blocks|key|3037764|text|getAccuracy上的文档说它返回以米为单位的精度。我猜想这意味着如果你得到的返回值是60，那么你就是在一个半径为60米的圆中。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3037765|entityMap|0|LINK|mutability|MUTABLE|url|http://developer.android.com/reference/android/location/Location.html#getAccuracy%2528%2529^0|0|B|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

The documentation on <a href="http://developer.android.com/reference/android/location/Location.html#getAccuracy%28%29" rel="noreferrer">getAccuracy</a> says that it returns the accuracy in meters. 
I would guess that this means if you get a return value of 60 you are somewhere in a circle with a 60 meter radius around the supplied position.

blocks|key|3037852|text|如果像文档中引用的那样，这是一个精确度，那么用户的实际位置在QUOTED_LOCATION+%2B/-精确度内。所以精确度定义了一个半径，你可以期望用户在那里。文档中没有提到的是，你能多确定用户在半径内--标准是95%25，所以我猜是这样的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3037853|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

If, as quoted it in the docs, it is an accuracy, then the users actual position is somewhere within QUOTED_LOCATION +/- ACCURACY. So the accuracy defines a radius where you could expect the user to be. What the docs don't say is how sure you can be that the user is within radius - the standard is 95% so I guess this it.

blocks|key|307626|text|定位是一项棘手的任务，当你的电池寿命有限，在建筑物和许多大型建筑物等区域没有GPS信号时，但Android让它变得容易得多。当你请求一个位置时，你只需要指定你需要的精度。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|307627|如果你指定你想要一个示例*100+meters*的accuracy，安卓将尝试获取位置，如果它可以获得精度为70米的位置，它将返回给你，但如果安卓可以获得一个精度高于100米的位置，你的应用程序将等待，不会收到任何信息，直到有这样的位置精度。|offset|length|style|CODE|307628|通常，Android将首先获取Cell+ID，然后将其发送到Google服务器，Google服务器将映射这些Cell+ID，服务器将返回纬度和经度，其精度对于1000米的示例来说是低的。到这个时候，Android还将尝试查看该区域内的所有WiFi网络，并将有关这些网络的信息发送到Google服务器，如果可能，Google服务器将以更高的精度返回一个新位置，例如800米。|307629|到这个时候GPS就会打开了。从冷启动开始，GPS设备至少需要30秒才能获得修复，因此如果可以修复，它将返回纬度和经度，但同样具有准确性，对于100米的示例来说，这将是最高的。GPS工作的时间越长，你得到的精度就越高。|307630|重要提示：前两种方法需要internet连接。如果没有数据连接，你将不得不等待GPS，但如果设备在建筑物内，你可能得不到位置。|BOLD|307631|entityMap^0|0|C|C|P|8|0|0|0|0|5|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]|$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|W|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|X|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|Y|8|@$D|Z|E|10|F|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|P|$]]

The location is a tricky task to do, when you have a limited battery life and when there is no GPS signal in buildings and in areas with many big building and etc. But Android makes it a much easier. When you request a location, you just have to specify what accuracy do you need.

If you specify that you want an <code>accuracy</code> for an example <code>*100 meters*</code>, Android will try to get the location and if it can get a location for accuracy 70 meters, it will return it to you, but if Android can get a location with an accuracy higher than 100 meters, your application will wait and will not receive anything till there is a location in such accuracy.

Typically Android will first get the Cell ID and then will send it to Google server, which maps such Cell IDs and the server will return a latitude and longitude with an accuracy which is low for an example 1000 meters. By this time Android will also try to see all WiFi networks in the area an will send information about them too to the Google server and if possible Google server will return a new location with higher accuracy for an example 800 meters.

By this time the GPS will be on. The GPS device needs at least 30 seconds from a cold start to get a fix, so if can get a fix it will return latitude and longitude but again with an accuracy, which will be the highest possible for an example 100 meters. The longer the GPS works, the better accuracy you will get.

Important notice: The first two methods requires an internet connection. If there is no data connection, you will have to wait for the GPS, but if the device is in a building, you will probably get no location.

blocks|key|3037988|text|据我对Android源代码的快速浏览可以看出，这取决于设备的硬件以及它选择返回的值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3037989|GpsLocationProvider.java文件有一个reportLocation+method，它由本机代码调用，并以值的形式传递精度。因此，至少在框架中似乎没有发生任何计算。|offset|length|style|CODE|3037990|qcom+(我相信是Qualcomm)+GPS+git+repo+is+passing+hor_unc_circular参数的准确性，这似乎意味着，至少，该实现正在使用CER。|3037991|entityMap|0|LINK|mutability|MUTABLE|url|http://android.git.kernel.org/?p=platform/frameworks/base.git;a=blob;f=services/java/com/android/server/location/GpsLocationProvider.java#l1026|1|http://android.git.kernel.org/?p=platform/hardware/qcom/gps.git;a=blob;f=loc_api/libloc_api/loc_eng.cpp#l798^0|0|0|O|T|E|T|L|0|0|18|G|X|A|1|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@$D|V|E|W|F|G]|$D|X|E|Y|F|G]]|9|@$D|Z|E|10|1|11]]|A|$]]|$1|H|3|I|5|6|7|12|8|@$D|13|E|14|F|G]]|9|@$D|15|E|16|1|17]]|A|$]]|$1|J|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]|R|$5|M|N|O|A|$P|S]]]]

As far as I can see from a quick glance at the Android source code, this is dependent on the hardware of the device and what value it chooses to return.

The <code>GpsLocationProvider.java</code> file has a <a href="http://android.git.kernel.org/?p=platform/frameworks/base.git;a=blob;f=services/java/com/android/server/location/GpsLocationProvider.java#l1026" rel="noreferrer"><code>reportLocation</code> method</a> which gets called by the native code and gets passed the accuracy as a value. Thus, no calculation seems to be happening in the framework at least.

The qcom (which I believe is the Qualcomm) GPS git repo <a href="http://android.git.kernel.org/?p=platform/hardware/qcom/gps.git;a=blob;f=loc_api/libloc_api/loc_eng.cpp#l798" rel="noreferrer">is passing</a> the <code>hor_unc_circular</code> parameter for accuracy which seems to imply that, at least, that implementation is using CER.

blocks|key|302498|text|我知道你要求的是概率方面的确切答案，但我认为这里有两件事需要考虑。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|302499|首先，这是由提供商决定他们想要在这个值中输入什么，所以根据提供商的不同，这可能只是一个错误的猜测。|302500|其次，将其视为潜在的舍入问题可能会有所帮助。如果我试图根据一些输入来计算您的位置，而其中一些输入仅适用于特定数量的有效数字，那么只能计算具有给定数量的有效数字的位置。|302501|这样想--“一加”和“一百”有什么关系？可能是100左右，因为100的精度可能小于1的量级。如果我突然说答案大约是101，那么我可能最终会暗示一个不存在的精度水平。但是，如果我实际指定了准确度，那么我可以说100加或减10加1加或减.1是101加或减10。我知道这通常指的是95%25的置信度(标准误差)，但再次强调，所有这些都假设提供商理解统计数据，而不仅仅是猜测。|302502|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

I get that you're asking for a definite answer in terms of probability, but I think there are two things to consider here. 

First off, it's up to the provider to decide what they want to put in this value, so depending on the provider, it may just be a bad guess. 

Secondly, it may help to think of this as a potential rounding problem. If I'm trying to calculate your location based on a number of inputs, and some of those inputs are only available to a certain number of significant digits, then it's only possible to calculate a location with a given number of significant digits. 

Think of it this way -- what's "about" one plus "about" one hundred. Probably about one hundred, because the accuracy of one hundred is likely less than the magnitude of 1. If I suddenly say the answer is about 101, then I may end up implying a level of accuracy that wasn't there. However, if I actually specify the accuracy, then I can say that it 100 plus or minus 10 plus 1 plus or minus .1 is 101 plus or minus 10. I get that this is generally referring to something like a 95% confidence level (standard error), but again, that all assumes the provider understands statistics and isn't just guessing.

blocks|key|3252394|text|为了回答问题的一部分，这个数字是the+radius+of+68%25+confidence,，这意味着真实位置有68%25的可能性在以米为单位的测量点的半径内。假设误差是正态分布的(正如文档所说，这不一定是真的)，这意味着这是一个标准差。例如，如果Location.getAccuracy返回10，那么有68%25的概率设备的真实位置在报告坐标的10米内。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3252395|http://developer.android.com/reference/android/location/Location.html#getAccuracy()|3252396|entityMap|0|LINK|mutability|MUTABLE|url^0|G|T|3D|K|0|0|2B|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@$9|U|A|V|1|W]]|E|$]]|$1|H|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|G]]]]

To answer one part of the question, the number is <code>the radius of 68% confidence,</code> 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 <code>Location.getAccuracy</code> returns 10, then there's a 68% chance the true location of the device is within 10 meters of the reported coordinates.

<a href="http://developer.android.com/reference/android/location/Location.html#getAccuracy()" rel="noreferrer">http://developer.android.com/reference/android/location/Location.html#getAccuracy()</a>

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

<ul>
<li>Circular Error Probability (meaning, if i understand correctly, radius of a 50% confidence circle)?</li>
<li>Radius of 95% confidence circle?</li>
<li>something else?</li>
</ul>

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)?

Many thanks for any advice you can give me.

How is location accuracy measured in Android?

有人知道getAccuracy()返回的精度度量的正确解释吗？例如，它们是否计算为：循环误差概率(如果我理解正确的话，表示50%置信度的半径或95%置信度的半径circle?something ？此外，实际使用的计算是什么，我们可以在多大程度上依赖它们？它依赖于位置估计的来源(GPS与网络)吗？非常感谢你能给我的任何建...

问Android中的定位精度是如何衡量的？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Android中的定位精度是如何衡量的？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Android中的定位精度是如何衡量的？
EN