) -999
(1,0,2) -999
(1,0,3) -999
(1,1,0) -999
(1,1,1) -999
(1,1,2) 5
(1,1,3) 4

Example 2

Determine all locations of a multi-dimensional array (x) where the elements match the minimum or maximum values. This is a more general approach than offered by the maxind and minind functions:

  xMax = max(x)
  xMin = min(x)
 
  x1D = ndtooned(x)     ; only do this once
  indMax = ind_resolve(ind(x1D.eq.xMax),dimsizes(x))  ; locations of max
  indMin = ind_resolve(ind(x1D.eq.xMin),dimsizes(x))  ; locations of min
  delete (x1D)

Example 3

Assume x is a three-dimensional array with coordinate variables (time,lat,lon). Print the coordinates where -3 < x < 9. Before executing various statements, check to make sure that at least some values of x satisfy the criteria. Use the num function to test if there are values between -3 and 9, then print the times, latitudes, and longitudes:

  if (num(x.gt.-3. .and. x.lt.9.) .gt. 0) then
      x1D = ndtooned(x)   
      i1D = ind(x1D.gt.-3. .and. x1D.lt.9.) 
      i   = ind_resolve(i1D, dimsizes(x) )     ; 2D  [npts,3]  (rank(x)=3)
      print(time(i(:,0)) +"  "+ lat(i(:,1)) +"  "+lon(i(:,2)) )

      delete(x1d)     ; no longer needed
      delete(i1d)
  end if