I read real valued data from an input file with list directed input. What is the best way to detect a NaN via iostat=iostatus?
As an example for the input for a large program I need to read a file like:
1 NaN 1.0
2 nan 2.0
where there are occasionally "NaN"s instead of real valued numbers.
with READ(ird, *, iostat=iostatus) nr, r1, r2
no iostatus is flagged for gfortran 7.4.0 on UBUNTU!
Apparently the input error is treated as a quiet NaN instead of a signalling NaN.
The gfortran compiler options do not seem to provide an option to turn this into a signalling NaN.
The fortran standard at
https://j3-fortran.org/doc/year/18/18-007r1.pdf
does not specify treatment of a Nan in list directed input.
In section 13.7.2.3.2 there is info for formatted input, and it specifically states that the Nan on input is treated as a quiet NaN.
I found a documentation on the NAG compiler and it states explicitely that a NaN on input is never flagged as a signalling NaN.
The philosophy behind these decisions in the standard beats me. I would have expected that this shoud be an error.
If the line is changed to:
read(ird, *, err=10, iostat=iostatus) nr, r1, r2
10 continue
the iostatus changes to +5001 ! But I do not really want to resort back to the ancient err= clause.
I am aware that I could read the line into a character string, analyze the string for the presence of any "NaN" and proceed. I could as in the example analyse ieee_is_nan(). That seems however unnecessarily tedious given the iostat= clause.
So is there a compiler option that I missed ?
PROGRAM main
USE ieee_arithmetic
IMPLICIT NONE
INTEGER, PARAMETER :: IRD = 12
INTEGER :: iostatus
REAL :: r1 = -1.0, r2 = -1.0
INTEGER :: nr = 0
CHARACTER(LEN=1024) :: line = ' '
OPEN(UNIT=IRD,FILE='data.data', STATUS='old')
READ(IRD, *, IOSTAT=iostatus, IOMSG=line) nr, r1, r2
WRITE(*,*) iostatus, nr, r1, r2, ieee_is_nan(r1), line(1:LEN_TRIM(line))
r1 = -1.0
r2 = -1.0
READ(IRD, ERR=10, IOSTAT=iostatus, IOMSG=line) nr, r1, r2
10 CONTINUE
WRITE(*,*) iostatus, nr, r1, r2, ieee_is_nan(r1), line(1:LEN_TRIM(line))
CLOSE(UNIT=IRD)
END PROGRAM main
The output of my program is: (Shortened insignificant digits and blanks)
0 1 NaN 1.00 T
5001 1 -1.00 -1.00 F
Missing format for FORMATTED data transfer
At the first read iostatus is set to 0, while on the second read its set to 5001!
Related
In 8 bits we can store 8 numbers from 0 to 1 each. We can also say that we can store 8 different piece of data in a range from 0 to 255.
0/1 0/1 0/1 0/1 0/1 0/1 0/1 0/1 → 8 different piece of data
↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓
[bit] [bit] [bit] [bit] [bit] [bit] [bit] [bit] → 8 bits total
If instead of storing 0 or 1 we need to store 0, 1 or 2, then we will end up using more bits, of course. In this condition, by the way, we can store 0, 1, 2 or 3. However, this is beyond what I need but technically it will use the same amount of data space.
0/1/2 0/1/2 0/1/2 0/1/2 → 4 different piece of data (expectative)
0/1/2/3 0/1/2/3 0/1/2/3 0/1/2/3 → 4 different piece of data (too much for me)
↓ ↓ ↓ ↓
[2bits] [2bits] [2bits] [2bits] → 8 bits total
We can agree that we are "losing" storage capacity. The solution is to do a radix conversion (I don't know if that's the right term) using the numerical base of 3, this way we will achieve the objective of storing exclusively 0, 1 or 2, and in the same 8 bits we can store more information (and, to be honest, I don't know how the bits organize themselves for this to work).
Examples:
00000₃ → int 0
01212₃ → int 50
11111₃ → int 121
12121₃ → int 151
22222₃ → int 242 (max)
In this conversion, we were able to store 5 pieces of information composed of 0, 1 or 2, instead of just 4. And there's still a "space" left, and that's what I'd like to talk about.
I was wondering if there is any way to store mixed base data instead of fixed base (eg. 3, as exemplified above). To be clearer, supposing that I wanted to store two pieces of data composed of 0, 1 or 2 each, and one piece of data composed of a number between 0 and 8. In binary, and still limited to 256, we can do it as follows:
0/1/2 0/1/2 0/1/2/3/4/5/6/7/8 → 3 piece of different data
↓ ↓ ↓
[2bits] [2bits] [ 4 bits ] → 8 bits total
But, it seems to me that we have empty "spaces" left again, since it doesn't seem to me to be using the full capacity possible in 8 bits, and maybe it's still possible to add more data if we use number base conversion instead.
The problem is: I don't know how to handle the data in this way, in a way to reliably merge and split it again.
In the real world: I need to transfer a lot of data that uses very little information (eg numbers 0 to 2, 0 to 5, 0 to 10, etc.). And I'm currently doing bitwise snapping of this data. And in my case, any optimized byte is a very nice gain (if this is really possible, maybe there is a 20~40% data saving).
I understand that it might consume more processing on rebasing, merge and split conversions, but that won't be an issue, because this processing will be done on the client side (merge when sending and split when receiving), and the server manages the same data already optimized (no additional processing).
--
Possible solutions:
Let's imagine that I have a set of three numbers that can be 0, 1 or 2, and another set of three numbers that can go from 0 to 8 (so it is 9 possibilities each). The goal is to merge all these numbers into a single integer (which should use about 2 bytes at best current method).
Solution 1: each number being stored in a single complete byte:
byte A from 0 to 2
byte B from 0 to 2
byte C from 0 to 2
byte D from 0 to 8
byte E from 0 to 8
byte F from 0 to 8
Problem: it will be necessary to consume 6 bytes to store these 6 numbers.
Solution 2: storage through bits:
2 bits to store A
2 bits to store B
2 bits to store C
4 bits to store F
4 bits to store G
4 bits to store H
6 unused bits to complete the last byte
Problems: in addition to 2 bits being "more than necessary" to store numbers from 0 to 2 (A, B and C), and same for 4 bits from 0 to 8, we will use a total of 3 bytes and still have 6 "dead" (unused) bits at end to complete the last byte.
Solution 3: Separate into two sets and convert their respective bases individually:
A, B and C to base 3 consumes 1 byte
D, E and F to base 9 consumes 2 bytes
Problem: despite being a great solution at the moment (and despite the example above, in more complex situations it can be more optimized than solution 2), and that's what I'm using, I believe there's a lot of "left over" space in this union, and maybe it's still possible to squeeze everything into 2 bytes only.
For example, the conversion of 222₃ consumes up to 5 bits, which means that in this byte we still have 3 unused bits. The 888₉ conversion consumes 10 bits. This makes me see the possibility of using only 15 bits with only 1 unused bit (2 bytes).
Then we come to the next solution.
Solution 4: move the bits to further optimize space:
higher bits stores A, B and C
lower bits stores D, E and F
Example:
[5 bits of numbers of base 3] +
[10 bits of numbers of base 9] +
[1 unused bit]
Problem: currently this solution is even better than the one I am currently using. However, I still see a possibility to improve in situations where more information may be available through number base conversion and union.
You pack your values in the following way:
Example:
possible values for
a = [0,2] -> 3 states
b = [0,2] -> 3 states
c = [0,8] -> 9 states
lets assume you have following values
a := 1
b := 2
c := 8
than you can calculate the final number the following way
int number = 0;
int multi = 1;
number = number.add(multi.multiply(a));
multi = multi.multiply(3));
number = number.add(multi.multiply(b));
multi = multi.multiply(3));
number = number.add(multi.multiply(c));
number holds now your packed numbers
to unpack you just need todo
a = number.mod(3)
number = number.divide(3)
b = number.mod(3)
number = number.divide(3)
c = number.mod(9)
I am working on vba code where I have data (for Slope Inclinometers) at various depths like so:
Depth A0 A180 Checksum B0 B180 Checksum
4.5 (-1256) 1258 2 (-394) 378 (-16)
4.5 (-1250) 1257 7 (-396) 376 (-20)
4.5 (-1257) 1257 0 (-400) 374 (-26)
Depth A0 A180 Checksum B0 B180 Checksum
5 (-1214) 1214 0 (-472) 459 (-13)
5 (-1215) 1212 -3 (-472) 455 (-17)
5 (-1216) 1211 -5 (-473) 455 (-18)
UNKNOWN AMOUNT OF DATA WILL BE PRESENT (depends how much the user transfers to this sheet)
Now I need to be able to calculate the A Axis Displacement, the B Axis Displacement, and the resultant which have formulas as followed:
A Axis Displacement = [((A0-A180)/2)-((A0*-A180*)/2))]*(constant/constant)
Where * is the initial readings which is always the first row of data at that specified depth.
B Axis Displacement = [((A0-A180)/2)-((A0*-A180*)/2))]*(constant/constant)
Where * is the initial readings which is always the first row of data at that specified depth.
Resultant = SQRT[(A Axis Displacement)^2 + (B Axis Displacement)^2]
I'm struggling to find examples of how I can implement this using vba as there will be various depths present (unknown amount) on the same sheet where the formula will need to start over at each new depth present.
Any helps/tips would be greatly appreciated!
how I can implement this using vba as there will be various depths present...
You still can do it purely with formulas and easy auto-fill, because the formula can find the the first occurrence of the current depth and perform all the necessary calculations, leaving blank at header rows or blank rows. For instance, you can enter these formulas at row 2 and fill down all the rows.
H2 (A Axis Displacement):
=IF(ISNUMBER($A2),0.5*(B2-C2-VLOOKUP($A2,$A:$F,2,0)+VLOOKUP($A2,$A:$F,3,0)), "")
I2 (B Axis Displacement):
=IF(ISNUMBER($A2),0.5*(E2-F2-VLOOKUP($A2,$A:$F,5,0)+VLOOKUP($A2,$A:$F,6,0)), "")
J2 (Resultant):
=IF(ISNUMBER($A2),SQRT(SUMSQ(H2,I2)),"")
p.s. in the displacements formulas I omitted the (constant/constant) factor as it is irrelevant to the answer, you can easily multiply the 0.5 factor by anything you need.
if we write 12wkd3, how to choose/filter 123 as integer in octave?
example in octave:
A = input("A?\n")
A?
12wkd3
A = 123
while 12wkd3 is user keyboard input and A = 123 is the expected answer.
assuming that the general form you're looking for is taking an arbitrary string from the user input, removing anything non-numeric, and storing the result it as an integer:
A = input("A? /n",'s');
A = int32(str2num(A(isdigit(A))));
example output:
A?
324bhtk.p89u34
A = 3248934
to clarify what's written above:
in the input statement, the 's' argument causes the answer to get stored as a string, otherwise it's evaluated by Octave first. most inputs would produce errors, others may be interpreted as functions or variables.
isdigit(A) produces a logical array of values for A with a 1 for any character that is a 0-9 number, and 0 otherwise.
isdigit('a1 3 b.') = [0 1 0 1 0 0 0]
A(isdigit(A)) will produce a substring from A using only those values corresponding to a 1 in the logical array above.
A(isdigit(A)) = 13
that still returns a string, so you need to convert it into a number using str2num(). that, however, outputs a double precision number. so finally to get it to be an integer you can use int32()
:N-remainder(dim(L1),N→ dim(L2)
:Fill(23,L2
:augment(L1, L2->L1
:{1,1→dim([A]
:For(x,1,dim(L1)/N
:augment([A],List▶matr(seq(L1(I),I,Nx-N+1,Nx),[B]
:End
I get a syntax error when running this Ti-basic code and I cannot figure out why (happens somewhere when List is being converted to matrix). Basically this code is suppose to take a L1 (add 23 until I dim(L1) is a multiply of N), then create a matrix with N rows and -int(-dim(L1)/n) columns.
Example:
Let N=3 and
L1 = {9,12,15,22,5,9,14,4,9,1,14,7,9,18,12,19}
dim(L1) = 16 which is not a multiply of 3 (18 is so add 23 to L1 twice)
L1 = {9,12,15,22,5,9,14,4,9,1,14,7,9,18,12,19,23,23}
dim(L1) = 18 which is a multiple of 3
Create a 3x6 matrix with Col1 = {9,12,15}, Col2 = {22,5,9}, ..., Col6 = {19,23,23}
http://tibasicdev.wikidot.com/forum/t-1039272/comments/show?from=activities#post-2131820
Read full convo. here
There are at least two issues with your code:
(1) For the augment command both matrices must share the same number of rows. In your program matrix [A] is set to dimension {1,1} (Why?), but the columns you want to append are of different size. So you'll get a "dimension error".
(2) The List▶matr command doesn't return a matrix (actually it doesn't return anything). So you can't use it as second parameter for the augment command. Instead you must run it first and then use something like augment([A],[B])▶[C].
I'm writing a task in pascal.
Everything is ok, just my result is not right.
I'm summing some numbers
Example: 2.3 + 3.4+ 3.3 = 9
But output shows: 9.000000 + EEE or something like that.
So- how to convert, to be only 9, not this REAL variable.
To actually convert:
var
i: integer;
...
i := round(floatVar);
To output only the integer part:
writeln(floatVar:9:0);
Let's consider this quite simpler equation:
3.5 + 2.5
What do you expect? 6, right? Let's try this code
write(3.5 + 2.5);
Unfortunately, it's a floating-point number, so it would produce a number represented in a scientific way:
6.00000000000E+00
or 6.0000000000 x 100, or 6 x 10o. Whatever, you only care about 6, who need this weird useless long number? So the idea is to cut off the decimal part and output to the console only the integer part, which can be done with this line of code:
write(3.5 + 2.5 : 0 : 0);
Ok, now it outputs a beautiful number as expected
6
Seems like the problem is solved, but you say that:
I'm summing some numbers
Example: 2.3 + 3.4+ 3.3 = 9
Ohh so that the evenly, beautiful integer is just randomly appeared? Here the problem comes, how do you expect this equation would output?
3.6 + 2.5
It should be 6.1, right? Let's try it with the worked line of code:
write(3.6 + 2.5 : 0 : 0);
And the output is...
6
Unexpected, right? So how about rounding to some decimal places, like 1?
write(3.5 + 2.5 : 0 : 1);
write(3.6 + 2.5 : 0 : 1);
Then, 3.5 + 2.5 = 6.0 and 3.6 + 2.5 = 6.1. But 6.0 may look quite long, so how to make it output 6 for 6.0 and 6.1 for 6.1?
Actually, you can't make the program auto-detect if a real variable contains an integer value because the way a real var is stored is completely different from an integer var (how different they are, please contact Google; but you can do it manually by making a function to do the job).
So my solution is, to be easy, making the output rounded to some decimal places, and that's it.
For purpose of showing pretty output on the screen you can use something like this:
Writeln(result:0:2);
Result on screen would be this:
9.00
What this means someone would ask? Well first number 0 means how wide filed is. if you say it's 0 then Pascal writes it at the very left side of screen. If you said writeln(result:5:2) result would be:
9.00
In other words i would print form the right side and leave 5 chars to do so.
Second number 2, in this example means you want that result printed with 2 decimal places. You can place it only if you want to print on screen value that is real, single, double, extended and so on.You can round to any number of decimals, and if you do writeln(result:0:0) you would get ouput:
9
If you are printing integer and want to have some length of field, lets sat 5 you would do: writeln(int:5). If you added :2 to the end you would get compile time error.
This all also works for something like this: writeln(5/3.5+sqrt(3):0:3),
You should know that this does not round variable itself but just formats output. This is also legal:
program test;
var
a:real;
n,m:integer;
begin
readln(a,m,n);
writeln(a:m:n);
end.
What i did here is i asked user if on how many decimals and with what length of field he wants to write entered number a. This can be useful so i'm pointing it out. Thank you for reading. I hope i helped
You can convert to string, get the int part, e convert to int number!
Or Float to Str than Str to Int:
nPage := StrToInt(FloatToStr(Int(nReg / nTPages))) + 1;