rounding

问题 This question already has answers here : Truncating floats in Python (29 answers) Closed 2 years ago . I want convert a number to a float number with 3 decimal point but I want it without rounding. For example: a = 12.341661 print("%.3f" % a) This code return this number: 12.342 but I need to original number,I need this: 12.341 I write a code that receive a number form user and convert it to a float number. I have no idea that what is the number entered with user. 回答1: My first thought was to

问题 This question already has answers here : Truncating floats in Python (29 answers) Closed 2 years ago . I want convert a number to a float number with 3 decimal point but I want it without rounding. For example: a = 12.341661 print("%.3f" % a) This code return this number: 12.342 but I need to original number,I need this: 12.341 I write a code that receive a number form user and convert it to a float number. I have no idea that what is the number entered with user. 回答1: My first thought was to

问题 This question already has answers here : Truncating floats in Python (29 answers) Closed 2 years ago . I want convert a number to a float number with 3 decimal point but I want it without rounding. For example: a = 12.341661 print("%.3f" % a) This code return this number: 12.342 but I need to original number,I need this: 12.341 I write a code that receive a number form user and convert it to a float number. I have no idea that what is the number entered with user. 回答1: My first thought was to

问题 I have a list of numbers which I need to round into integers before I continue using the list. Example source list: [25.0, 193.0, 281.75, 87.5, 80.5, 449.75, 306.25, 281.75, 87.5, 675.5,986.125, 306.25, 281.75] What would I do to save this list with all of the numbers rounded to an integer? 回答1: Simply use round function for all list members with list comprehension : myList = [round(x) for x in myList] myList # [25, 193, 282, 88, 80, 450, 306, 282, 88, 676, 986, 306, 282] If you want round

问题 I have a list of coordinates like this: [[(-88.99716274669669, 45.13003508233472), (-88.46889143213836, 45.12912220841379), (-88.47075415770517, 44.84090409706577), (-88.75033424251002, 44.84231949526811), (-88.75283245650954, 44.897062864942406), (-88.76794136151051, 44.898020801741716), (-88.77994787408718, 44.93415662283567), (-88.99624763048942, 44.93474749747682), (-88.99716274669669, 45.13003508233472)]] Or like this: [[(-88.99716274669669, 45.13003508233472)], [(-88.46889143213836, 45

问题 I have a list of coordinates like this: [[(-88.99716274669669, 45.13003508233472), (-88.46889143213836, 45.12912220841379), (-88.47075415770517, 44.84090409706577), (-88.75033424251002, 44.84231949526811), (-88.75283245650954, 44.897062864942406), (-88.76794136151051, 44.898020801741716), (-88.77994787408718, 44.93415662283567), (-88.99624763048942, 44.93474749747682), (-88.99716274669669, 45.13003508233472)]] Or like this: [[(-88.99716274669669, 45.13003508233472)], [(-88.46889143213836, 45

问题 Which is the best efficient way to round up a number and then truncate it (remove decimal places after rounding up)? for example if decimal is above 0.5 (that is, 0.6, 0.7, and so on), I want to round up and then truncate (case 1). Otherwise, I would like to truncate (case 2) for example: 232.98266601563 => after rounding and truncate = 233 (case 1) 232.49445450000 => after rounding and truncate = 232 (case 2) 232.50000000000 => after rounding and truncate = 232 (case 2) 回答1: There is no

问题 a = 1 for x in range(5): a += 0.1 print(a) This is the result: 1.1 1.2000000000000002 1.3000000000000003 1.4000000000000004 1.5000000000000004 How can I fix this? Is the round() function the only way? Can I set the precision of a variable before setting its value? 回答1: can i set the precision of a variable before setting the value? Use the decimal module which, unlike float(), offers arbitrary precision and can represent decimal numbers exactly: >>> from decimal import Decimal, getcontext >>>

问题 a = 1 for x in range(5): a += 0.1 print(a) This is the result: 1.1 1.2000000000000002 1.3000000000000003 1.4000000000000004 1.5000000000000004 How can I fix this? Is the round() function the only way? Can I set the precision of a variable before setting its value? 回答1: can i set the precision of a variable before setting the value? Use the decimal module which, unlike float(), offers arbitrary precision and can represent decimal numbers exactly: >>> from decimal import Decimal, getcontext >>>

问题 I'm looking for an output of 4.658227848101266 = 4.5 4.052117263843648 = 4.0 the closest I've gotten is rating = (Math.round(rating * 4) / 4).toFixed(1) but with this the number 4.658227848101266 = 4.8??? 回答1: (Math.round(rating * 2) / 2).toFixed(1) 回答2: It's rather simple, you should multiply that number by 2, then round it and then divide it by 2: var roundHalf = function(n) { return (Math.round(n*2)/2).toFixed(1); }; 回答3: This works for me! (Using the closest possible format to yours)