how to get floating point number upto 4 decimal points

hi i have an array of floating point numbers of type double...for example x=[ 0.8462 0.4906 0.9422 0.2054 0.6154 0.8923 0.3622 0.8710 0.4237 0.9206 0.2757 0.7528] how to limit the values upto 4 decimal points.

x11=(round(x*10000))/10000; does not work

Answer 1

[Walter Roberson]

vpa(sym(x), 4)

will give you 4 digit symbolic numbers.

arrayfun(@(X)  sprintf('%.4f',X), x, 'uniform', 0)

Or

cellstr(num2str(x(:), '%.4f')).'

will give you a cell array of strings representation.

round(x*10000)

will give you integers with an implied decimal point.

But if you are looking for floating point values with exactly 4 decimal places then the symbolic version is as close as you can get. The fraction 1/10 requires an infinite repeating value in binary, just like 1/7 requires an infinite repeating value in decimal. You just cannot represent 0.8462 exactly in binary.

[studentambitious]

vpa(sym(x), 4) solves the problem but i dont found any difference between vpa(sym(x), 4) and vpa(x, 4) both gives same result. Secondly can we directly read the values of a floating point number after decimal point .i dont want the integer part(number before the decimal point) without rounding or multiplying with 10000 for example the number is 23.098756..i want to read .0987 upto four decimal points. please help

[Walter Roberson]

vpa is not defined on numeric values, only on symbolic values.

mod(x, 1) gives you the fractional part for nonnegative values. For negative values you need to define whether you want the offset from the lower integer or if you want the offset from the higher integer. -0.25 is 0.75 above the integer below it, but you might be looking for the 0.25 part. If so then use mod(abs(x), 1)

You would then apply one of the above to get the number in the form you want, keeping in mind that it is not possible to exactly represent most fractions. Only 16 of the 4 digits decimal fractions can be exactly represented in binary.

[Walter Roberson]

More recently I have discovered that vpa() does not represent floating point numbers in decimal internally.

Answer 2

[Guillaume]

In what way does

x11 = round(x*10000)/10000  %could be simply replaced by round(x, 4)

not work?

>>round(pi*10000)/10000
ans = 

     3.1416

>>round(pi, 4)
ans = 

     3.1416

If you are talking about how the numbers are display, note that it has nothing to do with the actual value stored in the number. You use the format command to tell matlab how to display the numbers. Personally, I use

format shortg

[bhan stokes]

can we limit the precision value of floating point numbers upto two or three decimal value.. like if it is .237654...can be limited to .2300000(can or cannot be follow by zeroes)

[Steven Lord]

See description of the input argument named N in the documentation for the round function.

[Walter Roberson]

Also note that round() only finds the closest representable number, which will seldom be exactly the value being looked for. It is not possible to represent 0.23 (23/100) exactly in finite binary floating point.

[Jan]

[MOVED from flags] Vithal Bable on 2 Apr 2019 at 3:53.

i used 'shortg' before my string and i got the limited number string ... provided command was helpful for me . thank you !!!!!

@Vithal Bable: Please use flags only to inform admins and editors about inappropriate contents like spam or rudeness. Thanks.