How to remove trailing zeros while display any floating point number ?

format short g;
X=input ('any>');
%if input is 93.93
disp (X);

It display 93.930

How to disp 93.93

I can't remove

 format short g;

Bcz of another mathematical reasons and calculations...

Answer 1

[Walter Roberson]

The "format short g" statement does not affect calculations at all: it only affects displaying of data.

To display without trailing zeros you should use

fprintf('%g\n', X);

[Kelvin Viroja]

TVM...bro

Answer 2

[Azzi Abdelmalek]

sprintf('%g',X)

[Kelvin Viroja]

TVM...bro

Answer 3

[Muhammad Yasirroni]

I just found this out. You can remove the trailing zero by using floor.

format short g;
X=input ('any>');
%if input is 93.93
Xnew=floor(X);
disp (Xnew);

It works great. It even can work with matrices and even correcting the value if you want to save it to .mat file.

If you want to retain some value behind the point (.), you can multiply it first.

X=93.33*100;
Xnew=floor(X)/100;

[Walter Roberson]

floor rounds to negative infinity. floor(-93.93) would be -94. If you wanted -93 instead you would use trunc()

Trailing zeros on output of a numeric item depends on which "format" you currently have in effect, and sometimes also on the other values being displayed. For example

format short
[0 1e20]

is going to display trailing zeroes on the 0

[DGM]

The difference between floor() and fix() aside, this demonstrably doesn't do what the question asked to accomplish.

X = 1.1*10;
Xnew = floor(X)/10;
disp(Xnew)

    1.1000

X=93.33*100;
Xnew=floor(X)/100;
disp(Xnew)

   93.3300

Rounding the number to any given decimal place does not guarantee that it can be exactly represented in floating point, and it doesn't control how it will be displayed, either dumping to console directly with unsuppressed assignment, or by using disp(). You need to treat the number as text in order to have arbitrary control over text display formatting.

Answer 4

[Charles Kluepfel]

This function will convert trailing zeros in a string or character vector, including multiple numbers:

function btz=blankTrailZeros(str)
        sout=[char(str) ' '];
        trans=false; ppos=0;
        for i=1:length(sout)
          switch sout(i)
            case '.'
              trans=true; ppos=i;
            case ' '
             if trans==true
              trans=false; 
              for j=i-1:-1:ppos 
                if sout(j)>'0' && sout(j) <='9'
                  break
                end
                if sout(j)=='0'
                  sout(j)=' ';
                else
                  if sout(j)=='.'
                    sout(j)=' ';
                    break
                  end
                end
              end
              ppos=0; 
             end 
          end
        end
        btz=sout(1:end-1);
        if isequal(class(str),'string')
          btz=string(btz);
        end
end       

>> a=blankTrailZeros('12.3000 60 5.993 22.0000')

a =

'12.3 60 5.993 22 '

>> blankTrailZeros("12.3300")

ans =

"12.33 "

for use with character strings produced by sprintf.

[Walter Roberson]

might be easier to use regexprep.

Question about your code: if all digits after the decimal place are 0, is your code stripping the decimal place as well? So 60.0 would become 60 ?

Is there a reason why your code is leaving a variable number of blanks after the last entry even when no blanks originally occurred there?

[Charles Kluepfel]

As in the 22.0000, the decimal point is being stripped as well, so 60.0 would become 60. The 22.0000 has five spaces after it in the result: one for the decimal point that's been removed and four for the four zeros removed. In the font that the description appears in, spaces take up less room than other characters, but every removed character has been replaced by a space, so the length of the output matches that of the input and the decimal points are in the same position for alignment in a column of output.

The 60 in the original was to test, and show, that trailing zeros are not stripped from integers that have zero or a string of zeros all the way to the end.

The code replaces each zero stripped with a space. It doesn't necessarily look like that in the proportional font used for the regular text in the description, but if you copy the test cases to a text editor that uses a fixed width font, like Courier, you'll see there are four spaces between the 12.3 and the 60: one space for each of the three zeros removed and one that was originally there. The whole reason for this is that sprintf can be used rather than fprintf, then use this to transform the result, so that a simple fprintf('%s', result) can be used to keep decimal points aligned in a table of results.

Answer 5

[Art]

"strip" can do this for any amound of decimal places... just change to a str first:

X = 1.31230032000000;
XStr = sprintf('%1.15f', X);
XStripped = strip(XStr,'right','0');

If you're worried about having no trailing zeros after the decimal, a quick "if" statement:

if strcmp(XStripped(end),'.')
    XStripped = [XStripped '0'];
end