如何从变量中指定浮点小数精度? [英] How to specify floating point decimal precision from variable?

问题描述

我将以下重复的简单代码重复了几次，希望为其提供功能:

I have the following repetitive simple code repeated several times that I would like to make a function for:

for i in range(10):
    id  = "some id string looked up in dict"
    val = 63.4568900932840928 # some floating point number in dict corresponding to "id"
    tabStr += '%-15s = %6.1f\n' % (id,val)

我希望能够调用此函数:def printStr(precision)
它执行上面的代码，并返回带有val到precision小数点的tabStr.

I want to be able to call this function: def printStr(precision)
Where it preforms the code above and returns tabStr with val to precision decimal points.

例如:printStr(3)
将在tabStr中为val返回63.457.

For example: printStr(3)
would return 63.457 for val in tabStr.

有什么想法可以实现这种功能吗?

Any ideas how to accomplish this kind of functionality?

推荐答案

tabStr += '%-15s = %6.*f\n' % (id, i, val)  

其中i是小数位数.

顺便说一句，在最近的Python中，.format()取代了%，您可以使用

BTW, in the recent Python where .format() has superseded %, you could use

"{0:<15} = {2:6.{1}f}".format(id, i, val)

用于同一任务.

或者，为了清楚起见，使用字段名称:

Or, with field names for clarity:

"{id:<15} = {val:6.{i}f}".format(id=id, i=i, val=val)

如果您使用的是Python 3.6+，则可以简单地使用 f字符串:

If you are using Python 3.6+, you could simply use f-strings:

f"{id:<15} = {val:6.{i}f}"

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