有限的浮点小数点，不四舍五入 [英] limited float decimal point without rounding the number

问题描述

我想将数字转换为带小数点后3位的浮点数，但我不希望取整. 例如:

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)

此代码返回该数字:

12.342

但是我需要原始号码，我需要这个:

but I need to original number,I need this:

12.341

我编写了一个接收数字形式用户的代码，并将其转换为浮点数. 我不知道与用户输入的号码是什么.

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.

推荐答案

我的第一个念头是将print("%.3f" % a)更改为print("%.3f" % (a-0.0005))，但效果不佳:当输出a = 12.341661时，如果= 12.341则输出12.340，这显然是不正确的.

My first thought was to change print("%.3f" % a) to print("%.3f" % (a-0.0005)) but it does not quite work: while it outputs what you want for a=12.341661, if a=12.341 it outputs 12.340, which is obviously not right.

相反，我建议使用int()明确地进行铺地板:

Instead, I suggest doing the flooring explicitly using int():

a = 12.341661
b = int(a*1000)/1000.
print(b)

这将输出您想要的内容:

This outputs what you want:

12.341

要获得3个小数，即使输入较少，也可以设置输出格式:

To get 3 decimals out even if the input has fewer, you can format the output:

a = 3.1
b = int(a*1000)/1000.
print("%.3f" % b)

输出:

3.100

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