blocks|key|4398162|text|geosphere包的dist*函数是矢量化的，因此您只需要更好地准备您的输入。试试这个：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4398163|#prepare+a+matrix+with+coordinates+of+every+position
allCoords<-cbind(lon,rep(lat,each=length(lon)))
#call+the+dist+function+and+put+the+result+in+a+matrix
res<-matrix(distm(cbind(mylon,mylat),allCoords,fun=distHaversine)<=500000,nrow=length(lon))
#check+the+result
identical(res,dd2)
#[1]+TRUE|code-block|syntax|javascript|4398164|正如@Floo0答案所示，有很多不必要的计算。我们可以遵循另一种策略:我们首先确定可以比阈值更近的经度和经度范围，然后只使用它们来计算距离：|4398165|#initialize+the+return
res<-matrix(FALSE,nrow=length(lon),ncol=length(lat))
#we+find+the+possible+values+of+longitude+that+can+be+closer+than+500000
#How?+We+calculate+the+distances+between+us+and+points+with+our+same+lon+
longood<-which(distm(c(mylon,mylat),cbind(lon,mylat))<=500000)
#Same+for+latitude
latgood<-which(distm(c(mylon,mylat),cbind(mylon,lat))<=500000)
#we+build+the+matrix+with+only+those+values+to+exploit+the+vectorized
#nature+of+distm
allCoords<-cbind(lon[longood],rep(lat[latgood],each=length(longood)))
res[longood,latgood]<-distm(c(mylon,mylat),allCoords)<=500000|4398166|这样，您只计算lg%2Bln%2Blg*ln+(lg和ln是latgood和longood的长度)，即531个距离，而不是我之前方法的259200个距离。|4398167|entityMap^0|0|9|B|5|0|0|0|0|7|B|K|2|N|2|Q|7|Y|7|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|R|$]]

The <code>dist*</code> functions of the <code>geosphere</code> package are vectorized, so you only need to prepare better your inputs. Try this:

<pre><code>#prepare a matrix with coordinates of every position
allCoords&lt;-cbind(lon,rep(lat,each=length(lon)))
#call the dist function and put the result in a matrix
res&lt;-matrix(distm(cbind(mylon,mylat),allCoords,fun=distHaversine)&lt;=500000,nrow=length(lon))
#check the result
identical(res,dd2)
#[1] TRUE
</code></pre>

As the @Floo0 answer showed, there is a lot of unnecessary calculations. We can follow another strategy: we first determine the lon and lat range that can be closer than the threshold and then we use only them to calculate the distance:

<pre><code>#initialize the return
res&lt;-matrix(FALSE,nrow=length(lon),ncol=length(lat))
#we find the possible values of longitude that can be closer than 500000
#How? We calculate the distances between us and points with our same lon 
longood&lt;-which(distm(c(mylon,mylat),cbind(lon,mylat))&lt;=500000)
#Same for latitude
latgood&lt;-which(distm(c(mylon,mylat),cbind(mylon,lat))&lt;=500000)
#we build the matrix with only those values to exploit the vectorized
#nature of distm
allCoords&lt;-cbind(lon[longood],rep(lat[latgood],each=length(longood)))
res[longood,latgood]&lt;-distm(c(mylon,mylat),allCoords)&lt;=500000
</code></pre>

In this way, you calculate just <code>lg+ln+lg*ln</code> (<code>lg</code> and <code>ln</code> are the length of <code>latgood</code> and <code>longood</code>), i.e. 531 distances, opposed to the 259200 with my previous method.

blocks|key|4398447|text|我在下面使用spatialrisk包添加了一个解决方案。这个包中的关键函数是用C%2B%2B+(Rcpp)编写的，因此速度非常快。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4398448|首先，加载数据：|4398449|mylat+<-+47.9625
mylon+<-+-87.0431

lon+<-+seq(-179.75,179.75,+by+=+0.5)
lat+<-+seq(-89.75,89.75,+by+=+0.5)
df+<-+expand.grid(lon+=+lon,+lat+=+lat)|code-block|syntax|javascript|4398450|函数spatialrisk::points_in_circle()计算从中心点开始半径内的观测值。请注意，距离是使用Haversine公式计算的。|4398451|与@Hugh版本相比，spatialrisk方法的计时：|4398452|spatialrisk::points_in_circle(df,+mylon,+mylat,+radius+=+5e5)

Unit:+milliseconds
+++++++expr+++++++min++++++++lq++++++mean++++median++++++++uq+++++++max+neval+cld+
spatialrisk++3.071897++3.366256++5.224479++4.068124++4.809626++17.24378+++100+++a+
+++++hutils+17.507311+20.788525+29.470707+25.061943+31.066139+268.29375+++100+++b|4398453|结果可以很容易地转换为矩阵。|4398454|看看@philcolbourn关于如何测试一个点是否在圆内的优秀答案。请参阅：https://stackoverflow.com/a/7227057/5440749|offset|length|4398455|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/7227057/5440749^0|0|0|0|0|0|0|0|13|17|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|13|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|15|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|16|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|17|8|@]|9|@]|A|$G|H]]|$1|O|3|P|5|6|7|18|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|19|8|@]|9|@$S|1A|T|1B|1|1C]]|A|$]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

I add below a solution using the spatialrisk package. The key functions in this package are written in C++ (Rcpp), and are therefore very fast.

First, load the data:

<pre><code>mylat &lt;- 47.9625
mylon &lt;- -87.0431

lon &lt;- seq(-179.75,179.75, by = 0.5)
lat &lt;- seq(-89.75,89.75, by = 0.5)
df &lt;- expand.grid(lon = lon, lat = lat)
</code></pre>

The function spatialrisk::points_in_circle() calculates the observations within radius from a center point. Note that distances are calculated using the Haversine formula.

Timings for the spatialrisk approach compared to the @Hugh version:

<pre><code>spatialrisk::points_in_circle(df, mylon, mylat, radius = 5e5)

Unit: milliseconds
 expr min lq mean median uq max neval cld 
spatialrisk 3.071897 3.366256 5.224479 4.068124 4.809626 17.24378 100 a 
 hutils 17.507311 20.788525 29.470707 25.061943 31.066139 268.29375 100 b
</code></pre>

The result can be easily converted to a matrix. 

Have a look at the excellent answer by @philcolbourn on how to test if a point is inside a circle. See: <a href="https://stackoverflow.com/a/7227057/5440749">https://stackoverflow.com/a/7227057/5440749</a>

I have a gridded dataset, with data available at the following locations:

<pre><code>lon &lt;- seq(-179.75,179.75, by = 0.5)
lat &lt;- seq(-89.75,89.75, by = 0.5)
</code></pre>

I would like to find all of the data points that are within 500 km of the location:

<pre><code>mylat &lt;- 47.9625
mylon &lt;- -87.0431
</code></pre>

I aim to use the geosphere package in R, but the method I've currently written does not seem very efficient:

<pre><code>require(geosphere)
dd2 &lt;- array(dim = c(length(lon),length(lat)))
for(i in 1:length(lon)){
 for(ii in 1:length(lat)){
 clon &lt;- lon[i]
 clat &lt;- lat[ii]
 dd &lt;- as.numeric(distm(c(mylon, mylat), c(clon, clat), fun = distHaversine))
 dd2[i,ii] &lt;- dd &lt;= 500000
 }
}
</code></pre>

Here, I loop through each grid in the data and find if the distance is less than 500 km. I then store a variable with either TRUE or FALSE, which I can then use to average the data (other variable). From this method, I want a matrix with TRUE or FALSE for the locations within 500 km from the lat and lon shown. Is there a more efficient method for doing this?

find locations within certain lat/lon distance in r

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个网格数据集，数据位于以下位置：lon <- seq(-179.75,179.75, by = 0.5)lat <- seq(-89.75,89.75, by = 0.5)我希望找到该位置周围500公里内的所有数据点：mylat <- 47.9625mylon <- -87.0431我的目标是在R中使用geosp...

问在r中查找特定经度/经度距离内的位置
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