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how to generate integer random number in fortran 90 in the range [0,5]?

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8

5

I am kind of new in the fortran proramming.
Can anyone please help me out with the solution.
i am having a problem of generating integer random number
in the range [0,5] in fortran random number using
random_seed and rand

random numbers integer fortran

share|improve this question

asked Apr 14 '14 at 10:01

user3531410user3531410

4611 gold badge11 silver badge55 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  rand is a GNU extension. The Fortran 95 combination is random_seed and random_number.
  
  – M. S. B.
  Apr 14 '14 at 10:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Is a result of exactly 5 acceptable to you? Or did you mean to ask for the range [0,5) ? The suggestions below won't give you exactly the endpoint value (except possibly by rounding.)
  
  – Steve Lionel
  Apr 14 '14 at 19:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @SteveLionel I'm not quite sure what you mean... Could you expand on that?
  
  – Alexander Vogt
  Apr 15 '14 at 13:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You used the term "[0,5]". This is "interval notation" meaning that the value can be anywhere from exactly 0 to exactly 5. On reconsideration, that perhaps is what you want. francescalus' answer below explains in more detail.
  
  – Steve Lionel
  Apr 16 '14 at 15:38
  

  
  
  
  

add a comment | 

8

5

I am kind of new in the fortran proramming.
Can anyone please help me out with the solution.
i am having a problem of generating integer random number
in the range [0,5] in fortran random number using
random_seed and rand

random numbers integer fortran

share|improve this question

asked Apr 14 '14 at 10:01

user3531410user3531410

4611 gold badge11 silver badge55 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  rand is a GNU extension. The Fortran 95 combination is random_seed and random_number.
  
  – M. S. B.
  Apr 14 '14 at 10:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Is a result of exactly 5 acceptable to you? Or did you mean to ask for the range [0,5) ? The suggestions below won't give you exactly the endpoint value (except possibly by rounding.)
  
  – Steve Lionel
  Apr 14 '14 at 19:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @SteveLionel I'm not quite sure what you mean... Could you expand on that?
  
  – Alexander Vogt
  Apr 15 '14 at 13:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You used the term "[0,5]". This is "interval notation" meaning that the value can be anywhere from exactly 0 to exactly 5. On reconsideration, that perhaps is what you want. francescalus' answer below explains in more detail.
  
  – Steve Lionel
  Apr 16 '14 at 15:38
  

  
  
  
  

add a comment | 

I am kind of new in the fortran proramming.
Can anyone please help me out with the solution.
i am having a problem of generating integer random number
in the range [0,5] in fortran random number using
random_seed and rand

random numbers integer fortran

share|improve this question

asked Apr 14 '14 at 10:01

user3531410user3531410

4611 gold badge11 silver badge55 bronze badges

share|improve this question

asked Apr 14 '14 at 10:01

user3531410user3531410

4611 gold badge11 silver badge55 bronze badges

share|improve this question

asked Apr 14 '14 at 10:01

user3531410user3531410

4611 gold badge11 silver badge55 bronze badges

asked Apr 14 '14 at 10:01

user3531410user3531410

4611 gold badge11 silver badge55 bronze badges

asked Apr 14 '14 at 10:01

user3531410user3531410

4611 gold badge11 silver badge55 bronze badges

2 Answers
2

active

oldest

votes

4

What about:

program rand_test
 use,intrinsic :: ISO_Fortran_env
 real(REAL32) :: r(5)
 integer :: i(5)

 ! call init_random_seed() would go here

 call random_number(r)

 ! Uniform distribution requires floor: Thanks to @francescalus 
 i = floor( r*6._REAL32 )

 print *, i
end program

share|improve this answer

edited Apr 14 '14 at 10:18

answered Apr 14 '14 at 10:10

Alexander VogtAlexander Vogt

15.9k1313 gold badges3838 silver badges5555 bronze badges

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Use real :: ... and 6. then. But please, do yourself a favor and switch to explicit definitions!
  
  – Alexander Vogt
  Apr 14 '14 at 12:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i = floor( r*3. ) - 1
  
  – Alexander Vogt
  Apr 14 '14 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  that is ok but will it give uniform distribution .....?
  
  – user3531410
  Apr 15 '14 at 10:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  will it give uniform distribution for -1,0,1?
  
  – user3531410
  Apr 15 '14 at 10:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I don't see why not - you're just shifting the boundaries.
  
  – Alexander Vogt
  Apr 15 '14 at 12:58
  
  

  
  
  
  

 | 
show 3 more comments

4

To support the answer by Alexander Vogt, I'll generalize.

The intrinsic random_number(u) returns a real number u (or an array of such) from the uniform distribution over the interval [0,1). [That is, it includes 0 but not 1.]

To have a discrete uniform distribution on the integers n, n+1, ..., m-1, m carve the continuous distribution up into m+1-n equal sized chunks, mapping each chunk to an integer. One way could be:

call random_number(u)
j = n + FLOOR((m+1-n)*u) ! We want to choose one from m-n+1 integers

As you can see, for the initial question for 0, 1, 2, 3, 4, 5 this reduces to

call random_number(u)
j = FLOOR(6*u) ! n=0 and m=5

and for the other case in your comment -1, 0, 1

call random_number(u)
j = -1 + FLOOR(3*u) ! n=-1 and m=1

Of course, other transformations will be required for sets of non-contiguous integers, and one should pay attention to numerical issues.

share|improve this answer

edited May 23 '17 at 12:34

Community♦

111 silver badge

answered Apr 15 '14 at 19:33

francescalusfrancescalus

18.5k77 gold badges3737 silver badges6060 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  to avoid skew due to the limited number of bits in real mantissa (53 for IEEE 754 double-precision, 24 for single-precision), you might need to call random_number several times e.g., see how _randbelow(n) is implemented via random() in Python
  
  – jfs
  May 2 '17 at 12:15
  
  

  
  
  
  

add a comment | 

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4

What about:

program rand_test
 use,intrinsic :: ISO_Fortran_env
 real(REAL32) :: r(5)
 integer :: i(5)

 ! call init_random_seed() would go here

 call random_number(r)

 ! Uniform distribution requires floor: Thanks to @francescalus 
 i = floor( r*6._REAL32 )

 print *, i
end program

share|improve this answer

edited Apr 14 '14 at 10:18

answered Apr 14 '14 at 10:10

Alexander VogtAlexander Vogt

15.9k1313 gold badges3838 silver badges5555 bronze badges

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Use real :: ... and 6. then. But please, do yourself a favor and switch to explicit definitions!
  
  – Alexander Vogt
  Apr 14 '14 at 12:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i = floor( r*3. ) - 1
  
  – Alexander Vogt
  Apr 14 '14 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  that is ok but will it give uniform distribution .....?
  
  – user3531410
  Apr 15 '14 at 10:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  will it give uniform distribution for -1,0,1?
  
  – user3531410
  Apr 15 '14 at 10:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I don't see why not - you're just shifting the boundaries.
  
  – Alexander Vogt
  Apr 15 '14 at 12:58
  
  

  
  
  
  

 | 
show 3 more comments

4

What about:

program rand_test
 use,intrinsic :: ISO_Fortran_env
 real(REAL32) :: r(5)
 integer :: i(5)

 ! call init_random_seed() would go here

 call random_number(r)

 ! Uniform distribution requires floor: Thanks to @francescalus 
 i = floor( r*6._REAL32 )

 print *, i
end program

share|improve this answer

edited Apr 14 '14 at 10:18

answered Apr 14 '14 at 10:10

Alexander VogtAlexander Vogt

15.9k1313 gold badges3838 silver badges5555 bronze badges

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Use real :: ... and 6. then. But please, do yourself a favor and switch to explicit definitions!
  
  – Alexander Vogt
  Apr 14 '14 at 12:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i = floor( r*3. ) - 1
  
  – Alexander Vogt
  Apr 14 '14 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  that is ok but will it give uniform distribution .....?
  
  – user3531410
  Apr 15 '14 at 10:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  will it give uniform distribution for -1,0,1?
  
  – user3531410
  Apr 15 '14 at 10:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I don't see why not - you're just shifting the boundaries.
  
  – Alexander Vogt
  Apr 15 '14 at 12:58
  
  

  
  
  
  

 | 
show 3 more comments

What about:

program rand_test
 use,intrinsic :: ISO_Fortran_env
 real(REAL32) :: r(5)
 integer :: i(5)

 ! call init_random_seed() would go here

 call random_number(r)

 ! Uniform distribution requires floor: Thanks to @francescalus 
 i = floor( r*6._REAL32 )

 print *, i
end program

share|improve this answer

edited Apr 14 '14 at 10:18

answered Apr 14 '14 at 10:10

Alexander VogtAlexander Vogt

15.9k1313 gold badges3838 silver badges5555 bronze badges

program rand_test
 use,intrinsic :: ISO_Fortran_env
 real(REAL32) :: r(5)
 integer :: i(5)

 ! call init_random_seed() would go here

 call random_number(r)

 ! Uniform distribution requires floor: Thanks to @francescalus 
 i = floor( r*6._REAL32 )

 print *, i
end program

share|improve this answer

edited Apr 14 '14 at 10:18

answered Apr 14 '14 at 10:10

Alexander VogtAlexander Vogt

15.9k1313 gold badges3838 silver badges5555 bronze badges

answered Apr 14 '14 at 10:10

Alexander VogtAlexander Vogt

15.9k1313 gold badges3838 silver badges5555 bronze badges

answered Apr 14 '14 at 10:10

Alexander VogtAlexander Vogt

15.9k1313 gold badges3838 silver badges5555 bronze badges

4

To support the answer by Alexander Vogt, I'll generalize.

The intrinsic random_number(u) returns a real number u (or an array of such) from the uniform distribution over the interval [0,1). [That is, it includes 0 but not 1.]

To have a discrete uniform distribution on the integers n, n+1, ..., m-1, m carve the continuous distribution up into m+1-n equal sized chunks, mapping each chunk to an integer. One way could be:

call random_number(u)
j = n + FLOOR((m+1-n)*u) ! We want to choose one from m-n+1 integers

As you can see, for the initial question for 0, 1, 2, 3, 4, 5 this reduces to

call random_number(u)
j = FLOOR(6*u) ! n=0 and m=5

and for the other case in your comment -1, 0, 1

call random_number(u)
j = -1 + FLOOR(3*u) ! n=-1 and m=1

Of course, other transformations will be required for sets of non-contiguous integers, and one should pay attention to numerical issues.

share|improve this answer

edited May 23 '17 at 12:34

Community♦

111 silver badge

answered Apr 15 '14 at 19:33

francescalusfrancescalus

18.5k77 gold badges3737 silver badges6060 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  to avoid skew due to the limited number of bits in real mantissa (53 for IEEE 754 double-precision, 24 for single-precision), you might need to call random_number several times e.g., see how _randbelow(n) is implemented via random() in Python
  
  – jfs
  May 2 '17 at 12:15
  
  

  
  
  
  

add a comment | 

4

To support the answer by Alexander Vogt, I'll generalize.

The intrinsic random_number(u) returns a real number u (or an array of such) from the uniform distribution over the interval [0,1). [That is, it includes 0 but not 1.]

To have a discrete uniform distribution on the integers n, n+1, ..., m-1, m carve the continuous distribution up into m+1-n equal sized chunks, mapping each chunk to an integer. One way could be:

call random_number(u)
j = n + FLOOR((m+1-n)*u) ! We want to choose one from m-n+1 integers

As you can see, for the initial question for 0, 1, 2, 3, 4, 5 this reduces to

call random_number(u)
j = FLOOR(6*u) ! n=0 and m=5

and for the other case in your comment -1, 0, 1

call random_number(u)
j = -1 + FLOOR(3*u) ! n=-1 and m=1

Of course, other transformations will be required for sets of non-contiguous integers, and one should pay attention to numerical issues.

share|improve this answer

edited May 23 '17 at 12:34

Community♦

111 silver badge

answered Apr 15 '14 at 19:33

francescalusfrancescalus

18.5k77 gold badges3737 silver badges6060 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  to avoid skew due to the limited number of bits in real mantissa (53 for IEEE 754 double-precision, 24 for single-precision), you might need to call random_number several times e.g., see how _randbelow(n) is implemented via random() in Python
  
  – jfs
  May 2 '17 at 12:15
  
  

  
  
  
  

add a comment | 

To support the answer by Alexander Vogt, I'll generalize.

The intrinsic random_number(u) returns a real number u (or an array of such) from the uniform distribution over the interval [0,1). [That is, it includes 0 but not 1.]

To have a discrete uniform distribution on the integers n, n+1, ..., m-1, m carve the continuous distribution up into m+1-n equal sized chunks, mapping each chunk to an integer. One way could be:

call random_number(u)
j = n + FLOOR((m+1-n)*u) ! We want to choose one from m-n+1 integers

As you can see, for the initial question for 0, 1, 2, 3, 4, 5 this reduces to

call random_number(u)
j = FLOOR(6*u) ! n=0 and m=5

and for the other case in your comment -1, 0, 1

call random_number(u)
j = -1 + FLOOR(3*u) ! n=-1 and m=1

Of course, other transformations will be required for sets of non-contiguous integers, and one should pay attention to numerical issues.

share|improve this answer

edited May 23 '17 at 12:34

Community♦

111 silver badge

answered Apr 15 '14 at 19:33

francescalusfrancescalus

18.5k77 gold badges3737 silver badges6060 bronze badges

call random_number(u)
j = n + FLOOR((m+1-n)*u) ! We want to choose one from m-n+1 integers

call random_number(u)
j = FLOOR(6*u) ! n=0 and m=5

call random_number(u)
j = -1 + FLOOR(3*u) ! n=-1 and m=1

share|improve this answer

edited May 23 '17 at 12:34

Community♦

111 silver badge

answered Apr 15 '14 at 19:33

francescalusfrancescalus

18.5k77 gold badges3737 silver badges6060 bronze badges

edited May 23 '17 at 12:34

Community♦

111 silver badge

edited May 23 '17 at 12:34

Community♦

111 silver badge

answered Apr 15 '14 at 19:33

francescalusfrancescalus

18.5k77 gold badges3737 silver badges6060 bronze badges

answered Apr 15 '14 at 19:33

francescalusfrancescalus

18.5k77 gold badges3737 silver badges6060 bronze badges