I've pratically resolved my problem (the code is under that), but a last thing
is not perfect:
when I use the function plot to call after the function polygon, there is a
marge between my raster and the window. I think it's the axis of the function
"plot()", but I have not found how delete it. Someone have a solution please?
Pierre Bruyer
##smooth contour
contours <- contourLines(V2b,levels=paliers)
par(mar=c(0,0,0,0))
plot(1,col="white",main="polygon()", asp = 1, axes = FALSE, ann =
FALSE,xlim=c(0,1), ylim = c(0,1),type = "n", method = c("image"))
for (i in seq_along(contours)) {
x <- contours[[i]]$x
y <- contours[[i]]$y
c <- contours[[i]]$level
j <- 1
tmp <- 0
while(j < length(level[,1]) && tmp == 0){
if(level[j,1] == c){
tmp <- j
}
j <- j+1
}
polygon( spline( seq_along(x), x)$y, spline( seq_along(y), y)$y
,col = colgraph[tmp+1], border = NA)
}
Le 17 mai 2011 à 16:44, Pierre Bruyer a écrit :
> The result is good, thanks a lot, but how can I with this method fill my
> raster to color?
>
> Le 17 mai 2011 à 15:43, Duncan Murdoch a écrit :
>
>> I don't think filled.contour gives you access to the contour lines. If you
>> use contourLines() to compute them, then you can draw them using code like
>> this:
>>
>> contours <- contourLines(V2b,levels=paliers)
>> for (i in seq_along(contours)) {
>> x <- contours[[i]]$x
>> y <- contours[[i]]$y
>> lines( splines( seq_along(x), x)$y, splines( seq_along(y), y)$y )
>> }
>>
>> but as I said, you won't get great results. A better way is to use a finer
>> grid, e.g. by fitting a smooth surface to your set of points and using
>> predictions from the model to interpolate.
>>
>> Duncan Murdoch
>>
>>
>> On 17/05/2011 9:35 AM, Pierre Bruyer wrote:
>>> I work with large datasets (10000 points) so I can't post them , but my
>>> function is :
>>>
>>> create_map<- function(grd, level ,map_output, format = c("jpeg"), width_map
>>> = 150, height_map = 150,...)
>>> {
>>>
>>> ##sp<- spline(x = grd[,1], y = grd[,2])
>>>
>>> grd2<- matrix(grd[,3], nrow = sqrt(length(grd[,3])), ncol =
>>> sqrt(length(grd[,3])), byrow = FALSE)
>>>
>>> V2b<- grd2
>>>
>>>
>>> ##creation of breaks for colors
>>> i<-1
>>> paliers<- c(-1.0E300)
>>> while(i<=length(level[,1]))
>>> {
>>> paliers<- c(paliers,level[i,1])
>>> i<- i+1
>>> }
>>> paliers<- c(paliers, 1.0E300)
>>>
>>> ##scale color creation
>>> i<- 1
>>> colgraph<- c(rgb(255,255,255, maxColorValue = 255))
>>> while(i<=length(level[,2]))
>>> {
>>> colgraph<- c(colgraph, rgb(level[i,2],level[i,3],level[i,4],
>>> maxColorValue = 255))
>>> i<- i +1
>>> }
>>>
>>> ##user can choose the output format (default is jpeg)
>>> switch(format,
>>> png = png(map_output, width = width_map, height = height_map) ,
>>> jpeg = jpeg(map_output, width = width_map, height = height_map,
>>> quality = 100),
>>> bmp = bmp(map_output, width = width_map, height = height_map),
>>> tiff = tiff(map_output, width = width_map, height = height_map),
>>> jpeg(map_output, width = width_map, height = height_map))
>>>
>>> ## drawing map
>>>
>>> ##delete marge
>>> par(mar=c(0,0,0,0))
>>> filled.contour(V2b, col = colgraph, levels = paliers, asp = 1, axes =
>>> FALSE, ann = FALSE)
>>> dev.off()
>>>
>>> }
>>>
>>> where grd is a xyz data frame,
>>> map_output is the path+name of the output image file,
>>> and level is a matrix like this :
>>>
>>>
>>> level<- matrix(0,10,4)
>>> level[1,1]<- 1.0000E+00
>>> level[2,1]<- 3.0000E+00
>>> level[3,1]<- 5.0000E+00
>>> level[4,1]<- 1.0000E+01
>>> level[5,1]<- 1.5000E+01
>>> level[6,1]<- 2.0000E+01
>>> level[7,1]<- 3.0000E+01
>>> level[8,1]<- 4.0000E+01
>>> level[9,1]<- 5.0000E+01
>>> level[10,1]<- 7.5000E+01
>>>
>>>
>>> level[1,2]<- 102
>>> level[2,2]<- 102
>>> level[3,2]<- 102
>>> level[4,2]<- 93
>>> level[5,2]<- 204
>>> level[6,2]<- 248
>>> level[7,2]<- 241
>>> level[8,2]<- 239
>>> level[9,2]<- 224
>>> level[10,2]<- 153
>>>
>>> level[1,3]<- 153
>>> level[2,3]<- 204
>>> level[3,3]<- 204
>>> level[4,3]<- 241
>>> level[5,3]<- 255
>>> level[6,3]<- 243
>>> level[7,3]<- 189
>>> level[8,3]<- 126
>>> level[9,3]<- 14
>>> level[10,3]<- 0
>>>
>>> level[1,4]<- 153
>>> level[2,4]<- 204
>>> level[3,4]<- 153
>>> level[4,4]<- 107
>>> level[5,4]<- 102
>>> level[6,4]<- 33
>>> level[7,4]<- 59
>>> level[8,4]<- 63
>>> level[9,4]<- 14
>>> level[10,4]<- 51
>>>
>>> Le 17 mai 2011 à 15:17, Duncan Murdoch a écrit :
>>>
>>>> On 17/05/2011 8:24 AM, Pierre Bruyer wrote:
>>>>> Thank you for your answer, but the function spline() (and a lot of other
>>>>> function in R) can't take in its parameters the original contour which
>>>>> are define by a vector, i.e. :
>>>>>
>>>>
>>>> If you post some reproducible code to generate the contours, someone will
>>>> show you how to use splines to interpolate them.
>>>>
>>>> Duncan Murdoch
>>>>
>>>>> ##creation of breaks for colors
>>>>> i<-1
>>>>> paliers<- c(-1.0E300)
>>>>> while(i<=length(level[,1]))
>>>>> {
>>>>> paliers<- c(paliers,level[i,1])
>>>>> i<- i+1
>>>>> }
>>>>> paliers<- c(paliers, 1.0E300)
>>>>>
>>>>>
>>>>>
>>>>> Le 17 mai 2011 à 13:05, Duncan Murdoch a écrit :
>>>>>
>>>>>> On 11-05-17 5:58 AM, Pierre Bruyer wrote:
>>>>>>> I'm a French developer (so I am sorry if my english is not perfect). I
>>>>>>> have a problem to smooth the contours of a map. I have a dataset with 3
>>>>>>> columns, x, y and z, where x and y are the coordinates of my points and
>>>>>>> z is evaluate to a qualitative elevation and his representation is a
>>>>>>> set of colors, which is define by levels.
>>>>>>>
>>>>>>> The problem is the curve of my contour is so linear, and I would like
>>>>>>> a more continuous contour. I use the function fitted.contour to draw my
>>>>>>> map.
>>>>>>
>>>>>> If you use a finer grid of x,y values you'll get shorter segments and
>>>>>> they will look smoother.
>>>>>>
>>>>>> You might be able to use a smooth interpolator (e.g. spline()) rather
>>>>>> than linear interpolation, but those occasionally do strange things e.g.
>>>>>>
>>>>>> x<- c(1:4, 5.9, 6:10)
>>>>>> y<- c(1:4, 7, 6:10)
>>>>>> plot(spline(x,y, n=200), type="l")
>>>>>> points(x,y)
>>>>>>
>>>>>> where one point is out of line with the others, but the curve
>>>>>> overcompensates in order to stay smooth.
>>>>>>
>>>>>> Duncan Murdoch
>>>>>
>>>>
>>>
>>
>
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