rounding

问题 I want to The numbers are being rounded to the nearest 10 's place. For example, a number like 17.3 is being rounded to 20.0 . and want to be allow three significant digits. Meaning for round to the nearest tenth of a percent as the last step of the process. Sample : the number is 17.3 ,i want round to 20 , and this number is 13.3 , i want round to 10 ? How can i do this ? 回答1: Chris Charabaruk gives you your desired answer here To get to the core, here is his solution as an extension method:

问题 I want to The numbers are being rounded to the nearest 10 's place. For example, a number like 17.3 is being rounded to 20.0 . and want to be allow three significant digits. Meaning for round to the nearest tenth of a percent as the last step of the process. Sample : the number is 17.3 ,i want round to 20 , and this number is 13.3 , i want round to 10 ? How can i do this ? 回答1: Chris Charabaruk gives you your desired answer here To get to the core, here is his solution as an extension method:

问题 This question already has answers here : Rounding to nearest 100 (7 answers) Closed 7 years ago . I'm trying to round a number the 100. Example: 1340 should become 1400 1301 should become 1400 and 298 should become 300 200 should stay 200 I know about Math.round but it doesn't round to the 100. How can I do that ? 回答1: Try this... function roundUp(value) { return (~~((value + 99) / 100) * 100); } That will round up to the next hundred - 101 will return 200. jsFiddle example - http://jsfiddle

问题 Using GCC: printf("%i \n", -1 % (int)4); printf("%u \n", -1 % (unsigned int)4); Output: -1 3 Can I rely on this behaviour across platforms? Should I explicitly define MOD and REM macros to be sure this isn't altered? 回答1: From C99 onwards the result of % is required to be rounded toward 0 as quoted by Chris Dodd. Prior to C99 standard, % operator's behavior on negative number is implementation defined . When integers are divided and the division is inexact, if both operands are positive the

问题 Using GCC: printf("%i \n", -1 % (int)4); printf("%u \n", -1 % (unsigned int)4); Output: -1 3 Can I rely on this behaviour across platforms? Should I explicitly define MOD and REM macros to be sure this isn't altered? 回答1: From C99 onwards the result of % is required to be rounded toward 0 as quoted by Chris Dodd. Prior to C99 standard, % operator's behavior on negative number is implementation defined . When integers are divided and the division is inexact, if both operands are positive the

问题 Why does the following code: Console.WriteLine(String.Format("{0:C0}", 2170/ 20)); yield "$109", while doing Console.WriteLine(Math.Round(2170 / 20)); gives me 108? How can I get 2170 / 20 give me 109? 回答1: When you divide to values of integral type, such as 2170 and 20 , the runtime performs an integer division and discards (truncates) the decimal. If you change one of the operands to a float , double , or decimal (eg, 2170.0 / 20 , or 2170 / 20m ), it will perform a floating-point division,

问题 Why does the following code: Console.WriteLine(String.Format("{0:C0}", 2170/ 20)); yield "$109", while doing Console.WriteLine(Math.Round(2170 / 20)); gives me 108? How can I get 2170 / 20 give me 109? 回答1: When you divide to values of integral type, such as 2170 and 20 , the runtime performs an integer division and discards (truncates) the decimal. If you change one of the operands to a float , double , or decimal (eg, 2170.0 / 20 , or 2170 / 20m ), it will perform a floating-point division,

问题 I read that numpy is unbiased in rounding and that it works the way its designed. That "if you always round 0.5 up to the next largest number, then the average of a bunch rounded numbers is likely to be slightly larger than the average of the unrounded numbers: this bias or drift can have very bad effects on some numerical algorithms and make them inaccurate." Disregarding this information and assuming that I always want to round up, how can I do it in numpy? Assuming my array can be quite

问题 Consider the following: > x<-178379.4999999999999999999999999999999 > x [1] 178379.5 > round(x) [1] 178380 This seems to be a basic rounding error. Are there known rounding errors in R? Or is it because even in working memory R can only process up to 22 digits? 回答1: This is a combination of two extremely Frequently A'd Qs. finite floating-point precision : this R FAQ 7.31, see e.g. Why are these numbers not equal? . The value gets rounded to 178379.5. It won't help if you set options(digits

问题 This question already has an answer here : Golang Round to Nearest 0.05 (1 answer) Closed 2 years ago . I'm trying to unconventionally round all the digits in a float64 variable. For example: 3.4444445 --> 3.5 I want to do this without converting it into a string! 回答1: Golang's math library provides a Round function. However, it rounds the float64 to an int, meaning decimals are lost. A quick workaround around this would be to multiple the number by the number of decimals you want to save,