reduce precision of a number

I am having a problem trying to reduce precision output of a number

for example

x = 1.123456 I want x = 1.123 (only 3 values after a decimal and not a string)

I use x = round(x*1000)/1000 and get x = 1.1230 (I dont want the end zero)

I use x = sprintf('%.3f',x) and get a string '1.123' (i dont want a string)

so i use x = str2num(sprintf('%.3f',x)) and get x = 1.1230 (again with the zero)

please help me remove a zero at the end. I am not concerned with precision.

[David DeTar]

Try using fprintf instead of sprintf.

[Walter Roberson]

fprintf() sends to the command window output or a file, instead of constructing a string to be returned. The return value from fprintf() is the number of characters printed, not the string. fprintf() and sprintf() have exactly the same format codes.

[CYRUS SAFI]

any other alternative opetions for rounding

[Walter Roberson]

The options are

round(X)

or

round(X, N)

or

round(X*10^N)/10^N

or

vpa(X, N)   %requires symbolic toolbox

For positive N, the output of round() will not generally happen to be an exact multiple of a power of 2, so the output() of round() will not generally happen to be exactly representable in binary. (It will happen sometimes -- for example round(0.25319,2) will happen to be exactly 0.25 which is exactly representable in binary -- but round(0.26319,2) would be approximately 0.26 and exactly 0.2600000000000000088817841970012523233890533447265625 decimal.)

The internal value is different from how the values are displayed.

If format short is in effect, then a complicated set of rules is used. Values with absolute value between 0.001 and 999.9999 that do not happen to be integers are generally displayed with 4 digits after the decimal place. Values outside that range are displayed in scientific notation with 5 signficant digits (including the value before the decimal point.) If you are using format short then there is no way to get rid of trailing 0s

If format long g is in effect and the absolute value is between 1e-4 and (1e15 -1e15*eps) then decimal representation will be used, and trailing zeros will be stripped away.

If you want other display rules, then you will need to code them yourself using fprintf() or sprintf() or compose()

[Stephen23]

This question is a very good example of

https://xyproblem.info/

https://en.wikipedia.org/wiki/XY_problem

The actual problem is that the OP was attempting to test for exact equivalence of binary floating point numbers.

The actual solution is given here:

https://www.mathworks.com/matlabcentral/answers/79928-reduce-precision-of-a-number#comment_156753

Answer 1

[Azzi Abdelmalek]

When you use x for calculation use

 out = str2num(sprintf('%.3f',x))  % when x is not displayed, the 0 does not appear!
%or
x = round(x*1000)/1000

When you want to display it use

sprintf('%.3f',x)

[Dwight]

this still gives 4 decimal places. this is what I did originally.

[Dwight]

is there a way to regexp read and output a value

from x = anynumberofwho;enumbers.1230

to give me x = anynumberofwholenumbers.123?

thanks

[Azzi Abdelmalek]

You have to explain, If you want to display a number 0.123, it's not important if it's double or char.

[Dwight]

i dont need to display per se. I want to use the value in a loop to find an index of the same value.

using x(:,1) as my matrix

x(2500,1) = 1.123456 %my row of interest

b = 1.1230 %inherent data ony significant up to 1/1000th

xindex = find(x(:,1)== b)

i get xindex = [] because 1.1230 is not found in matrix x

Thus my issue is i need to reduce the precision of b so i can get an index for my loop

thanks for suggestions

[Walter Roberson]

Not using regexp, No. Instead use

find( (x(:,1)-b) < 1/1000 )

[Iain]

 find( abs(x(:,1)-b) < 1/1000 )

would be better.

Answer 2

[Steven Lord]

This is an old discussion, but if you're using release R2014b or later you can round to a specified number of digits.

[Faez Alkadi]

Steven, I'm using R2017a. and i used round function as

x=2.123456789123456789

y=round(x,3),

so i get

y=2.123000000000000000

where i want it to be (saved) as

y=2.123

its odd that mat lab doesn't have this. Just dumb whatever comes after certain decimal number.

[Walter Roberson]

MATLAB uses IEEE 754 Binary Double Precision to represent floating point numbers. All floating point scheme that uses binary mantissas cannot exactly represent 1/10, just like decimal representation schemes cannot exactly represent 1/3 or 1/7 .

IEEE 754 also defined a Decimal Double Precision representation scheme, which can represent 2.123 exactly. However, computing those values in software is much slower. The only systems I know of that implement IEEE 754 Decimal Double Precision in hardware are the IBM z90 series.

If you need a certain specific number of decimal places to be stored, then use rationals with a power-of-10 denominator.

Answer 3

[Walter Roberson]

At the command prompt give the command

format short g

Note: it is not possible to represent 1.123 exactly as a binary double precision number. The closest representable number is 1.1229999999999999982236431605997495353221893310546875

[Dwight]

I did this as well, but i want the answer given to the thousandths place with no zero at the end

x = 1.123456 so instead of x = 1.1230

i need x = 1.123

thanks

[Walter Roberson]

Use "format short g" together with the rounding that Azzi shows.

Note that there is a difference between the internal value stored and the printable representation of the internal value. When you display a number using disp() then the printable representation used depends on which "format" command is in effect. The default is "format short" which shows the value rounded to 4 decimal places unless the value is close enough to an integer that it shows it in integer form instead. But that form is not how the value is stored internally, just how it is printed out. "format short g" will show up to 4 decimal places, dropping trailing 0's from the display.

None of the "format" commands allow you to select the number of decimal places to display. If you want a display behavior different than what "format" can give you, you will need to use code such as sprintf() to format the number into a character string and then display the string. Note that this does not affect the internal representation of the number.

Only a small number of decimal fractions can be stored exactly in binary floating point -- e.g., .5, .25, .75, .125, .375, and so on. The integer multiples of negative powers of 2. Anything else, including 0.1, cannot be represented exactly and will use all 53 bits of precision internally, such as that long string of digits I showed above being necessary to represent "as close as practical" to 0.123

[Mayank Ghugretkar]

use

digits(4)
Answer= vpa(1.123456)

[Walter Roberson]

vpa() requires the Symbolic Toolbox

Answer 4

[Francis Agbali]

I would try using

format short

1.23400007787666

[mohamed kandil]

format bank