Related
This is a small portion of the data I am trying to read:
01/06/2009,Tom Sanders,,264,220,73,260
01/08/2009,Adam Apple,158,,260,,208
01/13/2009,Lori Freeman,230,288,218,282,234
01/15/2009,Diane Greenberg,170,,250,321,197
01/20/2009,Adam Apple,257,,263,256,190
01/21/2009,Diane Greenberg,201,,160,195,142
01/27/2009,Tom Sanders,267,,143,140,206
01/29/2009,Tina Workman,153,,124,155,140
02/03/2009,Tina Workman,233,,115,,163
02/03/2009,Adam Apple,266,130,310,,310
the numbers between each comma are from a different location
Where two commas would represent missing data and a trailing comma would mean the fifth data point is missing
My goal is to organize the data into a table after calculating the average of each site and person, hence my two dim arrays
I want my output to look something like the following:
(obviously neater formatting but a table nonetheless)
Average Observed TDS (mg/l)
Name Site 1 Site 2 Site 3 Site 4 Site 5
------------------------------------------------------
Tom Sanders 251.0 172.5 251.7 160.0 229.0
Adam Apple 227.0 130.0 277.7 256.0 236.0
Lori Freeman 194.0 288.0 216.7 279.0 202.7
Diane Greenberg 185.5 190.0 205.0 258.0 169.5
Tina Workman 193.0 140.0 119.5 155.0 163.0
This is my program so far:
program name_finder
implicit none
integer, parameter :: wp = selected_real_kind(15)
real(wp) :: m, tds
real(wp), dimension(20,5) :: avg_site, site_sum
integer, dimension(20) :: nobs
integer, dimension(5) :: x
integer :: ierror, i, nemp, cp, non, ni, n
character(len=40), dimension(20) :: names
character(len=200) :: line, aname
character(len=20) :: output, filename
character(len=3), parameter :: a = "(A)"
do
write(*,*) "Enter file to open."
read(*,*) filename
open(unit=10,file = filename, status = "old", iostat = ierror)
if (ierror==0) exit
end do
write(*,*) "File, ",trim(filename)," has been opened."
non = 0
outer: do
read(10,a, iostat = ierror) line
if (ierror/=0) exit
cp = index(line(12:),",") + 11
aname = line(12:cp-1)
n=0
middle: do
read(line,'(Tcp,f4.2)') tds
write(*,*) "tds=", tds
n=n+1
if (n>10) exit
i = 1
inner: do
if (i > non) then
non = non +1
names(non) = trim(aname)
!ni = non
exit
end if
if (aname == names(i)) then
!ni = i
!cycle outer
exit inner
end if
i = i + 1
end do inner
end do middle
end do outer
write(*,*)
write(*,*) "Names:"
do i = 1,non
write(*,*) i, names(i)
end do
close(10)
close(20)
STOP
end program name_finder
TLDR; I am having trouble reading the data from the file shown at the top of each site after the names.
Suggestions? Thanks!
I hope the following is helpful. I have omitted any easily assumed declarations or any further data manipulation or writing to another file. The code is used just to read the data line by line.
character(150) :: word
read(fileunit, '(A)') word ! read the entire line
comma_ind = index(word,',') ! find the position of first comma
! Find the position of next comma
data_begin = index(word(comma_ind+1:),',')
! Save the name
thename = word(comma_ind+1:comma_ind+data_begin-1)
! Define next starting point
data_begin = comma_ind+data_begin
! Read the rest of the data
outer: do
if (word(data_begin+1:data_begin+1) == ',') then
! decide what to do when missing an entry
data_begin = data_begin + 1
cycle outer
else if (word(data_begin+1:data_begin+1) == ' ') then
! Missing last entry
exit outer
else
! Use it to find the length of current entry
st_ind = index(word(data_begin+1:),',')
if (st_ind == 0) then
! You reached the last entry, read it and exit
read(word(data_begin+1:), *) realData
exit outer
else
! Read current entry
read(word(data_begin+1: data_begin+st_ind-1),*) realData
end if
! Update starting point
data_begin = data_begin + st_ind
end if
end do outer
There could be a more elegant way to do it but I cannot think of any at the moment.
I have a few 110-element vectors. They sometimes have a value from 0 to 9, but their default value is -1. I'd like to print a blank if a cell's value is -1; print their value otherwise.
I'm printing several things in an output line so I can't use an if with two writes. Passing the values to a character vector worked but I can't help but think there must be a better way.
My attempt:
program integer_print_blank_test
implicit none
integer, dimension(9) :: longint
character(len=3), dimension(9) :: longchar
integer :: i, j
do i = 0, 2
write(*,*) (longint(3*i+j), j = 1, 3)
end do
longint = -1
longint(1) = 1
longint(4) = 3
longint(9) = 7
write(*,*) "longint"
do i = 0, 2
write(*,*) (longint(3*i+j), j = 1, 3)
end do
do i = 1, 9
write(longchar(i),"(I3)") longint(i)
end do
write(*,*) "longchar"
do i = 0, 2
write(*,*) (longchar(3*i+j), j = 1, 3)
end do
write(*,*) "only positives in longchar"
longchar = " "
do i = 1, 9
if (longint(i) > -1) then
write(longchar(i),"(I3)") longint(i)
end if
end do
do i = 0, 2
write(*,*) (longchar(3*i+j), j = 1, 3)
end do
end program integer_print_blank_test
You might think this is a better way. Define a function such as
ELEMENTAL FUNCTION borf(int) RESULT(str)
INTEGER, INTENT(in) :: int
CHARACTER(len=2) :: str
str = ' '
IF (int>-1) WRITE(str,'(i2)') int
END FUNCTION borf
and use it like this
WRITE(*,*) borf(longint)
I'm learning how to programming with fortran90 and i need receive data from a txt file by the command prompt (something like that:
program.exe"<"data.txt).
at the Input txt file I'll always have a single line with at least 6 numbers till infinity.
if the data was wrote line by line it runs fine but as single line I'm receiving the error: "traceback:not available,compile with - ftrace=frame or - ftrace=full fortran runtime error:end file"
*note: i'm using Force fortran 2.0
here is example of data:
0 1 0.001 5 3 1 0 -9 3
edit: just clarifying: the code is working fine itself except for the read statement, which is a simple "read*,". I want know how To read a entire line from a txt once the entrance will be made by the promt command with stream direction.
( you can see more about that here: https://www.microsoft.com/resources/documentation/windows/xp/all/proddocs/en-us/redirection.mspx?mfr=true).
there is no need to read the code, i've posted it just for knowledge.
I'm sorry about the whole inconvenience.
here is the code so far:
program bissecao
implicit none
integer::cont,int,e,k,intc,t1,t2,t3
doubleprecision::ii,is,pre,prec,erro,somaa,somab,xn
doubleprecision,dimension(:),allocatable::co
t1=0
t2=0
t3=0
! print*,"insira um limite inf da funcao"
read*,ii
!print*,"insira o limite superior da func"
read*,is
! print*,"insira a precisÆo admissivel"
read*,pre
if (erro<=0) then !elimina criterio de parada negativo ou zero
Print*,"erro"
go to 100
end if
!print*,"insira a qtd iteracoes admissiveis"
read*,int
!print*,"insira o grau da f(x)"
read*,e
if (e<=0) then ! elimina expoente negativo
e=(e**2)**(0.5)
end if
allocate(co(e+1))
!print*, "insira os coeficientes na ordem:&
! &c1x^n+...+(cn-1)x^1+cnx^0"
read(*,*)(co(k),k=e+1,1,-1)
somab=2*pre
intc=0
do while (intc<int.and.(somab**2)**0.5>pre.and.((is-ii)**2)**0.5>pre)
somab=0
somaa=0
xn =(ii+is)/2
do k=1,e+1,1
if (ii /=0) then
somaa=ii**(k-1)*co(k)+somaa
else
somaa=co(1)
end if
! print*,"somaa",k,"=",somaa
end do
do k=1,(e+1),1
if (xn/=0) then
somab=xn**(k-1)*co(k)+somab
else
somab=co(1)
end if
!print*,"somab",k,"=",somab
end do
if ((somaa*somab)<0) then
is=xn
else if((somaa*somab)>0)then
ii=xn
else if ((somaa*somab)==0) then
xn=(ii+is)/2
go to 100
end if
intc =intc+1
prec=is-ii
if ((((is-ii)**2)**.5)< pre) then
t3=1
end if
if (((somab**2)**.5)< pre) then
t2=1.
end if
if (intc>=int) then
t1=1
end if
end do
somab=0
xn=(ii+is)/2
do k=1,(e+1),1
if (xn/=0) then
somab=xn**(k-1)*co(k)+somab
else
somab=co(1)
end if
end do
100 write(*,'(A,F20.15,A,F20.15,A,A,F20.15,A,F20.15,A,I2)'),"I:[",ii,",",is,"]","raiz:",xn,"Fraiz:",somab,"Iteracoes:",intc
end program !----------------------------------------------------------------------------
In your program, you are using the "list-directed input" (i.e., read *, or read(*,*))
read *, ii
read *, is
read *, pre
read *, int
read *, e
read *, ( co( k ), k = e+1, 1, -1 )
which means that the program goes to the next line of the data file after each read statement (by neglecting any remaining data in the same line). So, the program works if the data file (say "multi.dat") consists of separate lines (as suggested by OP):
0
1
0.001
5
3
1 0 -9 3
But now you are trying to read an input file containing only a single line (say "single.dat")
0 1 0.001 5 3 1 0 -9 3
In this case, we need to read all the values with a single read statement (if list-directed input is to be used).
A subtle point here is that the range of array co depends on e, which also needs to be read by the same read statement. A workaround might be to just pre-allocate co with a sufficiently large number of elements (say 100) and read the data in a single line, e.g.,
integer :: k
allocate( co( 100 ) )
read *, ii, is, pre, int, e, ( co( k ), k = e+1, 1, -1 )
For completeness, here is a test program where you can choose method = 1 or 2 to read "multi.dat" or "single.dat".
program main
implicit none
integer :: int, e, k, method
double precision :: ii, is, pre
double precision, allocatable :: co(:)
allocate( co( 1000 ) )
method = 1 !! 1:multi-line-data, 2:single-line-data
if ( method == 1 ) then
call system( "cat multi.dat" )
read*, ii
read*, is
read*, pre
read*, int
read*, e
read*, ( co( k ), k = e+1, 1, -1 )
else
call system( "cat single.dat" )
read*, ii, is, pre, int, e, ( co( k ), k = e+1, 1, -1 )
endif
print *, "Input data obtained:"
print *, "ii = ", ii
print *, "is = ", is
print *, "pre = ", pre
print *, "int = ", int
print *, "e = ", e
do k = 1, e+1
print *, "co(", k, ") = ", co( k )
enddo
end program
You can pass the input file from standard input as
./a.out < multi.dat (for method=1)
./a.out < single.dat (for method=2)
Please note that "multi.dat" can also be read directly by using "<".
I have put a check in error made in the input as:
integer :: lsp
chksp:do
write(*,*) "#Enter Number"
read(*,*,iostat=istat)lsp
if (istat==0) then
exit chksp
else
write(*,*)"Number can only be integer. Re-enter!"
end if
end do chksp
The problem is, it can detect error if a character value in enteres, instead of a numeric value; but it cannot detect error, if a real value is entered, instead of a integer.
Any way to force it detect integer only?
NB: May be problem with ifort; gfortran is happy with the code.
You can specify the format to request an integer:
program enter_int
implicit none
integer :: ierror, intVal
do
write(*,*) "Enter an integer number"
read(*,'(i10)',iostat=ierror) intval
if ( ierror == 0 ) then
exit
endif
write(*,*) 'An error occured - please try again'
enddo
write(*,*) 'I got: ', intVal
end program
Then, providing a float fails.
Something like the following?
ian#ian-pc:~/test/stackoverflow$ cat read.f90
Program readit
Integer :: val
Integer :: iostat
val = -9999
Do
Read( *, '( i20 )', iostat = iostat ) val
If( iostat == 0 ) Then
Write( *, * ) 'val = ', val
Else
Write( *, * ) 'oh dear!!'
End If
End Do
End Program readit
ian#ian-pc:~/test/stackoverflow$ nagfor -o read read.f90
NAG Fortran Compiler Release 5.3.1(907)
[NAG Fortran Compiler normal termination]
ian#ian-pc:~/test/stackoverflow$ ./read
10
val = 10
10.0
oh dear!!
safs
oh dear!!
123dfs23
oh dear!!
^C
ian#ian-pc:~/test/stackoverflow$ gfortran -o read read.f90
ian#ian-pc:~/test/stackoverflow$ ./read
10
val = 10
10.0
oh dear!!
dsfs
oh dear!!
^C
ian#ian-pc:~/test/stackoverflow$
I'm attempting my first program in Fortran, trying to solve a quadratic equation. I have double and triple checked my code and don't see anything wrong. I keep getting "Invalid character in name at (1)" and "Unclassifiable statement at (1)" at various locations. What is wrong with the code?
! This program solves quadratic equations
! of the form ax^2 + bx + c = 0.
! Record:
! Name: Date: Notes:
! Damon Robles 4/3/10 Original Code
PROGRAM quad_solv
IMPLICIT NONE
! Variables
REAL :: a, b, c
REAL :: discrim, root1, root2,
COMPLEX :: comp1, comp2
CHARACTER(len=1) :: correct
! Prompt user for coefficients.
WRITE(*,*) "This program solves quadratic equations "
WRITE(*,*) "of the form ax^2 + bx + c = 0. "
WRITE(*,*) "Please enter the coefficients a, b, and "
WRITE(*,*) "c, separated by commas:"
READ(*,*) a, b, c
WRITE(*,*) "Is this correct: a = ", a, " b = ", b
WRITE(*,*) " c = ", c, " [Y/N]? "
READ(*,*) correct
IF correct = N STOP
IF correct = Y THEN
! Definition
discrim = b**2 - 4*a*c
! Calculations
IF discrim > 0 THEN
root1 = (-b + sqrt(discrim))/(2*a)
root2 = (-b - sqrt(discrim))/(2*a)
WRITE(*,*) "This equation has two real roots. "
WRITE(*,*) "x1 = ", root1
WRITE(*,*) "x2 = ", root2
IF discrim = 0 THEN
root1 = -b/(2*a)
WRITE(*,*) "This equation has a double root. "
WRITE(*,*) "x1 = ", root1
IF discrim < 0 THEN
comp1 = (-b + sqrt(discrim))/(2*a)
comp2 = (-b - sqrt(discrim))/(2*a)
WRITE(*,*) "x1 = ", comp1
WRITE(*,*) "x2 = ", comp2
PROGRAM END quad_solv
Re the additional question of how to loop back to redo input -- an example program demonstrating loop features of Fortran >= 90. The loop is apparently infinite -- exit controlled by the IF statement exits the loop and makes the loop finite. Also shown is the cycle control, which is used if invalid input is encountered, which would otherwise crash the program. (For example, type "A" as input to the first read, instead of a number.) In this case, the iostat variable acquires a non-zero value, the IF statement activates the cycle, and the rest of the DO loop is skipped, so that the loop cycles anew.
program test_readdata
real :: a, b, c
integer :: ReturnCode
character (len=1) :: correct
ReadData: do
write (*, '( "This program solves quadratic equations of the form ax^2 + bx + c = 0. " )' )
write (*, '( "Please enter the coefficients a, b, and c, separated by commas: " )', advance='no' )
read (*,*, iostat=ReturnCode) a, b, c
if ( ReturnCode /= 0 ) cycle ReadData
write (*,*) "Is this correct: a = ", a, " b = ", b, " c = ", c
write (*, '( "Enter Y or N: " )', advance='no' )
read (*,*, iostat=ReturnCode) correct
if ( ReturnCode /= 0 ) cycle ReadData
if ( correct == 'Y' .or. correct == 'y' ) exit ReadData
end do ReadData
stop
end program test_readdata
My book recommendation: Fortran 95/2003 explained by Metcalf, Reid and Cohen.
The first thing I noticed with your code is the syntax error at the end of your program.
END
should have preceded the word program
Doesn't your editor highlight your syntax errors?
You can get the student version of ELF 90 pretty inexpensively and it's a good place to start. I would then upgrade to Lahey ELF 95 with the Visual Studio and algorithm flow chart generator which color codes the pathways of the passing of the values.
So many errors... It seems you don't know Fortran basics...
With minimal corrections
! This program solves quadratic equations
! of the form ax^2 + bx + c = 0.
! Record:
! Name: Date: Notes:
! Damon Robles 4/3/10 Original Code
PROGRAM quad_solv
IMPLICIT NONE
! Variables
REAL :: a, b, c
REAL :: discrim, root1, root2
COMPLEX :: comp1, comp2
CHARACTER(len=1) :: correct
! Prompt user for coefficients.
WRITE(*,*) "This program solves quadratic equations "
WRITE(*,*) "of the form ax^2 + bx + c = 0. "
WRITE(*,*) "Please enter the coefficients a, b, and "
WRITE(*,*) "c, separated by commas:"
READ(*,*) a, b, c
WRITE(*,*) "Is this correct: a = ", a, " b = ", b
WRITE(*,*) " c = ", c, " [Y/N]? "
READ(*,*) correct
IF (correct == 'N') STOP
IF (correct == 'Y') THEN
! Definition
discrim = b**2 - 4*a*c
! Calculations
IF (discrim > 0) THEN
root1 = (-b + sqrt(discrim))/(2*a)
root2 = (-b - sqrt(discrim))/(2*a)
WRITE(*,*) "This equation has two real roots. "
WRITE(*,*) "x1 = ", root1
WRITE(*,*) "x2 = ", root2
ELSEIF (discrim == 0) THEN
root1 = -b/(2*a)
WRITE(*,*) "This equation has a double root. "
WRITE(*,*) "x1 = ", root1
ELSE
comp1 = (-b + sqrt(discrim))/(2*a)
comp2 = (-b - sqrt(discrim))/(2*a)
WRITE(*,*) "x1 = ", comp1
WRITE(*,*) "x2 = ", comp2
END IF
END IF
END PROGRAM quad_solv
P.S. I didn't check the correctness.
P.P.S. Always indent your code to make it readable and don't use STOP statement. Each program (or subroutine) should have one entry and one exit. The only right place for STOP is exactly before END PROGRAM statement where it is redundant. So don't use STOP at all.
Some changes: correct IF syntax; constants to be real numbers rather than integers, when appropriate:
IF (CORRECT == "N" ) stop
if ( discrim > 0.0 ) then
Apply at additional locations, and you should be very close. Good luck.
It can be a poor algorithm to test a floating point number for having an exact value. What if rounding error makes disrim have the value 1.0E-30, when perfect arithmetic would give the value zero and indicate a double root?