如何将浮点数打印到n个小数位(包括尾随0)? [英] How to print float to n decimal places including trailing 0s?

问题描述

我需要将浮点数打印或转换为小数点后15位的字符串，即使结果中包含许多尾随的0，例如:

I need to print or convert a float number to 15 decimal place string even if the result has many trailing 0s eg:

1.6变成1.6000000000000000

1.6 becomes 1.6000000000000000

我尝试了舍入(6.2,15)，但它返回6.2000000000000002，并添加了舍入错误

I tried round(6.2,15) but it returns 6.2000000000000002 adding a rounding error

我还在线上看到各种各样的人，他们将浮点数放入字符串中，然后手动添加尾随0，但这似乎很糟糕...

I also saw various people online who put the float into a string and then added trailing 0's manually but that seems bad...

做到这一点的最佳方法是什么?

What is the best way to do this?

推荐答案

适用于2.6+和3.x版的Python版本

您可以使用 str.format 方法.例子:

For Python versions in 2.6+ and 3.x

You can use the str.format method. Examples:

>>> print('{0:.16f}'.format(1.6))
1.6000000000000001

>>> print('{0:.15f}'.format(1.6))
1.600000000000000

请注意，第一个示例末尾的1是舍入错误；之所以会发生这种情况，是因为精确表示十进制数1.6要求使用无数个二进制数字.由于浮点数的位数是有限的，因此该数字将四舍五入为一个相邻的值，但不相等.

Note the 1 at the end of the first example is rounding error; it happens because exact representation of the decimal number 1.6 requires an infinite number binary digits. Since floating-point numbers have a finite number of bits, the number is rounded to a nearby, but not equal, value.

您可以使用模格式"语法(这也适用于Python 2.6和2.7):

You can use the "modulo-formatting" syntax (this works for Python 2.6 and 2.7 too):

>>> print '%.16f' % 1.6
1.6000000000000001

>>> print '%.15f' % 1.6
1.600000000000000

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