I have a file to read whose structure is:
26.0 24773 -55.4 -86.8 1 0.01 60 27 617.7 617.8 617.
24.2 25230 -55.7 -86.7 1 0.01 64 22 629.7 629.8 629.
Station information and sounding indices
I'm interested in reading only the number. I used the following format:
read(333,10) a(i),b(i)...
10 format(1x,f6.1,2x,i5,...
read(333,iostat=stat) Station
if(stat/=0)then
else
endif
if(stat==0)exit
from this the error appears:
forrtl: severe (64): input conversion error, unit 333, file
Related
I have a solution for solving an MIP problem of graph which works fine and gives the following output when I run it for smaller graphs. I'm using Gurobi solver with Pyomo.
Problem:
- Name: x73
Lower bound: 192.0
Upper bound: 192.0
Number of objectives: 1
Number of constraints: 10
Number of variables: 37
Number of binary variables: 36
Number of integer variables: 36
Number of continuous variables: 1
Number of nonzeros: 37
Sense: minimize
Solver:
- Status: ok
Return code: 0
Message: Model was solved to optimality (subject to tolerances), and an optimal solution is available.
Termination condition: optimal
Termination message: Model was solved to optimality (subject to tolerances), and an optimal solution is available.
Wall time: 0.03206682205200195
Error rc: 0
Time: 0.09361410140991211
Solution:
- number of solutions: 0
number of solutions displayed: 0
But I am getting the following error while running the code with larger graphs.
ERROR: Solver (gurobi) returned non-zero return code (137)
ERROR: Solver log: Using license file /opt/shared/gurobi/gurobi.lic Set
parameter TokenServer to value gurobi.lm.udel.edu Set parameter TSPort to
value 40100 Read LP format model from file /tmp/tmpaud9ogrn.pyomo.lp
Reading time = 0.01 seconds x1101: 56 rows, 551 columns, 551 nonzeros
Changed value of parameter TimeLimit to 600.0
Prev: inf Min: 0.0 Max: inf Default: inf
Gurobi Optimizer version 9.0.1 build v9.0.1rc0 (linux64) Optimize a model
with 56 rows, 551 columns and 551 nonzeros Model fingerprint: 0xafe0319a
Model has 15400 quadratic objective terms Variable types: 1 continuous,
550 integer (550 binary) Coefficient statistics:
Matrix range [1e+00, 1e+00] Objective range [0e+00, 0e+00]
QObjective range [4e+00, 8e+01] Bounds range [1e+00, 1e+00] RHS
range [1e+00, 1e+00]
Found heuristic solution: objective 22880.000000 Presolve removed 1 rows
and 1 columns Presolve time: 0.01s Presolved: 55 rows, 550 columns, 550
nonzeros Presolved model has 15400 quadratic objective terms Variable
types: 0 continuous, 550 integer (550 binary)
Root simplex log...
Iteration Objective Primal Inf. Dual Inf. Time
130920 9.8490000e+02 1.610955e+03 0.000000e+00 5s 263917
1.0999000e+03 1.710649e+03 0.000000e+00 10s 397157
1.0999000e+03 2.243077e+03 0.000000e+00 15s 529512
1.0999000e+03 1.910603e+03 0.000000e+00 20s 662404
1.0999000e+03 1.584650e+03 0.000000e+00 25s 791296
1.0999000e+03 1.812443e+03 0.000000e+00 30s 906473
1.3475000e+03 0.000000e+00 0.000000e+00 34s
Root relaxation: objective 1.347500e+03, 906473 iterations, 34.32 seconds
Nodes | Current Node | Objective Bounds | Work
Expl Unexpl | Obj Depth IntInf | Incumbent BestBd Gap | It/Node
Time
H 0 0 1730.0000000 0.00000 100% - 52s
H 0 0 1654.0000000 0.00000 100% - 52s
H 0 0 1578.0000000 0.00000 100% - 52s
0 0 1347.50000 0 137 1578.00000 1347.50000 14.6% - 52s
0 0 1347.50000 0 137 1578.00000 1347.50000 14.6% - 53s
H 0 0 1540.0000000 1347.50000 12.5% - 53s
0 2 1347.50000 0 145 1540.00000 1347.50000 12.5% - 55s
101 118 1396.92351 10 140 1540.00000 1347.50000 12.5% 157 61s
490 591 1416.40484 18 128 1540.00000 1347.50000 12.5% 63.6 65s
2136 2347 1440.09938 42 100 1540.00000 1347.50000 12.5% 42.9 70s
3847 3402 1461.55736 81 80 1540.00000 1347.50000 12.5% 37.0 82s
/opt/shared/gurobi/9.0.1/bin/gurobi.sh: line 17: 23890 Killed
$PYTHONHOME/bin/python3.7 "$#"
Traceback (most recent call last):
File "/home/2925/EdgeColoring/main.py", line 91, in <module>
qubo_coloring, qubo_time = qubo(G, colors, edge_list, solver)
File "/home/2925/EdgeColoring/qubo.py", line 59, in qubo
result = solver.solve(model)
File "/home/2925/.conda/envs/qubo/lib/python3.9/site-packages/pyomo/opt/base/solvers.py", line 596, in solve
raise ApplicationError(
pyomo.common.errors.ApplicationError: Solver (gurobi) did not exit normally
Using TimeLimit upto 2 minutes breaks the model early without any error but doesn't always give any optimal solution for larger graphs. Memory or processing power is not an issue here. I need to run the code without any interruption for at least 10 minutes if not for hours.
I have got a .dat file with 2 columns and rows between 14000 to 36000 saved in file like below:
0.00 0.00
2.00 1.00
2.03 1.01
2.05 1.07
.
.
.
79.03 23.01
The 1st column is extension, the 2nd is strain. When I want to count gradient to designate Hooks Law of the plot, I use below code.
CCCCCC
Program gradient
REAL S(40000),E(40000),GRAD(40000,1)
open(unit=300, file='Probka1A.dat', status='OLD')
open(unit=321, file='result.out', status='unknown')
write(321,400)
400 format('alfa')
260 DO 200 i=1, 40000
read(300,30) S(i),E(i)
30 format(2F7.2)
GRAD(i,1)=(S(i)-S(i-1))/(E(i)-E(i-1))
write(321,777) GRAD(i,1)
777 Format(F7.2)
200 Continue
END
But after I executed it I got the warning
PGFIO-F-231/formatted read/unit=300/error on data conversion.
File name = Probka1A.dat formatted, sequential access record = 1
In source file gradient1.f, at line number 9
What can I do to count gradient by this or other way in Fortran 77?
You are reading from file without checking for the end of the file. Your code should be like this:
260 DO 200 i=1, 40000
read(300,*,ERR=400,END=400) S(i),E(i)
if (i>1) then
GRAD(i-1,1)=(S(i)-S(i-1))/(E(i)-E(i-1))
write(321,777) GRAD(i-1,1)
end if
777 Format(F7.2)
200 Continue
400 continue
Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 7 years ago.
Improve this question
Hi all I am working on a school beginners project using files I/O in C++,
This program consist of two parts:
1) reading and processing a student data file, and writing the results to a student report file
2) modifying part 1 to calculate some statistics and writing them to another file.
For this assignment, you will be reading one input file and writing out two other files.
Your program will be run using the referenced student data file.
Part 1 Detail
Read in the student data file. This 50 record file consists of a (8-digit numeric) student id, 8 assignment's points, midterm points, final points and lab exercise points. You must again follow the syllabus specifications for the determination of the letter grade, this time, processing 50 student grades. Extra credit points are not applicable for this assignment. You will write the input student data and the results of the processing into a student report file that looks like the output shown below. In addition to the input student data, the report should contain the "total" of the assignment grades, the total and percent of all points achieved, and the letter grade. You may assume that the input data file does not contain any errant data.
The file looks like the one below:
The file that we need to read from is hyperlinked here
The student report output file should look like this:
The Student Report Output File
Student --- Asignment Grades -- Ass Mid Fin LEx Total Pct Gr
-------- -- -- -- -- -- -- -- -- --- --- --- --- ----- --- --
56049257 16 16 20 16 12 15 12 20 115 58 123 59 355 89 B+
97201934 19 15 13 19 16 12 13 18 113 72 101 55 341 85 B
93589574 13 16 19 19 18 12 6 14 111 58 108 50 327 82 B
85404010 17 19 19 19 19 10 17 19 129 70 102 58 359 90 A-
99608681 11 15 19 19 17 10 16 19 116 42 117 57 332 83 B
84918110 11 20 18 17 12 8 12 19 109 46 122 31 308 77 C
89307179 16 16 19 18 14 17 15 19 120 56 117 52 345 86 B
09250373 15 15 18 18 11 18 17 19 120 44 106 51 321 80 B-
91909583 12 14 16 19 20 11 20 16 117 66 92 50 325 81 B-
...
Part 2 Detail
Write a summary report file that contains the average total points and average percent for all students. Also, display the number of A's, B's, C's, D's and F's for the students. Your summary output file should look something like this:
The average total points = ???
The average percent total = ??
The number of A's = ??
The number of B's = ??
The number of C's = ??
The number of D's = ??
The number of F's = ??
Additional requirements
All files must be checked for a successful open. They should also be closed when you are finished with them.
Make sure you write the student id with a leading 0, if appropriate (i.e. the 8th id).
Add headings to your output report file. They should be aligned and correctly identify the column data.
Do not use global variables, except for constants, in your solution.
For part 1 How do I duplicate the file and format it to add the headings above it and the grades at the end of each file into the new duplicated file??
Any help in this matter would be appreciated
thanks in advance.
Engineering is all about converting a large complex problem into many smaller, easy to solve, problems.
Here is how I would start.
1.) Open input file.
2.) Read one line from input file.
3.) Break the input string from one line into values.
4.) Close input file.
5.) Open output file.
6.) Write results to output file.
References:
1.)File I/O
2.)std::string
3.)File I/O C
Now you're pretty much there. Take it one step at a time.
I am updating my RRD file with some counts...
For example:
time: value:
12:00 120
12:05 135
12:10 154
12:20 144
12:25 0
12:30 23
13:35 36
here my RRD is updating as below logic:
((current value)-(previous value))/((current time)-(previous time))
eg. ((135-120))/5 = 15
but my problem is when it comes 0 the reading will be negative:
((0-144))/5
Here " 0 " value comes with system failure only( from where the data is fetched)..It must not display this reading graph.
How can I configure like when 0 comes it will not update the "RRD graph" (skip this reading (0-144/5)) and next time it will take reading like ((23-0)/5) but not (23-144/10)
When specifying the data sources when creating the RRD, you can specify which range of values is acceptable.
DS:data_source:GAUGE:10:1:U will only accept values above 1.
So if you get a 0 during an update, rrd will replace it with unknown and i assume it can find a way to discard it.
I have a Fortran code which reads a txt file with seperate lines of characters and digits and then write them in a 1D array with 20 elements.
This code is not compatible with Fortran 77 compiler Force 2.0.9. My question is that how we can apply the aformenetioned procedure using a Fortran 77 compiler;i.e defining a 1D array nd then write the txt file line by line into elements of the array?
Thank you in advance.
The txt file follows:
Case 1:
10 0 1 2 0
1.104 1.008 0.6 5.0
25 125.0 175.0 0.7 1000.0
0.60
1 5
Advanced Case
15 53 0 10 0 1 0 0 1 0 0 0 0
0 0 0 0
0 0 1500.0 0 0 .03
0 0.001 0
0.1 0 0.125 0.08 0.46
0.1 5.0 0.04
# Jason:
I am a beginner and still learning Fortran. I guess Force 2 uses g77.
The followings are the correspond part of the original code. Force 2 editor returns an empty txt file as a result.
DIMENSION CARD(20)
CHARACTER*64 FILENAME
DATA XHEND / 4HEND /
OPEN(UNIT=3,FILE='CON')
OPEN(UNIT=4,FILE='CON')
OPEN(UNIT=7,STATUS='SCRATCH')
WRITE(3,9000) 'PLEASE ENTER THE INPUT FILE NAME : '
9000 FORMAT (A)
READ(4,9000) FILENAME
OPEN(UNIT=5,FILE=FILENAME,STATUS='OLD')
WRITE(3,9000) 'PLEASE ENTER THE OUTPUT FILE NAME : '
READ(4,9000) FILENAME
OPEN(UNIT=6,FILE=FILENAME,STATUS='NEW')
FILENAME = '...'
IR = 7
IW = 6
IP = 15
5 REWIND IR
I = 0
2 READ (5,7204,END=10000) CARD
IF (I .EQ. 0 ) WRITE (IW,7000)
7000 FORMAT (1H1 / 10X,15HINPUT DECK ECHO / 10X,15(1H-))
I= I + 1
WRITE (IW,9204) I,CARD
IF (CARD(1) .EQ. XHEND ) GO TO 7020
WRITE (IR,7204) CARD
7204 FORMAT (20A4)
9204 FORMAT (1X,I4,2X,20A4)
GO TO 2
7020 REWIND IR
It looks that CARD is being used as a to hold 20 4-character strings. I don't see the declaration as a character variable, only as an array, so perhaps in extremely old FORTRAN style a non-character variable is being used to hold characters? You are using a 20A4 format, so the values have to be positioned in the file precisely as 20 groups of 4 characters. You have to add blanks so that they are aligned into groups of 4 columns.
If you want to read numbers it would be much easier to read them into a numeric type and use list-directed IO:
real values (20)
read (5, *) values
Then you wouldn't have to worry about precision positioning of the values in the file.
This is really archaic FORTRAN ... even pre-FORTRAN-77 in style. I can't remember the last time that I saw Hollerith (H) formats! Where are you learning this from?
Edit: While I like Fortran for many programming tasks, I wouldn't use FORTRAN 66! Computers are supposed to make things easier ... there is no reason to have to count characters. Instead of
7000 FORMAT (1H1 / 10X,15HINPUT DECK ECHO / 10X,15(1H-))
You can use
7000 FORMAT ( / 10X, "INPUT DECK ECHO" / 10X, 15("-") )
I can think of only two reasons to use a Hollerith code: not bothering to change legacy source code (it is remarkable that a current Fortran compiler can process a feature that was obsolete 30 years ago! Fortran source code never dies!), or studying the history of computing languages. The name honors a great computing pioneer, whose invention accomplished the 1890 US Census in one year, when the 1880 Census took eight years: http://en.wikipedia.org/wiki/Herman_Hollerith
I much doubt that you will see the "1" in the first column performing "carriage control" today. I had to look up that "1" was the code for page eject. You are much more likely to see it in your output. See Are Fortran control characters (carriage control) still implemented in compilers?