rounding

问题 I have an API call which returns a double. The double's decimal length can variate from many decimal places to a few (it all comes down to the state of an actuator). This double represents the current position on the range radius of an actuator. I am not interested in such a detailed number, because it adds alot of noise to the system. I've been using floats to save space, but still I have floats which have 6-8 decimal length. I am not interested in floor or ceil, simply because it doesn't do

问题 I have an API call which returns a double. The double's decimal length can variate from many decimal places to a few (it all comes down to the state of an actuator). This double represents the current position on the range radius of an actuator. I am not interested in such a detailed number, because it adds alot of noise to the system. I've been using floats to save space, but still I have floats which have 6-8 decimal length. I am not interested in floor or ceil, simply because it doesn't do

问题 Here is my function to roundoff a number upto two decimals but when the rounded off number is 1.50 it seems to ignore the trailing zero and just returns 1.5 public static double roundOff(double number) { double accuracy = 20; number = number * accuracy; number = Math.ceil(number); number = number / accuracy; return number; } So if I send 1.499 it returns 1.5 where as I want 1.50 回答1: This is a printing poblem: double d = 1.5; System.out.println(String.format("%.2f", d)); // 1.50 回答2: 1.5 is,

问题 I am currently trying to split one value with percentage column. But as most of percentages values are 1/3, I am not able to get aboslute 100% with two decimal points in the value. For example: Product Supplier percentage totalvalue customer_split decimal(15,14) (decimal(18,2) decimal(18,2) -------- -------- ------------ --------------- --------------- Product1 Supplier1 0.33 10.00 3.33 Product1 Supplier2 0.33 10.00 3.33 Product1 Supplier3 0.33 10.00 3.33 So, here we are missing 0.01 in the

问题 I am currently trying to split one value with percentage column. But as most of percentages values are 1/3, I am not able to get aboslute 100% with two decimal points in the value. For example: Product Supplier percentage totalvalue customer_split decimal(15,14) (decimal(18,2) decimal(18,2) -------- -------- ------------ --------------- --------------- Product1 Supplier1 0.33 10.00 3.33 Product1 Supplier2 0.33 10.00 3.33 Product1 Supplier3 0.33 10.00 3.33 So, here we are missing 0.01 in the

问题 Consider two very simple multiplications below: double result1; long double result2; float var1=3.1; float var2=6.789; double var3=87.45; double var4=234.987; result1=var1*var2; result2=var3*var4; Are multiplications by default done in a higher precision than the operands? I mean in case of first multiplication is it done in double precision and in case of second one in x86 architecture is it done in 80-bit extended-precision or we should cast operands in expressions to the higher precision

问题 My Java program is centered around high precision calculations, which need to be accurate to at least 120 decimal places. Consequentially, all non-integer numbers will be represented by BigDecimals in the program. Obviously I need to specify the accuracy of the rounding for the BigDecimals, to avoid infinite decimal expressions etc. Currently, I find it a massive nuisance to have to specify the accuracy at every instantiation or mathematical operation of a BigDecimal. Is there a way to set a

问题 My Java program is centered around high precision calculations, which need to be accurate to at least 120 decimal places. Consequentially, all non-integer numbers will be represented by BigDecimals in the program. Obviously I need to specify the accuracy of the rounding for the BigDecimals, to avoid infinite decimal expressions etc. Currently, I find it a massive nuisance to have to specify the accuracy at every instantiation or mathematical operation of a BigDecimal. Is there a way to set a

问题 Let's look at the ever-shocking round statement: >>> round(2.675, 2) 2.67 I know why round "fails"; it's because of the binary representation of 2.675: >>> import decimal >>> decimal.Decimal(2.675) Decimal('2.67499999999999982236431605997495353221893310546875') What I do not understand is: why does NumPy not fail ? >>> import numpy >>> numpy.round(2.675, 2) 2.6800000000000002 Thinking Do not mind the extra zeros; it's an artefact from Python's printing internal rounding. If we look at the

问题 I'm looking to group dates in R by an arbitrary level of precision. It's pretty straightforward to do this to the nearest hour or minute, using e.g. lubridate : library(lubridate) nearest_hour = floor_date(now(), 'hour') You can then group a list of such dates with e.g. a simple summarise ddply from plyr . What I'd like to do is round dates with arbitrary precision, e.g. to the nearest 15 minutes or every 3 hours: nearest_three_hours = floor_date(now(), '3 hours') There's a discussion of such