trailing

What is the function that removes trailing zeros from doubles? var double = 3.0 var double2 = 3.10 println(func(double)) // 3 println(func(double2)) // 3.1 Grzegorz R. Kulesza In Swift 4 you can do it like that: extension Double { func removeZerosFromEnd() -> String { let formatter = NumberFormatter() let number = NSNumber(value: self) formatter.minimumFractionDigits = 0 formatter.maximumFractionDigits = 16 //maximum digits in Double after dot (maximum precision) return String(formatter.string(from: number) ?? "") } } example of use: print (Double("128834.567891000").removeZerosFromEnd())

I have a string with variable length and I want to give a format to strptime in order for the rest of the string to be ignored. Let me exemplify. I have something like 9/4/2013,00:00:00,7.8,7.4,9.53 10/4/2013,00:00:00,8.64,7.4,9.53 and I want a format that makes the command strptime(line,format) work to read those lines. Something like format='%d/%m/%Y,%H:%M:%S*' , although I know that doesn't work. I guess my question is kind of similar to this one , but no answer there could help me and my problem is a little worse because the full length of my string can vary. I have a feeling that dateutil

This question already has an answer here: How to nicely format floating numbers to String without unnecessary decimal 0? 22 answers For example I need 5.0 to become 5 , or 4.3000 to become 4.3 . soniccool Use DecimalFormat double answer = 5.0; DecimalFormat df = new DecimalFormat("###.#"); System.out.println(df.format(answer)); Marcin Szymczak You should use DecimalFormat("0.#") For 4.3000 Double price = 4.3000; DecimalFormat format = new DecimalFormat("0.#"); System.out.println(format.format(price)); output is: 4.3 In case of 5.000 we have Double price = 5.000; DecimalFormat format = new

So I have a script which returns a price for a product. However the price may or may not include trailing zeros so sometimes I might have: 258.22 and other times I might have 258.2 In the later case I need to add the trailing zero with jQuery. How would I go about doing this? rosscj2533 You can use javascript's toFixed method ( source ), you don't need jQuery. Example: var number = 258.2; var rounded = number.toFixed(2); // rounded = 258.20 Edit: Electric Toolbox link has succumbed to linkrot and blocks the Wayback Machine so there is no working URL for the source. Javascript has a function -

I have several alphanumeric strings like these listOfNum = ['000231512-n','1209123100000-n00000','alphanumeric0000', '000alphanumeric'] The desired output for removing trailing zeros would be: listOfNum = ['000231512-n','1209123100000-n','alphanumeric', '000alphanumeric'] The desired output for leading trailing zeros would be: listOfNum = ['231512-n','1209123100000-n00000','alphanumeric0000', 'alphanumeric'] The desire output for removing both leading and trailing zeros would be: listOfNum = ['231512-n','1209123100000-n00000','alphanumeric0000', 'alphanumeric'] For now i've been doing it the

I am curious why in Python a trailing comma in a list is valid syntax, and it seems that Python simply ignores it: >>> ['a','b',] ['a', 'b'] It makes sense when its a tuple since ('a') and ('a',) are two different things, but in lists? Raymond Hettinger The main advantages are that it makes multi-line lists easier to edit and that it reduces clutter in diffs. Changing: s = ['manny', 'mo', 'jack', ] to: s = ['manny', 'mo', 'jack', 'roger', ] involves only a one-line change in the diff: s = ['manny', 'mo', 'jack', + 'roger', ] This beats the more confusing multi-line diff when the trailing comma