由网友(满船清梦压星河)分享简介：有谁知道（）由getAccuracy返回的测量精度的正常跨pretation？例如，他们在计算公式为：Does anyone know the proper interpretation of the accuracy measurements returned by getAccuracy()? For insta...

有谁知道（）由getAccuracy返回的测量精度的正常跨pretation？例如，他们在计算公式为：

Does anyone know the proper interpretation of the accuracy measurements returned by getAccuracy()? For instance, are they calculated as:

圆概​​率误差（意思是，如果我理解正确的，50％的置信圆半径）？

Circular Error Probability (meaning, if i understand correctly, radius of a 50% confidence circle)?

半径？

别的东西？

此外，什么是用于我们多少可以依靠他们的实际计算？这是否依赖于位置估计（全球定位系统与网络）的来源？

In addition, what are the actual calculations that are used and how much can we rely on them? Does it depend on the source of the location estimate (GPS vs. network)?

非常感谢什么建议可以给我。

Many thanks for any advice you can give me.

推荐答案

要回答这个问题的一部分，这个数字是 68％的置信半径，这意味着有68％的机会的真实位置是在米测量点的该半径范围内的。假设误差是正态分布的（其中，作为文档说，不一定是真的），这意味着，这是一个标准差。例如，如果 Location.getAccuracy 返回1​​0，然后有一个68％的机会设备的真实位置是10米范围内的报道坐标。

To answer one part of the question, the number is the radius of 68% confidence, meaning that there is a 68% chance that the true location is within that radius of the measured point in meters. Assuming that errors are normally distributed (which, as the docs say, is not necessarily true), that means that this is one standard deviation. For example, if Location.getAccuracy returns 10, then there's a 68% chance the true location of the device is within 10 meters of the reported coordinates.

的http://developer.android.com/reference/android/location/Location.html#getAccuracy()