Python 由其错误确定的值的有效数字

我需要一个函数，只返回与给定错误相关的值的有效部分。意思是这样的：

def (value, error):
    """ This function takes a value and determines its significant
        accuracy by its error.
        It returns only the scientific important part of a value and drops the rest. """

    magic magic magic....

    return formated value as String.

到目前为止，我写了些什么来表达我的意思：

import numpy as np

def signigicant(value, error):
    """ Returns a number in a scintific format. Meaning a value has an error
        and that error determines  how many digits of the
        value are signifcant. e.g. value = 12.345MHz, 
        error = 0.1MHz => 12.3MHz because the error is at the first digit.
        (in reality drop the MHz its just to show why.)"""


    xx = "%E"%error    # I assume this is most ineffective.
    xx = xx.split("E")
    xx = int(xx[1])

    if error <= value: # this should be the normal case
        yy = np.around(value, -xx)

        if xx >= 0: # Error is 1 or bigger
            return "%i"%yy

        else: # Error is smaller than 1
            string = "%."+str(-xx) +"f"
            return string%yy

    if error > value: # This should not be usual but it can happen.
        return "%g"%value

例如：

>>> errors = [0.2,1.123,1.0, 123123.1233215,0.123123123768]
>>> values = [12.3453,123123321.4321432, 0.000321 ,321321.986123612361236,0.00001233214 ]
>>> map(round_on_error, values, errors)
[12.3, 123123321.0, 0.0, 300000.0, 0.0]

如果你想保持一个低于其误差的值

if (value <  error)
    return value
else
    def round_on_error(value, error):
    significant_digits = 10**math.floor(math.log(error, 10))
    return value // significant_digits * significant_digits

if（值<错误）
返回值
其他的
def四舍五入打开错误（值，错误）：
有效数字=10**数学下限（数学日志（错误，10））
返回值//有效数字*有效数字

AFAIK，对于科学计算来说，您不仅仅是放弃错误。如果某个值不精确，通常表示为12.345MHz±0.1MHz。

>>> errors = [0.2,1.123,1.0, 123123.1233215,0.123123123768]
>>> values = [12.3453,123123321.4321432, 0.000321 ,321321.986123612361236,0.00001233214 ]
>>> map(round_on_error, values, errors)
[12.3, 123123321.0, 0.0, 300000.0, 0.0]

if (value <  error)
    return value
else
    def round_on_error(value, error):
    significant_digits = 10**math.floor(math.log(error, 10))
    return value // significant_digits * significant_digits