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Manual Reference Pages  - findloc (3)

NAME

FINDLOC(3f) - [FORTRAN:INTRINSIC] Location of the first element of ARRAY identified by MASK along dimension DIM having a value

CONTENTS

Synopsis
Description
Options
Result
Example

SYNOPSIS

 
FINDLOC (ARRAY, VALUE, DIM [, MASK, KIND, BACK])


or


FINDLOC (ARRAY, VALUE [, MASK, KIND, BACK])

DESCRIPTION
Location of the first element of ARRAY identified by MASK along dimension DIM having a value equal to VALUE.

Class. Transformational function.

If both ARRAY and VALUE are of type logical, the comparison is performed with the .EQV. operator; otherwise, the comparison is performed with the == operator. If the value of the comparison is true, that element of ARRAY matches VALUE.

If only one element matches VALUE, that element’s subscripts are returned. Otherwise, if more than one element matches VALUE and BACK is absent or present with the value false, the element whose subscripts are returned is the first such element, taken in array element order. If BACK is present with the value true, the element whose subscripts are returned is the last such element, taken in array element order.

OPTIONS

ARRAY shall be an array of intrinsic type.
VALUE shall be scalar and in type conformance with ARRAY, as specified in Table 7.3 for relational intrinsic operations 7.1.5.5.2).
DIM shall be an integer scalar with a value in the range 1 DIM n, where n is the rank of ARRAY. The corresponding actual argument shall not be an optional dummy argument.
MASK (optional) shall be of type logical and shall be conformable with ARRAY.
KIND (optional) shall be a scalar integer initialization expression.
BACK (optional) shall be a logical scalar.

RESULT

4 Result Characteristics. Integer. If KIND is present, the kind type parameter is that specified by the value of KIND; otherwise the kind type parameter is that of default integer type. If DIM does not appear, the result is an array of rank one and of size equal to the rank of ARRAY; otherwise, the result is of rank n - 1 and shape [d1 , d2 , . . . , dDIM-1 , dDIM+1 , . . . , dn ], where [d1 , d2 , . . . , dn ] is the shape of ARRAY.

5 Result Value.

Case (i):
  The result of FINDLOC (ARRAY, VALUE) is a rank-one array whose element values are the values of the subscripts of an element of ARRAY whose value matches VALUE. If there is such a value, the ith subscript returned lies in the range 1 to ei , where ei is the extent of the ith dimension of ARRAY. If no elements match VALUE or ARRAY has size zero, all elements of the result are zero.
Case (ii):
  The result of FINDLOC (ARRAY, VALUE, MASK = MASK) is a rank-one array whose element values are the values of the subscripts of an element of ARRAY, corresponding to a true element of MASK, whose value matches VALUE. If there is such a value, the ith subscript returned lies in the range 1 to ei , where ei is the extent of the ith dimension of ARRAY. If no elements match VALUE, ARRAY has size zero, or every element of MASK has the value false, all elements of the result are zero.
Case (iii):
  If ARRAY has rank one, the result of 
                  FINDLOC (ARRAY, VALUE, DIM=DIM [, MASK = MASK])
  is a scalar whose value is equal to that of the first element of 

                  FINDLOC (ARRAY, VALUE [, MASK = MASK])

.Otherwise, the value

of element (s1 , s2 , . . . , sDIM-1 , sDIM+1 , . . . , sn ) of the result is equal to 

                   FINDLOC (ARRAY (s1, s2, ..., sDIM-1, :, sDIM+1, ..., sn ), &
                   VALUE, DIM=1 [, MASK = MASK (s1, s2, ..., sDIM-1, :, sDIM+1 ,... , sn )]).

EXAMPLE
Case (i):
  The value of 
                  FINDLOC ([2, 6, 4, 6,], VALUE = 6)
  is [2], and the value of 

                  FINDLOC ([2, 6, 4, 6], VALUE = 6, BACK = .TRUE.)
is [4].
 
0 -5 7 7
  T T F T
Case (ii):
  If A has the value 3 4 -1 2 , and M has the value T T F T , FINDLOC (A, 7,
1 5 6 7 T T F T MASK = M) has the value [1, 4] and FINDLOC (A, 7, MASK = M, BACK = .TRUE.) has the value [3, 4]. This is independent of the declared lower bounds for A.
Case (iii):
  The value of FINDLOC ([2, 6, 4], VALUE = 6, DIM = 1) is 2. If B has the value 1 2 -9 , FINDLOC (B, VALUE = 2, DIM = 1) has the value [2, 1, 0] and FINDLOC (B, 2 2 6 VALUE = 2, DIM = 2) has the value [2, 1]. This is independent of the declared lower bounds for B.

findloc (3) March 18, 2019
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