Python-сообщество

Здравствуйте. С коллегой скиптик пишем и возникла такая интересная ситуация. У меня python 2.7.1, а у коллеги 2.6.5.
Ставим такой эксперимент:

Python 2.7.1 (r271:86832, Jan  6 2011, 11:45:30) 
[GCC 4.5.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> round (4.3111111, 2)
4.31

Python 2.6.5 (r265:79063, Apr 16 2010, 13:57:41) 
[GCC 4.4.3] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> round (4.3111111, 2)
4.3099999999999996

regall

>>> 4.3099999999999996 == 4.31
True

Для вывода пользуйтесь форматированием строк

Dr.Ziko
Т.е. если я запишу это число в БД, например, то там будет 4.31 вместо 4.30999999999996?

Это относится только к repr у float, а не к самому числу:

Python now uses David Gay’s algorithm for finding the shortest floating point representation that doesn’t change its value. This should help mitigate some of the confusion surrounding binary floating point numbers.

The significance is easily seen with a number like 1.1 which does not have an exact equivalent in binary floating point. Since there is no exact equivalent, an expression like float('1.1') evaluates to the nearest representable value which is 0x1.199999999999ap+0 in hex or 1.100000000000000088817841970012523233890533447265625 in decimal. That nearest value was and still is used in subsequent floating point calculations.

What is new is how the number gets displayed. Formerly, Python used a simple approach. The value of repr(1.1) was computed as format(1.1, ‘.17g’) which evaluated to ‘1.1000000000000001’. The advantage of using 17 digits was that it relied on IEEE-754 guarantees to assure that eval(repr(1.1)) would round-trip exactly to its original value. The disadvantage is that many people found the output to be confusing (mistaking intrinsic limitations of binary floating point representation as being a problem with Python itself).

The new algorithm for repr(1.1) is smarter and returns ‘1.1’. Effectively, it searches all equivalent string representations (ones that get stored with the same underlying float value) and returns the shortest representation.

The new algorithm tends to emit cleaner representations when possible, but it does not change the underlying values. So, it is still the case that 1.1 + 2.2 != 3.3 even though the representations may suggest otherwise.

The new algorithm depends on certain features in the underlying floating point implementation. If the required features are not found, the old algorithm will continue to be used. Also, the text pickle protocols assure cross-platform portability by using the old algorithm.

(Contributed by Eric Smith and Mark Dickinson; issue 1580)

regall

Dr.Ziko
Т.е. если я запишу это число в БД, например, то там будет 4.31 вместо 4.30999999999996?

А попробовать? =)

Dr.Ziko
Вывод мне не нужен в принципе, я лишь проверяю корректность данных в списке. Т.е. если я запишу это число в БД, например, то там будет 4.31 вместо 4.30999999999996?

Отредактировано (Фев. 4, 2011 16:09:07)

Dr.Ziko, Isem.
Вы различаете двоичную запись и текстовое представление для float?
И то, что многие дроби можно записать двумя способами, 4.31 одна из них?

Андрей Светлов
Вы различаете двоичную запись и текстовое представление для float?
И то, что многие дроби можно записать двумя способами, 4.31 одна из них?

Отредактировано (Фев. 4, 2011 17:38:46)

Вариант - использовать свой round:

    def round(num):
        num = float(num)
        x = num - math.floor(num)
        if x == 0: return int(num)
        elif x < 0.5: return int(math.floor(num))
        else: return int(math.ceil(num))