I am creating a program for a project that will help me learn visual basic. But when I work out how much a certain material is worth with dimensions. It come out with more than two decimal places. I know how to use math.floor function but it doesn't round to 2 decimal places.
Does anyone know what function I should use to round to two decimal places?
You can use;
Math.Round([DECIMAL], 2)
If you want to round up you can use;
MidpointRounding.AwayFromZero
As an additional parameter to make;
Math.Round([DECIMAL], 2, MidpointRounding.AwayFromZero)
Hope it helps!
The Decimal.Round method will do the trick for you.
You can specify the number of decimal places.
Related
I'm trying to collect some SMA data. Currently, the decimal place is auto limited to
0.000062
I need 2 more decimal places:
0.00006283
Here is my code:
symbol_df['5sma'] = symbol_df['close'].rolling(5).mean()
I've tried round, but no matter how high I set it, it's still limited to (6) decimal places.
This helped me out. I just placed it before the code and the correct decimal places were shown.
pd.set_option("display.precision", 8)
I can find a workaround, but it is really annoying and I may certainly be missing something. Redshift's ROUND function doesn't round to the number of decimals specified.
For example,
select round(cast(176 as float)/cast(492 as float),4) as result;
Above select statement will return 0.35769999999999996.
However, this statement:
select round(cast(229 as float)/cast(491 as float),4) as result;
... will return 0.4664.
Why? I can work around this, but seems like it should work and return only four decimal places.
If your issues is all those 9999s, then the issue is floating point representation. Convert to a decimal to get fixed-point precision:
select round(cast(176 as float)/cast(492 as float), 4)::decimal(10, 4) as result;
Elaborating more on Gordon's answer -
So you’ve written some absurdly simple code, say for example:
0.1 + 0.2
and got a really unexpected result:
0.30000000000000004
Because internally, computers use a format (binary floating-point) that cannot accurately represent a number like 0.1, 0.2 or 0.3 at all.
When the code is compiled or interpreted, your “0.1” is already rounded to the nearest number in that format, which results in a small rounding error even before the calculation happens.
What can I do to avoid this problem?
That depends on what kind of calculations you’re doing.
If you really need your results to add up exactly, especially when you work with money: use a decimal datatype.
If you just don’t want to see all those extra decimal places: simply format your result rounded to a fixed number of decimal places when displaying it.
Shamelessly stolen from : Floating Point
try multiplying by 10 to the power of your desired places after the decimal point, rounding, and then dividing it out again:
-- exclude decimal point inside ROUND(), include outside ROUND()
SELECT ROUND(10000 * 176 / 492) / 10000.0
which will return the expected 0.3577.
I'm trying to extract some numbers with decimals but need to remove the decimal part. There's no fixed length on any of both sides.
I have already tried:
INT()
INTR()
ROUND()
Usually INT() should solve this but sometimes it doesn't return the correct number (for example, INT(3) returns 2).
Did you try the function TRUNC(number, [, decimal_places]) ?
EDIT
For the round precision you can see this link
I ended up solving it on my own! ROUND didn't work and FLOOR was also undefined...
I ended up needing to use two INTs with a FORMAT in one of them.
Thank you anyway.
I know about the Math.Round and Math.Ceiling methods, but they return a Double and a Decimal. Does VB.NET have any built-in functions which always round a floating point number up, not down, with a return type of Integer? I know there's CInt, but this can round down if it's below 6.5.
From the comments I understood you wanted to round 6.1 to 7.
Just add 1 and truncate.
If it looks awkward, create a method for it.
Correction:
Unless the number already is truncated.
Addendum:
Note here that doing == with floats is not without problem; you should always have some sort of precision when trying == with floats.
Now when you have decided on this precision - then you can rewrite your code to add 0.999999 (according to precision) and the first add-0.9999-and-truncat works.
Note again that adding 0.9999 is does not really mean you add 0.9999 with our "normal" float.
So if you really want to add 0.9999 you have to work with BND and/or some monetary arithmetics.
Which you "always" should do when calculating money (or any exact decimal stuff)
I am writing a class for handling decimal numbers, e.g. "123.456". I want one function for extracting the digits before the decimal point (123), and one function to extract the digits after the decimal point (0.456). My question is not how to do the programming, but how to name the functions? Do you have any better idea than digits_before_point() and digits_after_point()?
integralPart() and decimalPart() or integers() and decimals()
How about integral_part and decimal_part?
integer_part() and fraction_part()