Python浮点数到十进制转换

本文介绍了Python浮点数到十进制转换的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

Python Decimal不支持从float构造。它期望您必须先将float转换为字符串。

Python Decimal doesn't support being constructed from float; it expects that you have to convert float to a string first.

这非常不方便，因为float的标准字符串格式器要求您指定小数位数而不是有效位数。因此，如果您的数字最多可以保留15个小数位，则需要将其格式设置为Decimal（％。15f％my_float），如果您之前还有任何有效数字，也将在第15个小数位处产生垃圾

This is very inconvenient since standard string formatters for float require that you specify number of decimal places rather than significant places. So if you have a number that could have as many as 15 decimal places you need to format as Decimal(""%.15f"% my_float), which will give you garbage at the 15th decimal place if you also have any significant digits before decimal.

有人可以建议一种在用户输入时从float转换为Decimal保留值的好方法吗，也许会限制可以支持的有效位数？

Can someone suggest a good way to convert from float to Decimal preserving value as the user has entered, perhaps limiting number of significant digits that can be supported?

推荐答案

Python< 2.7

Python <2.7

"%.15g" % f

或者在Python 3.0中：

Or in Python 3.0:

format(f, ".15g")

Python 2.7 +，3.2 +

只需将浮点数直接传递给 Decimal 构造函数，如下所示：

Python 2.7+, 3.2+

Just pass the float to Decimal constructor directly, like this:

from decimal import Decimal Decimal(f)