Evaluating a floating point variable in Scheme language - variables

I want to read multiple data files (10 in total) in Ansys Fluent. I wrote a journal file which uses scheme language
(Do ((count 11.100 (+ count 0.100))) ((>= count 12.000))
(ti-menu-load-string (format #f "file read-data data-~a.dat" count)))
The format of the file name is like data-11.200.dat, but the program reads it as data-11.2.dat. How I can force it to read the floating point numbers after the decimal point?
Of course I can rename the data files, but that is not useful for I have to use the code many times.
I have tried data-~03d.dat, but that didn't work!

Try this:
(do ((count 111/10 (+ count 1/10))) ((>= count 12))
(ti-menu-load-string
(format #f "file read-data data-~2,3F.dat" count)))
I assume format is from SRFI-48 Intermediate Format Strings. I've changed the numbers to rational numbers because adding 0.1 gives you rounding errors in floating point.

I think I have figured that out:
data-~.3f.dat

Related

Delimiting output in Lisp

I'm sure this is a very basic question, but how to output delimited values?
(defun q (n)
(if (<= n 2)
1
(+
(q (- n (q (- n 1))))
(q (- n (q (- n 2)))))))
(loop for x from 1 to 25
do (
write (q x)))
The above outputs the first 25 terms of the Hofstadter Q-sequence, but concatenated:
11233455666888109101111121212121614
Can be comma-, space- or tab-delimited. I tried playing with the format function, but it's way over my head for a Lisp newbie like me.
You can write the delimiter yourself, e. g. (write ", "). You can also write the (platform dependent) line separator with (terpri).
As for format, you can get by with some basic usage at first:
(format <stream> <template> <args…>)
<stream> is where the output should go. For standard output, use t. In order to write to a file, you would use the stream created by with-open-file here.
<template> is a template string. That's just a string, but the ~ character is special. For starters, just use ~a wherever you want to insert an argument.
<args…> are exactly as many further arguments as you used ~a above.
Using this simple toolbox, you could do: (format t "~a, " (q x)) for each item.
Format has a lot of other possibilities, e. g. it can iterate a list by itself, do different output formats and escaping, or even be extended by user functions. Look that up in the hyperspec (e. g. at clhs.lisp.se).
use write-char or, indeed, format:
(format t "~D: ~:D~%" x (q x))

How to print a number with number of fixed characters in Smalltalk/Pharo

Actual task: I want to print the matrix (of my own implementation) in humanly readable format. As a pre-requisite, I figured I need to be able to specify "fit the number representation into X characters". I found #printShowingDecimalPlaces: and #printPaddedWith:to: in Float and Integer classes (the first method is in more general Number class). Individually, they work, but the former works on fractional part only and the the latter on part before fractional, e.g.:
10.3 printPaddedWith: Character space to: 5.
"' 10.3'"
-10.3 printPaddedWith: Character space to: 5.
"' -10.3'"
10.3 printShowingDecimalPlaces: 3.
"'10.300'"
Also, their action on very large (or equally small numbers) in scientific form is not ideal:
12.3e9 printShowingDecimalPlaces: 3.
"'12300000000.000'"
12.3e9 printPaddedWith: Character space to: 5.
"' 1.23e10'"
So, I would like something like Common Lisp's (FORMAT T "~10g" 12.3d9) or C's printf("%10g", 12.3e9), that (a) restricts the whole width to 10 characters and (b) chooses the most suitable format depending on the size of the number. Is there something like this in Pharo?
For versatile printing options, I suggest loading NumberPrinter package from
http://ss3.gemstone.com/ss/NumberPrinter/
(FloatPrinter fixed) digitCount: 2; print: 10.3.
-> '10.30'
I did not try it in recent Pharo versions though.
EDIT:
Ah, but I see no format for handling exponents multiple of 3, maybe you would have to create a subclass for such format.
EDIT:
Or I missunderstood: you don't want it to print as '12.3e9' but rather '1.23e10'? note that apart significand digitCount, you need extra size for at worst 1 for sign + 1 for fraction separator + 1 for exponent letter + 1 for exponent sign + 3 for exponent (worst case for double precision floating point).
The more or less equivalent to g format would be something like this:
(FloatPrinter freeFormat)
totalWidth: 13; "size of the generated string"
digitCount: 6; "number of significant figures"
print: -12.3e-205.
->' -1.23e-204'

Fortran does'nt end when obtain unexpected value?

I've got a program, which compute a several variables and then these variables are writing in to the output file.
Is it possilbe, that when my program can't get a correct results for my formula, it does'nt terminate?
To clarify what I do, here is part of my code, where the variable of my interest are compute:
dx=x(1,i)-x(nk,i)
dy=y(1,i)-y(nk,i)
dz=z(1,i)-z(nk,i)
call PBC(dx,dy,dz)
r2i=dx*dx+dy*dy+dz*dz
r2=r2+r2i
r2g0=0.0d0
r2gx=0.0d0
dx=x(1,i)-x(2,i)
call PBC(dx,dy,dz)
rspani=dsqrt(dx*dx)
do ii=1,nk-1
rx=x(ii,i)
ry=y(ii,i)
rz=z(ii,i)
do jj=ii+1,nk
dx=x(jj,i)-rx
dy=y(jj,i)-ry
dz=z(jj,i)-rz
call PBC(dx,dy,dz)
r21=dx*dx+dy*dy+dz*dz
r21x=dx*dx
r2g=r2g+r21
r2gx=r2gx+r21x
r2g0=r2g0+r21
rh=rh+1.0d0/dsqrt(r21)
rh1=rh1+1.0d0
ir21=dnint(dsqrt(r21)/dr)
p(ir21)=p(ir21)+2.0D0
dxs=dsqrt(r21x)
if(dxs.gt.rspani) rspani=dxs
end do
and then in to the output I just write these variables:
write(12,870)r2i,sqrt(r2i),r2g0,r2gx/(nk*nk)
870 FORMAT(3(f15.7,3x),f15.7)
The x, y, z are actully generate via a random number generator.
The problem is that my output contains, correct values for lets say 457 lines, and then a one line is just "*********" when I use mc viewer and then the output continues with correct values, but let's say 12 steps form do cycle which compute these variables is missing.
So my questions are basic:
Is it possible, that my program can't get a correct numbers, and that's why the result is not writing in to the program?
or
Could it this been caused due to wrong output formating or something related with formating?
Thank you for any suggestion
********* is almost certainly the result of trying to write a number too large for the field specified in a format string.
For example, a field specified as f15.7 will take 1 spot for the decimal point, 1 spot for a leading sign (- will always be printed if required, + may be printed if options are set), 7 for the fractional digits, leaving 6 digits for the whole part of the number. There may therefore be cases where the program won't fit the number into the field and will print 15 *s instead.
Programs compiled with an up to date Fortran compiler will write a string such as NaN or -Inf if they encounter a floating-point number which represents one of the IEEE special values

Write data to file in columns (Fortran)

I need to write some data to file in Fortran 90. How should I use WRITE (*,*) input to have the values grouped in columns? WRITE always puts a new line after each call, that's the problem.
code example:
open (unit = 4, file = 'generated_trajectories1.dat', form='formatted')
do time_nr=0, N
write (4,*) dble(time_nr)*dt, initial_traj(time_nr)
end do
And now the point is to have it written in separate columns.
You can use implied DO loops to write values as single records. Compare the following two examples:
integer :: i
do i=1,10
write(*,'(2I4)') i, 2*i
end do
It produces:
1 2
2 4
3 6
...
Using implied DO loops it can rewritten as:
integer :: i
write(*, '(10(2I4))') (i, 2*i, i=1,10)
This one produces:
1 2 2 4 3 6 ...
If the number of elements is not fixed at compile time, you can either use the <n> extension (not supported by gfortran):
write(*, '(<n>(2I4))') (i, 2*i, i=1,n)
It takes the number of repetitions of the (2I4) edit descriptor from the value of the variable n. In GNU Fortran you can first create the appropriate edit descriptor using internal files:
character(len=20) :: myfmt
write(myfmt, '("(",I0,"(2I4))")') n
write(*, fmt=myfmt) (i, 2*i, i=1,n)
Of course, it also works with list directed output (that is output with format of *):
write(*, *) (i, 2*i, i=1,10)
This really depends on what data you are trying to write to file (i.e. whether you have a scalar within a loop or an array...). Can you include a description of this in your question?
If your are trying to write a scalar multiple times to the same row then try using non-advancing I/O, passing the keyword argument advance="no" to the write statement, e.g.
integer :: x
do x = 1,10
write(*, fmt='(i4,1x)', advance="no") x
end do
However, be aware of a suprise with non-advancing I/O.
The answer depends on your answer to Chris's question. If you want a single line, then you will have to use non-advancing IO as described by Chris. Without this, with multiple formatted write statement you will always get multiple lines.
Also, you will likely need to use formatted IO instead of list-directed (*) IO. The rules are loose for list-directed IO. Different compilers may produce different output. With many output items, line breaks are likely to keep the output lines from being too long.
Here a format that should work if all of your variables are reals:
write (4, '( *(2X, ES14.6) )', advance="no" )
how about the good old $ edit descriptor:
write(*, fmt='(i4,$)') x
remember to do a write(*,*) after your loop...

Fortran: How do I read the first character from each line of a text file?

this is my first time trying to program in Fortran. I'm trying to write a program that prints the first 1476 terms of the Fibonacci sequence, then examines the first digit of each term and stores the number of 1s, 2s, 3s, ..., 9s that occur in an array.
The problem that I can't seem to figure out is how to read the first digit of each term. I've tried several things but am having difficulty with my limited knowledge of Fortran techniques. I write the terms to a text file and the idea is to read the first digit of each line and accumulate the respective number in the array. Does anyone have any suggestions of how to do this?
Here is my code so far:
(edit: I included the code I have for reading the file. Right now it just prints out 3.60772951994415996E-313,
which seems like an address of some sort, because it's not one of the Fibonacci numbers. Also, it is the only thing printed, I expected that it would print out every line of the file...)
(edit edit: After considering this, perhaps there's a way to format the writing to the text file to just the first digit. Is there a way to set the number of significant digits of a real number to one? :P)
subroutine writeFib(n)
integer :: i
real*8 :: prev, current, newFib
prev = 0
current = 1
do i = 1, n
newFib = prev + current
prev = current
current = newFib
write(7,*) newFib
end do
return
end subroutine
subroutine recordFirstDigits(a)
integer :: openStat, inputStat
real*8 :: fibNum
open(7, file = "fort.7", iostat = openStat)
if (openStat > 0) stop "*** Cannot open the file ***"
do
read(7, *, iostat = inputStat) fibNum
print *,fibNum
if (inputStat > 0) stop "*** input error ***"
if (inputStat < 0) exit ! end of file
end do
close(7)
end subroutine
program test
integer :: k, a(9)
k = 1476
call writeFib(k)
call recordFirstDigits(a)
end program
Although the suggestions were in place, there were also several things that were forgotten. Range of the REAL kind, and some formatting problems.
Anyways, here's one patched up solution, compiled and working, so try to see if this will work for you. I've took the liberty of choosing my own method for fibonacci numbers calculation.
program SO1658805
implicit none
integer, parameter :: iwp = selected_real_kind(15,310)
real(iwp) :: fi, fib
integer :: i
character(60) :: line
character(1) :: digit
integer :: n0=0, n1=0, n2=0, n3=0, n4=0, n5=0, n6=0, n7=0, n8=0, n9=0
open(unit=1, file='temp.txt', status='replace')
rewind(1)
!-------- calculating fibonacci numbers -------
fi = (1+5**0.5)/2.
do i=0,1477
fib = (fi**i - (1-fi)**i)/5**0.5
write(1,*)fib,i
end do
!----------------------------------------------
rewind(1)
do i=0,1477
read(1,'(a)')line
line = adjustl(line)
write(*,'(a)')line
read(line,'(a1)')digit
if(digit.eq.' ') n0=n0+1
if(digit.eq.'1') n1=n1+1
if(digit.eq.'2') n2=n2+1
if(digit.eq.'3') n3=n3+1
if(digit.eq.'4') n4=n4+1
if(digit.eq.'5') n5=n5+1
if(digit.eq.'6') n6=n6+1
if(digit.eq.'7') n7=n7+1
if(digit.eq.'8') n8=n8+1
if(digit.eq.'9') n9=n9+1
end do
close(1)
write(*,'("Total number of different digits")')
write(*,'("Number of digits 0: ",i5)')n0
write(*,'("Number of digits 1: ",i5)')n1
write(*,'("Number of digits 2: ",i5)')n2
write(*,'("Number of digits 3: ",i5)')n3
write(*,'("Number of digits 4: ",i5)')n4
write(*,'("Number of digits 5: ",i5)')n5
write(*,'("Number of digits 6: ",i5)')n6
write(*,'("Number of digits 7: ",i5)')n7
write(*,'("Number of digits 8: ",i5)')n8
write(*,'("Number of digits 9: ",i5)')n9
read(*,*)
end program SO1658805
Aw, ... I just read you need the number of digits stored in to an array. While I just counted them.
Oh well, ... "left as an exercise for the reader ..." :-)
Can you read with a FORMAT(A1)? It's been 20 years so I don't remember the exact syntax.
I wonder why the open statement succeeds when file 7 hasn't been closed. I think you need an endfile statement and/or a rewind statement in between writing and reading.
Paul Tomblin posted what you have to do after you solve your problem in getting reads to work in the first place.
I am getting an 'end of line' runtime error
You don't show the ! code to read here... which makes it kind of difficult to guess what you are doing wrong :-)
Perhaps you need a loop to read each line and then jump out of the loop to a continue statement when there are no more lines.
Something like this:
do
read(7,*,end=10) fibNumber
end do
10 continue
Better still - look at the more modern style used in this revcomp program.
here are some hints:
You don't need to use characters,
much less file i/o for this problem
(unless you forgot to state that a
file must be created).
Therefore, use math to find your statistics. There are lots of resources on Fibonacci numbers that might provide a simplifying insight or at least a way to independently spot check your answers.
Here is a complicated hint in non-Fortran lingo:
floor(10^(frac(log_10(7214989861293412))))
(Put this in Wolfram Alpha to see what it does.)
A simpler hint (for a different approach) is that you can do very
well in Fortran with simple
arithmetic inside of looping
constructs--at least for a first pass at the solution.
Accumulate your statistics as you
go. This advice would even apply to your character-driven approach. (This problem is ideally suited
for coming up with a cute indexing
scheme for your statistics, but some
people hate cute schemes in
programming. If you don't fear
cuteness ... then you can have associative
arrays in Fortran as long as your
keys are integers ;-)
The most important aspect of this
problem is the data type you will
use to calculate your answers. For
example, here's the last number you
will have to print.
Cheers, --Jared