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How to convert numbers to string with optimal number of digits?

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Hi.

 

I am trying to do something that seems quite simply, and I am surprised at how much trouble I am having with this.  I am pretty sure that there is a way to get LabVIEW to do this.

 

I have an array of numbers (double), and I am using the "Array To Spreadsheet String" function to generate a string that I can write to a file.

 

The problem I am having is that I want to string to omit unnecessary trailing zeros after the decimal point, but I don't want to limit the precision of numbers that do require it.

 

For example, if my data is this:

0.02

0.4476

-0.0244570001

 

I would like for the string that I get to be exactly like what I wrote above.  However, depeding on the format string I use, this is what I get:

 

format string: %.10f

0.0200000000

0.4476000000

-0.0244570001

 

<OR>

 

format string: %f  (this seems to default to 6 digits)

0.020000

0.447600

-0.024457

 

<OR>

format string: %.4f

0.0200

0.4476

-0.0244 <= all trailing digits for this number have been lost!!

 

Does someone know what the format string should be to get the string I want (which is this:)

0.02

0.4476

-0.0244570001

 

Thanks in advance,

 

Alejandro

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Accepted by AlejandroZepeda

Hello Alejandro:

 

Try %#f

 

 

Message 2 of 4
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AlejandroZepeda wrote:

Hi.

 

I am trying to do something that seems quite simply,


 

For example, if my data is this:

0.02

0.4476

-0.0244570001

 

I would like for the string that I get to be exactly like what I wrote above.  However, depeding on the format string I use, this is what I get:

 

format string: %.10f

0.0200000000

0.4476000000

-0.0244570001

 

Does someone know what the format string should be to get the string I want (which is this:)

0.02

0.4476

-0.0244570001

 

Thanks in advance,

 

Alejandro


Hi Alejandro,

 

remember that if  you used a %20f formatter you most likely would see a digit appearing in the .4476 value. Your computer is binary, but your display is decimal, there allways will be rounding errors.

 

Ton

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Message 3 of 4
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Hello.

 

Thanks for your responses.

 

I tried using %#f, but I got the same result as %f.

Then I tried %#.10f, and that worked exactly as I wanted it to.

 

Thanks a lot for the tip, DonPogi!

 

Alejandro

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