blocks|key|963042|text|您可以利用这样一个事实:在内部，由map()函数返回的map对象可以重新计算，并在map()函数中再次使用。我会创建一个包含单个多边形的列表，检查哪些多边形具有非常不同的经度值，并重新排列这些多边形。我在下面的函数*中给出了这种方法的一个例子，它允许这样的事情：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|963043|plot.map("world",+center=180,+col="white",bg="gray",
+++fill=TRUE,ylim=c(-60,90),mar=c(0,0,0,0))|code-block|syntax|javascript|963044|要获得|963045|​|963046|📷|atomic|963047|963048|如果我是你，我会把所有的东西都移动一点，就像下面这样：|963049|plot.map("world",+center=200,+col="white",bg="gray",
+++fill=TRUE,ylim=c(-60,90),mar=c(0,0,0,0))|963050|963051|963052|963053|函数：|963054|plot.map<-+function(database,center,...){
++++Obj+<-+map(database,...,plot=F)
++++coord+<-+cbind(Obj[[1]],Obj[[2]])

++++#+split+up+the+coordinates
++++id+<-+rle(!is.na(coord[,1]))
++++id+<-+matrix(c(1,cumsum(id$lengths)),ncol=2,byrow=T)
++++polygons+<-+apply(id,1,function(i){coord[i[1]:i[2],]})

++++#+split+up+polygons+that+differ+too+much
++++polygons+<-+lapply(polygons,function(x){
++++++++x[,1]+<-+x[,1]+%2B+center
++++++++x[,1]+<-+ifelse(x[,1]>180,x[,1]-360,x[,1])
++++++++if(sum(diff(x[,1])>300,na.rm=T)+>0){
++++++++++id+<-+x[,1]+<+0
++++++++++x+<-+rbind(x[id,],c(NA,NA),x[!id,])
+++++++}
+++++++x
++++})
++++#+reconstruct+the+object
++++polygons+<-+do.call(rbind,polygons)
++++Obj[[1]]+<-+polygons[,1]
++++Obj[[2]]+<-+polygons[,2]

++++map(Obj,...)
}|963055|*请注意，此函数仅采用正中心值。它很容易调整，以允许两个方向的中心值，但我不再麻烦，因为这是微不足道的。|963056|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/f60963584c03597d43639424390810dd.png|imageAlt|1|https://ask.qcloudimg.com/http-save/yehe-900000/0fb899ee1b6070756808bb9e2a288154.png^0|H|5|R|3|15|5|0|0|0|0|0|1|0|0|0|0|0|0|0|1|1|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1N|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1O|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|1P|8|@]|D|@]|E|$]]|$1|O|3|P|5|Q|7|1Q|8|@]|D|@$9|1R|A|1S|1|1T]]|E|$]]|$1|R|3|N|5|6|7|1U|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|1V|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1W|8|@]|D|@]|E|$I|J]]|$1|W|3|N|5|6|7|1X|8|@]|D|@]|E|$]]|$1|X|3|P|5|Q|7|1Y|8|@]|D|@$9|1Z|A|20|1|21]]|E|$]]|$1|Y|3|N|5|6|7|22|8|@]|D|@]|E|$]]|$1|Z|3|10|5|6|7|23|8|@]|D|@]|E|$]]|$1|11|3|12|5|H|7|24|8|@]|D|@]|E|$I|J]]|$1|13|3|14|5|6|7|25|8|@]|D|@]|E|$]]|$1|15|3|-4|5|6|7|26|8|@]|D|@]|E|$]]]|16|$17|$5|18|19|1A|E|$1B|1C|1D|-4]]|1E|$5|18|19|1A|E|$1B|1F|1D|-4]]]]

You could use the fact that internally, a <code>map</code> object returned by the <code>map()</code> function can be recalculated and used again in the <code>map()</code> function. I'd create a list with individual polygons, check which ones have very different longitude values, and rearrange those ones. I gave an example of this approach in the function below*, which allows something like :

<pre><code>plot.map("world", center=180, col="white",bg="gray",
 fill=TRUE,ylim=c(-60,90),mar=c(0,0,0,0))
</code></pre>

to get

<img src="https://i.stack.imgur.com/DbcO5.png" alt="Corrected map center 180">

If I were you, I'd shift everything a bit more, like in :

<pre><code>plot.map("world", center=200, col="white",bg="gray",
 fill=TRUE,ylim=c(-60,90),mar=c(0,0,0,0))
</code></pre>

<img src="https://i.stack.imgur.com/YviUz.png" alt="Corrected map center 200">

The function :

<pre><code>plot.map&lt;- function(database,center,...){
 Obj &lt;- map(database,...,plot=F)
 coord &lt;- cbind(Obj[[1]],Obj[[2]])

 # split up the coordinates
 id &lt;- rle(!is.na(coord[,1]))
 id &lt;- matrix(c(1,cumsum(id$lengths)),ncol=2,byrow=T)
 polygons &lt;- apply(id,1,function(i){coord[i[1]:i[2],]})

 # split up polygons that differ too much
 polygons &lt;- lapply(polygons,function(x){
 x[,1] &lt;- x[,1] + center
 x[,1] &lt;- ifelse(x[,1]&gt;180,x[,1]-360,x[,1])
 if(sum(diff(x[,1])&gt;300,na.rm=T) &gt;0){
 id &lt;- x[,1] &lt; 0
 x &lt;- rbind(x[id,],c(NA,NA),x[!id,])
 }
 x
 })
 # reconstruct the object
 polygons &lt;- do.call(rbind,polygons)
 Obj[[1]] &lt;- polygons[,1]
 Obj[[2]] &lt;- polygons[,2]

 map(Obj,...)
}
</code></pre>

*Note that this function only takes positive center values. It's easily adapted to allow for center values in both directions, but I didn't bother anymore as that's trivial.

blocks|key|963112|text|有点晚了，但您也可以使用投影创建一个移动的贴图(需要mapproj包)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|963113|++map("world",+projection="rectangular",+parameter=0,+
++++++orientation=c(90,0,180),+wrap=TRUE,+fill=T,+resolution=0,col=0)|code-block|syntax|javascript|963114|这将会移动180度。但与'world2‘的不同之处在于，经度坐标将不同(-pi，pi)。这个包的所有预测都将0放在中心位置。在这种情况下，'wrap‘选项会正确地检测跳转。|963115|'resolution=0‘有助于获得更清晰的边框。|963116|您可以通过更改投影描述中的'180‘值轻松地更改中心经度。|963117|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

A bit late, but you can also create a shifted map by using a projection (requires the mapproj package):

<pre><code> map("world", projection="rectangular", parameter=0, 
 orientation=c(90,0,180), wrap=TRUE, fill=T, resolution=0,col=0)
</code></pre>

This will shift by 180 degrees. But the difference with 'world2' is that the longitude co-ordinate will be different ([-pi,pi]). All projections of this package put 0 at the centre. And in that case, the 'wrap' option detects the jump correctly.

'resolution=0' helps to get cleaner borders.

You can easily change the centre longitude by changing the '180' value in the projection description.

blocks|key|747293|text|安装最新版本的地图(3.2.0)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|747294|执行以下操作：|747295|d$lon2+<-+ifelse(d$lon+<+-25,+d$lon+%2B+360,+d$lon)+#+where+d+is+your+df
mapWorld+<-+map_data('world',+wrap=c(-25,335),+ylim=c(-55,75))

ggplot()+%2B
geom_polygon(data+=+mapWorld,+aes(x=long,+y+=+lat,+group+=+group))+%2B
geom_point(data+=+d,+aes(x+=+lon2,+y+=+lat))|code-block|syntax|javascript|747296|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

install the latest version of maps (3.2.0).

do this:

<pre><code>d$lon2 &lt;- ifelse(d$lon &lt; -25, d$lon + 360, d$lon) # where d is your df
mapWorld &lt;- map_data('world', wrap=c(-25,335), ylim=c(-55,75))

ggplot() +
geom_polygon(data = mapWorld, aes(x=long, y = lat, group = group)) +
geom_point(data = d, aes(x = lon2, y = lat))
</code></pre>

blocks|key|747301|text|这个解决方案怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|747302|xlims+=+c(0,+359)+#+these+are+the+limits+you+want+to+change
ylims+=+c(-55,75)

mapWorld+<-+map_data('world',+wrap=xlims,+ylim=ylims)
head(mapWorld)

g1+<-+ggplot()+%2B
++++geom_polygon(data+=+mapWorld,+aes(x=long,+y+=+lat,+group+=+group))+%2B
++++coord_map("rectangular",+lat0=0,+xlim=xlims,+ylim=ylims)

g1|code-block|syntax|javascript|747303|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

What about this solution?
<pre><code>xlims = c(0, 359) # these are the limits you want to change
ylims = c(-55,75)

mapWorld &lt;- map_data('world', wrap=xlims, ylim=ylims)
head(mapWorld)

g1 &lt;- ggplot() +
 geom_polygon(data = mapWorld, aes(x=long, y = lat, group = group)) +
 coord_map(&quot;rectangular&quot;, lat0=0, xlim=xlims, ylim=ylims)

g1
</code></pre>

I'm plotting some points on a map of the world using the R <code>maps</code> package, something like:

<img src="https://i.stack.imgur.com/vmtDo.png" alt="Map of the world, -180° to 180° longitude">

The command to draw the base map is:

<pre><code>map("world", fill=TRUE, col="white", bg="gray", ylim=c(-60, 90), mar=c(0,0,0,0))
</code></pre>

But I need to display Pacific centred map. I use <code>map("world2",</code> etc to use the Pacific centred basemap from the maps package, and convert the coordinates of the data points in my dataframe (<code>df</code>) with:

<pre><code>df$longitude[df$longitude &lt; 0] = df$longitude[df$longitude &lt; 0] + 360
</code></pre>

This works if I don't use the <code>fill</code> option, but with <code>fill</code> the polygons which cross 0° cause problems.

<img src="https://i.stack.imgur.com/H2Goo.png" alt="Map of the world, 0° to 360° longitude">

I guess I need to transform the polygon data from the <code>maps</code> library somehow to sort this out, but I have no idea how to get at this.

My ideal solution would be to draw a maps with a left boundary at -20° and a right boundary at -30° (i.e. 330°). The following gets the correct points and coastlines onto the map, but the crossing-zero problem is the same

<pre><code>df$longitude[df$longitude &lt; -20] = df$longitude[d$longitude &lt; -20] + 360
map("world", fill=TRUE, col="white", bg="gray", mar=c(0,0,0,0),
 ylim=c(-60, 90), xlim=c(-20, 330))
map("world2", add=TRUE, col="white", bg="gray", fill=TRUE, xlim=c(180, 330))
</code></pre>

Any help would be greatly appreciated.

Fixing maps library data for Pacific centred (0°-360° longitude) display

我正在使用R maps包在世界地图上绘制一些点，如下所示：​​绘制底图的命令是：map("world", fill=TRUE, col="white", bg="gray", ylim=c(-60, 90), mar=c(0,0,0,0))但我需要显示以太平洋为中心的地图。我使用map("world2",等来使用地图包...

问修复以太平洋为中心(0°-360°经度)显示的地图库数据
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