如何根据位置接近度平均值(How to average values based on location proximity)
我有一个带有地理标记值(经度,纬度,值)的SQL表。 该表快速累积,有数千个条目。 因此,在表中查询某些区域中的值会返回非常大的数据集。
我想知道平均值与近距离接近一个值的方法,这里是一个例子:
表:
Long lat value 10.123001 53.567001 10 10.123002 53.567002 12 10.123003 53.567003 18 10.124003 53.568003 13
比方说我目前的位置是10.123004,53.567004。 如果我在查询附近的值,我会得到值为10,12,18和13的四个原始数据。如果数据集相对较小,则可以使用。 如果数据很大我想查询sql的舍入位置(10.123,53.567)并需要sql来返回类似的东西
Long lat value 10.123 53.567 10 (this is the average of 10, 12, and 18) 10.124 53.568 13
这可能吗? 我们如何根据位置平均大数据集?
sql数据库首先是正确的选择吗?
I have an SQL table with geo-tagged values (Longitude, Latitude, value). The table is accumulated quickly and has thousands entries. Therefore, querying the table for values in some area return very large data-set.
I would like to know the way to average value with close location proximity to one value, here is an illustration:
Table:
Long lat value 10.123001 53.567001 10 10.123002 53.567002 12 10.123003 53.567003 18 10.124003 53.568003 13
lets say my current location is 10.123004, 53.567004. If I am querying for the values near by I will get the four raws with values 10, 12, 18, and 13. This works if the data-set is relatively small. If the data is large I would like to query sql for rounded location (10.123, 53.567) and need sql to return something like
Long lat value 10.123 53.567 10 (this is the average of 10, 12, and 18) 10.124 53.568 13
Is this possible? how we can average large data set based on locations?
Is sql database is the right choice in the first place?
原文:https://stackoverflow.com/questions/10384076
满意答案
最后我可以使用我的苹果ID登录。 我从应用商店获得更新,所以我更新了软件(现在的版本是OS X El Capitan版本10.11.6)。 现在我可以登录了。 希望这对其他人有所帮助
Finally i am able to login using my apple id. I got update from app store so i have updated the software(Now version is OS X El Capitan version 10.11.6 ). Now i am able to login. Hope this will help to others
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