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From: Tim M. <tim...@gm...> - 2008-07-14 09:37:53
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Dear Matplotlib-Users, I am tryring to create a contour plot over a basemap. My main problem is creating the array for the Z values as a basis for the plt.contour command from a CSV file where latitude, longitude and value are stored column-wise: lat; lon; value 50; 10; 6 ... The data represents a regular spaced grid with a datapoint each 0.25 degrees. I tried various possibilities but didn't have success: 1) following simpletest.py from the basemap examples: X, Y = meshgrid(data[:,1], data[:,0]) Z = data[:,2] m.contourf(x,y, Z) => Error: Z must be a 2D array -> How do I get Z to be a 2D array? 2) using the griddata package Here I was nearly without orientation how to call griddata correctly. 3) Using the python bindings of ogr Any examples on this one? >From my above demonstrated methods the following questions arrise: What is the preferred way to plot - Points stored in the above descripbed format (lat, lon, value)? - Interpolate a grid of data points by using different interpolation methods like inverse distance wheighting, natural neighbor interpolation, etc. to get a contour map? Thanks in advance for your help & kind regards, Timmie |
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From: Jeff W. <js...@fa...> - 2008-07-14 12:19:22
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Tim Michelsen wrote: > Dear Matplotlib-Users, > I am tryring to create a contour plot over a basemap. > > My main problem is creating the array for the Z values as a basis for the > plt.contour command from a CSV file where latitude, longitude and value are > stored column-wise: > > lat; lon; value > 50; 10; 6 > ... > > The data represents a regular spaced grid with a datapoint each 0.25 degrees. > > I tried various possibilities but didn't have success: > > 1) following simpletest.py from the basemap examples: > X, Y = meshgrid(data[:,1], data[:,0]) > > Z = data[:,2] > Timmie: Try: X, Y = meshgrid(data[:,1], data[:,0]) Z = data[:,2] nlons = X.shape[1]; nlats = X.shape[0] Z = Z.reshape(nlats,nlons) > m.contourf(x,y, Z) > > => Error: Z must be a 2D array > -> How do I get Z to be a 2D array? > > 2) using the griddata package > Here I was nearly without orientation how to call griddata correctly. > You don't need to use griddata since you have regularly gridded data. > > 3) Using the python bindings of ogr > Any examples on this one? > Again, no need. A simple reshape will get you the 2d lat/lon array you need. > >From my above demonstrated methods the following questions arrise: > What is the preferred way to plot > - Points stored in the above descripbed format (lat, lon, value)? > - Interpolate a grid of data points by using different interpolation methods > like inverse distance wheighting, natural neighbor interpolation, etc. to get a > contour map? > For interpolation of irregular, randomly distributed data points see http://www.scipy.org/Cookbook/Matplotlib/Gridding_irregularly_spaced_data. However, if there is some structure to the data grid then it's probably better not to use these approaches. -Jeff -- Jeffrey S. Whitaker Phone : (303)497-6313 NOAA/OAR/CDC R/PSD1 FAX : (303)497-6449 325 Broadway Boulder, CO, USA 80305-3328 |
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From: Jeff W. <js...@fa...> - 2008-07-14 12:27:22
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Jeff Whitaker wrote: > Tim Michelsen wrote: > >> Dear Matplotlib-Users, >> I am tryring to create a contour plot over a basemap. >> >> My main problem is creating the array for the Z values as a basis for the >> plt.contour command from a CSV file where latitude, longitude and value are >> stored column-wise: >> >> lat; lon; value >> 50; 10; 6 >> ... >> >> The data represents a regular spaced grid with a datapoint each 0.25 degrees. >> >> I tried various possibilities but didn't have success: >> >> 1) following simpletest.py from the basemap examples: >> X, Y = meshgrid(data[:,1], data[:,0]) >> >> Z = data[:,2] >> >> > Timmie: Try: > > X, Y = meshgrid(data[:,1], data[:,0]) > Z = data[:,2] > nlons = X.shape[1]; nlats = X.shape[0] > Z = Z.reshape(nlats,nlons) > Timmie: Sorry, but upon further reflection I don't think this will work. You'll need to know the number of lats and the number of lons on the grid beforehand. Then you should be able to do X = X.reshape(nlats,nlons) Y = Y.reshape(nlats,nlons) Z = Z.reshape(nlats,nlons) after reading the data in. (skip the meshgrid call, that's only useful when X is a vector with length nlons and Y is a vector with length nlats). If you still have problems, send us a full example. -Jeff > >> m.contourf(x,y, Z) >> >> => Error: Z must be a 2D array >> -> How do I get Z to be a 2D array? >> >> 2) using the griddata package >> Here I was nearly without orientation how to call griddata correctly. >> >> > You don't need to use griddata since you have regularly gridded data. > >> 3) Using the python bindings of ogr >> Any examples on this one? >> >> > Again, no need. A simple reshape will get you the 2d lat/lon array you > need. > > >> >From my above demonstrated methods the following questions arrise: >> What is the preferred way to plot >> - Points stored in the above descripbed format (lat, lon, value)? >> - Interpolate a grid of data points by using different interpolation methods >> like inverse distance wheighting, natural neighbor interpolation, etc. to get a >> contour map? >> >> > > For interpolation of irregular, randomly distributed data points see > http://www.scipy.org/Cookbook/Matplotlib/Gridding_irregularly_spaced_data. > > However, if there is some structure to the data grid then it's probably > better not to use these approaches. > > -Jeff > > > > -- Jeffrey S. Whitaker Phone : (303)497-6313 NOAA/OAR/CDC R/PSD1 FAX : (303)497-6449 325 Broadway Boulder, CO, USA 80305-3328 |
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From: Tim M. <tim...@gm...> - 2008-07-14 13:21:17
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Hello Jeff, > >> - Points stored in the above descripbed format (lat, lon, value)? This one I solved using a m.scatter() function > >> - Interpolate a grid of data points by using different interpolation >>> methods like inverse distance wheighting, natural neighbor interpolation, etc. to get a contour map? > > For interpolation of irregular, randomly distributed data points see > > http://www.scipy.org/Cookbook/Matplotlib/ Gridding_irregularly_spaced_data. > > > > However, if there is some structure to the data grid then it's probably > > better not to use these approaches. The problem is that although regular spaced the grid is still to large to countour to a nice map. I will play a bit more with contour and other interpolation functions. I tried griddata: > 2) using the griddata package > Here I was nearly without orientation how to call griddata correctly. I tried again. Here is what I got: x = data[:,1] y = data[:,0] z = data[:,2] X, Y = mlab.meshgrid(x, y) X, Z = mlab.meshgrid(x, y) # zi = griddata(x,y,z,xi,yi,**kwargs) Z = grid.griddata(x,y,z, X, Y) plt.contour(X,Y, Z) => ValueError: output grid defined by xi,yi must be monotone increasing The coordinates are stored in a way that first longitude (x) increases and then the latitude (y) increases. 10 6.0 4 10 6.25 3 10 6.50 2 10 6.75 1 10 6.0 6 11 6.25 7 11 6.50 6 11 6.75 9 12 6.0 4 What how do I need to arrange my data to get it monotone increasing for griddata? Thanks for your help. One settled I will send you another example for the examples package. Kind regards, Timmie |
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From: Jeff W. <js...@fa...> - 2008-07-14 13:28:03
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Tim Michelsen wrote: > Hello Jeff, > > >>>> - Points stored in the above descripbed format (lat, lon, value)? >>>> > This one I solved using a m.scatter() function > > >>>> - Interpolate a grid of data points by using different interpolation >>>> methods like inverse distance wheighting, natural neighbor >>>> > interpolation, etc. to get a contour map? > >>> For interpolation of irregular, randomly distributed data points see >>> http://www.scipy.org/Cookbook/Matplotlib/ >>> > Gridding_irregularly_spaced_data. > >>> However, if there is some structure to the data grid then it's probably >>> better not to use these approaches. >>> > The problem is that although regular spaced the grid is still to large to > countour to a nice map. I will play a bit more with contour and other > interpolation functions. > > I tried griddata: > > >> 2) using the griddata package >> Here I was nearly without orientation how to call griddata correctly. >> > I tried again. > > Here is what I got: > x = data[:,1] > y = data[:,0] > z = data[:,2] > X, Y = mlab.meshgrid(x, y) > X, Z = mlab.meshgrid(x, y) > # zi = griddata(x,y,z,xi,yi,**kwargs) > Z = grid.griddata(x,y,z, X, Y) > plt.contour(X,Y, Z) > > => ValueError: output grid defined by xi,yi must be monotone increasing > > The coordinates are stored in a way that first longitude (x) increases and > then the latitude (y) increases. > 10 6.0 4 > 10 6.25 3 > 10 6.50 2 > 10 6.75 1 > 10 6.0 6 > 11 6.25 7 > 11 6.50 6 > 11 6.75 9 > 12 6.0 4 > > What how do I need to arrange my data to get it monotone increasing for > griddata? > > Thanks for your help. One settled I will send you another example for the > examples package. > > Kind regards, > Timmie > > Timmie: You shouldn't use griddata. You have a regular lat/lon grid, so it's just a matter of loading the data into the proper 2-d array. Please send a self-contained script (and post the data somewhere) and then we can help you. -Jeff -- Jeffrey S. Whitaker Phone : (303)497-6313 NOAA/OAR/CDC R/PSD1 FAX : (303)497-6449 325 Broadway Boulder, CO, USA 80305-3328 |
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From: Tim M. <tim...@gm...> - 2008-07-14 15:11:07
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Hello, thanks. I checked again from contour_demo.py of the basemap distribution. There lats, lons are uniquely monoton increasing from 0-360 and from -90 to 90. In my case data is written row-by-row: * increasing from lowest latitude western most longitude to easternmost longitude and then increasing each rows in the same manner to the northermost latitude (see below). So, as you said, it's a question of re-aranging the data. that it fits the to the way m.contour expects the 2-D array. Also, since the grid is still coarse, I would need to apply some smoothing afterwards. What do you recommend for that? I don't know how I can do this easily by hand. May you give me some guidance here, please? But I may just convert it to a shape file using GIS then load it with the shapefile interface you wrote. What would you see as most convenient way? If I produce maps with a GIS but want to use matplotlib for the map plotting, what would be the preferred export format? Any gdal format? Many thanks in advance, Timmie ### data example Latitude Longitude value 45 7 7.65251434 45 7.25 6.841345477 45 7.5 3.923153289 45 7.75 3.644313708 45 8 3.550977951 45 8.25 3.352525137 45 8.5 3.080082094 45 8.75 2.971992657 45 9 2.998723785 45 9.25 3.080082094 45 9.5 3.185687405 45 9.75 3.102075854 45 10 3.185687405 45 10.25 3.213960325 45 10.5 3.32326373 45 10.75 3.465643983 45 11 3.612980369 45 11.25 3.644313708 45 11.5 3.701277511 45 11.75 3.923153289 45 12 3.797848342 45 12.25 3.612980369 45 12.5 3.435577844 45 12.75 3.294210812 45 13 3.26536503 45.25 7 6.485050223 45.25 7.25 6.343081631 45.25 7.5 3.856783573 45.25 7.75 3.405725407 45.25 8 3.550977951 45.25 8.25 3.294210812 45.25 8.5 3.294210812 45.25 8.75 3.185687405 45.25 9 3.15761656 45.25 9.25 3.213960325 45.25 9.5 3.15761656 45.25 9.75 3.32326373 45.25 10 3.405725407 45.25 10.25 3.495925216 45.25 10.5 3.465643983 45.25 10.75 3.550977951 45.25 11 3.465643983 45.25 11.25 3.765429652 45.25 11.5 3.95669157 45.25 11.75 3.797848342 45.25 12 3.923153289 45.25 12.25 3.733239867 45.25 12.5 3.550977951 45.25 12.75 3.520306012 45.25 13 3.376085288 45.5 7 6.383367092 45.5 7.25 6.383367092 45.5 7.5 6.009422688 45.5 7.75 4.679469855 45.5 8 3.435577844 45.5 8.25 3.435577844 45.5 8.5 3.236725042 45.5 8.75 3.236725042 45.5 9 3.185687405 45.5 9.25 3.102075854 45.5 9.5 3.102075854 45.5 9.75 3.185687405 45.5 10 3.352525137 45.5 10.25 3.405725407 45.5 10.5 3.376085288 45.5 10.75 3.612980369 45.5 11 3.520306012 45.5 11.25 3.352525137 45.5 11.5 3.823949103 45.5 11.75 3.856783573 45.5 12 3.856783573 45.5 12.25 3.765429652 45.5 12.5 3.669541114 45.5 12.75 3.550977951 45.5 13 3.435577844 45.75 7 5.309043916 45.75 7.25 6.057519881 45.75 7.5 5.030958443 45.75 7.75 4.836570243 45.75 8 4.836570243 45.75 8.25 2.724965001 45.75 8.5 2.607751091 45.75 8.75 3.26536503 45.75 9 2.898163214 45.75 9.25 2.872155245 45.75 9.5 1.893252754 45.75 9.75 2.043669061 45.75 10 1.75488883 45.75 10.25 2.004264146 45.75 10.5 2.971992657 45.75 10.75 1.804949998 45.75 11 2.846334614 45.75 11.25 5.519419657 45.75 11.5 2.517818813 45.75 11.75 3.733239867 45.75 12 3.376085288 45.75 12.25 3.550977951 45.75 12.5 3.612980369 45.75 12.75 3.520306012 45.75 13 3.495925216 46 7 5.06399168 46 7.25 4.949174095 46 7.5 5.266087828 46 7.75 5.352298328 46 8 4.757472437 46 8.25 2.800325674 46 8.5 3.612980369 46 8.75 3.185687405 46 9 2.323282473 46 9.25 1.671485743 46 9.5 3.856783573 46 9.75 4.572079662 46 10 4.679469855 46 10.25 4.679469855 46 10.5 5.309043916 46 10.75 3.294210812 46 11 3.405725407 46 11.25 3.669541114 46 11.5 3.495925216 46 11.75 4.255093726 46 12 3.495925216 46 12.25 3.185687405 46 12.5 3.213960325 46 12.75 3.550977951 46 13 3.520306012 46.25 7 1.969297411 46.25 7.25 4.908706364 46.25 7.5 3.052767233 46.25 7.75 3.765429652 46.25 8 3.95669157 46.25 8.25 5.06399168 46.25 8.5 5.266087828 46.25 8.75 3.669541114 46.25 9 3.185687405 46.25 9.25 3.797848342 46.25 9.5 3.352525137 46.25 9.75 5.439709782 46.25 10 5.69098301 46.25 10.25 4.949174095 46.25 10.5 5.736883145 46.25 10.75 5.105542055 46.25 11 4.255093726 46.25 11.25 3.701277511 46.25 11.5 4.255093726 46.25 11.75 4.572079662 46.25 12 3.98369323 46.25 12.25 4.148941623 46.25 12.5 3.129746478 46.25 12.75 3.236725042 46.25 13 3.550977951 46.5 7 2.872155245 46.5 7.25 3.701277511 46.5 7.5 3.15761656 46.5 7.75 3.765429652 46.5 8 5.18951259 46.5 8.25 6.105948261 46.5 8.5 5.266087828 46.5 8.75 5.69098301 46.5 9 6.009422688 46.5 9.25 5.147381739 46.5 9.5 5.829636932 46.5 9.75 5.654489904 46.5 10 6.243327668 46.5 10.25 5.395852976 46.5 10.5 5.736883145 46.5 10.75 6.057519881 46.5 11 5.147381739 46.5 11.25 3.520306012 46.5 11.5 3.856783573 46.5 11.75 4.148941623 46.5 12 4.71833512 46.5 12.25 4.71833512 46.5 12.5 3.701277511 46.5 12.75 3.889851131 46.5 13 3.32326373 46.75 7 1.859766825 46.75 7.25 2.198852355 46.75 7.5 2.345277833 46.75 7.75 2.517818813 46.75 8 3.856783573 46.75 8.25 3.856783573 46.75 8.5 5.06399168 46.75 8.75 4.184077131 46.75 9 5.829636932 46.75 9.25 3.644313708 46.75 9.5 3.765429652 46.75 9.75 5.309043916 46.75 10 6.009422688 46.75 10.25 5.147381739 46.75 10.5 5.609155594 46.75 10.75 5.783100444 46.75 11 5.147381739 46.75 11.25 3.581868928 46.75 11.5 4.908706364 46.75 11.75 3.465643983 46.75 12 3.465643983 46.75 12.25 4.148941623 46.75 12.5 3.98369323 46.75 12.75 3.581868928 46.75 13 3.644313708 |
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From: Jeff W. <js...@fa...> - 2008-07-14 16:00:55
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Tim Michelsen wrote:
> Hello,
> thanks.
> I checked again from contour_demo.py of the basemap distribution.
>
> There lats, lons are uniquely monoton increasing from 0-360 and from -90 to 90.
> In my case data is written row-by-row:
> * increasing from lowest latitude western most longitude to easternmost
> longitude and then increasing each rows in the same manner to the northermost
> latitude (see below).
>
> So, as you said, it's a question of re-aranging the data. that it fits the to
> the way m.contour expects the 2-D array.
> Also, since the grid is still coarse, I would need to apply some smoothing
> afterwards. What do you recommend for that?
>
Timme: Here's one way to do it
from matplotlib.mlab import load
import matplotlib.pyplot as plt
import numpy as np
data = load("data.txt")
# need to know nlons and nlats beforehand!
nlons = 8; nlats = 25
X = data[0::nlats,0]
Y = data[0:nlats,1]
# data is in nlons,nlats order in file, need to transpose
Z = data[:,2].reshape(nlons,nlats).transpose()
X,Y = np.meshgrid(X,Y)
CS = plt.contourf(X,Y,Z,20)
plt.show()
I don't have any recommendations for smoothing - why don't you plot the
raw data first and see if you really need it?
> I don't know how I can do this easily by hand. May you give me some guidance
> here, please?
>
> But I may just convert it to a shape file using GIS then load it with the
> shapefile interface you wrote.
> What would you see as most convenient way?
> If I produce maps with a GIS but want to use matplotlib for the map plotting,
> what would be the preferred export format? Any gdal format?
>
I prefer netCDF format for gridded data (basemap contains a function for
reading netCDF files - NetCDFFile).
-Jeff
> Many thanks in advance,
> Timmie
>
> ### data example
>
> Latitude Longitude value
> 45 7 7.65251434
> 45 7.25 6.841345477
> 45 7.5 3.923153289
> 45 7.75 3.644313708
> 45 8 3.550977951
> 45 8.25 3.352525137
> 45 8.5 3.080082094
> 45 8.75 2.971992657
> 45 9 2.998723785
> 45 9.25 3.080082094
> 45 9.5 3.185687405
> 45 9.75 3.102075854
> 45 10 3.185687405
> 45 10.25 3.213960325
> 45 10.5 3.32326373
> 45 10.75 3.465643983
> 45 11 3.612980369
> 45 11.25 3.644313708
> 45 11.5 3.701277511
> 45 11.75 3.923153289
> 45 12 3.797848342
> 45 12.25 3.612980369
> 45 12.5 3.435577844
> 45 12.75 3.294210812
> 45 13 3.26536503
> 45.25 7 6.485050223
> 45.25 7.25 6.343081631
> 45.25 7.5 3.856783573
> 45.25 7.75 3.405725407
> 45.25 8 3.550977951
> 45.25 8.25 3.294210812
> 45.25 8.5 3.294210812
> 45.25 8.75 3.185687405
> 45.25 9 3.15761656
> 45.25 9.25 3.213960325
> 45.25 9.5 3.15761656
> 45.25 9.75 3.32326373
> 45.25 10 3.405725407
> 45.25 10.25 3.495925216
> 45.25 10.5 3.465643983
> 45.25 10.75 3.550977951
> 45.25 11 3.465643983
> 45.25 11.25 3.765429652
> 45.25 11.5 3.95669157
> 45.25 11.75 3.797848342
> 45.25 12 3.923153289
> 45.25 12.25 3.733239867
> 45.25 12.5 3.550977951
> 45.25 12.75 3.520306012
> 45.25 13 3.376085288
> 45.5 7 6.383367092
> 45.5 7.25 6.383367092
> 45.5 7.5 6.009422688
> 45.5 7.75 4.679469855
> 45.5 8 3.435577844
> 45.5 8.25 3.435577844
> 45.5 8.5 3.236725042
> 45.5 8.75 3.236725042
> 45.5 9 3.185687405
> 45.5 9.25 3.102075854
> 45.5 9.5 3.102075854
> 45.5 9.75 3.185687405
> 45.5 10 3.352525137
> 45.5 10.25 3.405725407
> 45.5 10.5 3.376085288
> 45.5 10.75 3.612980369
> 45.5 11 3.520306012
> 45.5 11.25 3.352525137
> 45.5 11.5 3.823949103
> 45.5 11.75 3.856783573
> 45.5 12 3.856783573
> 45.5 12.25 3.765429652
> 45.5 12.5 3.669541114
> 45.5 12.75 3.550977951
> 45.5 13 3.435577844
> 45.75 7 5.309043916
> 45.75 7.25 6.057519881
> 45.75 7.5 5.030958443
> 45.75 7.75 4.836570243
> 45.75 8 4.836570243
> 45.75 8.25 2.724965001
> 45.75 8.5 2.607751091
> 45.75 8.75 3.26536503
> 45.75 9 2.898163214
> 45.75 9.25 2.872155245
> 45.75 9.5 1.893252754
> 45.75 9.75 2.043669061
> 45.75 10 1.75488883
> 45.75 10.25 2.004264146
> 45.75 10.5 2.971992657
> 45.75 10.75 1.804949998
> 45.75 11 2.846334614
> 45.75 11.25 5.519419657
> 45.75 11.5 2.517818813
> 45.75 11.75 3.733239867
> 45.75 12 3.376085288
> 45.75 12.25 3.550977951
> 45.75 12.5 3.612980369
> 45.75 12.75 3.520306012
> 45.75 13 3.495925216
> 46 7 5.06399168
> 46 7.25 4.949174095
> 46 7.5 5.266087828
> 46 7.75 5.352298328
> 46 8 4.757472437
> 46 8.25 2.800325674
> 46 8.5 3.612980369
> 46 8.75 3.185687405
> 46 9 2.323282473
> 46 9.25 1.671485743
> 46 9.5 3.856783573
> 46 9.75 4.572079662
> 46 10 4.679469855
> 46 10.25 4.679469855
> 46 10.5 5.309043916
> 46 10.75 3.294210812
> 46 11 3.405725407
> 46 11.25 3.669541114
> 46 11.5 3.495925216
> 46 11.75 4.255093726
> 46 12 3.495925216
> 46 12.25 3.185687405
> 46 12.5 3.213960325
> 46 12.75 3.550977951
> 46 13 3.520306012
> 46.25 7 1.969297411
> 46.25 7.25 4.908706364
> 46.25 7.5 3.052767233
> 46.25 7.75 3.765429652
> 46.25 8 3.95669157
> 46.25 8.25 5.06399168
> 46.25 8.5 5.266087828
> 46.25 8.75 3.669541114
> 46.25 9 3.185687405
> 46.25 9.25 3.797848342
> 46.25 9.5 3.352525137
> 46.25 9.75 5.439709782
> 46.25 10 5.69098301
> 46.25 10.25 4.949174095
> 46.25 10.5 5.736883145
> 46.25 10.75 5.105542055
> 46.25 11 4.255093726
> 46.25 11.25 3.701277511
> 46.25 11.5 4.255093726
> 46.25 11.75 4.572079662
> 46.25 12 3.98369323
> 46.25 12.25 4.148941623
> 46.25 12.5 3.129746478
> 46.25 12.75 3.236725042
> 46.25 13 3.550977951
> 46.5 7 2.872155245
> 46.5 7.25 3.701277511
> 46.5 7.5 3.15761656
> 46.5 7.75 3.765429652
> 46.5 8 5.18951259
> 46.5 8.25 6.105948261
> 46.5 8.5 5.266087828
> 46.5 8.75 5.69098301
> 46.5 9 6.009422688
> 46.5 9.25 5.147381739
> 46.5 9.5 5.829636932
> 46.5 9.75 5.654489904
> 46.5 10 6.243327668
> 46.5 10.25 5.395852976
> 46.5 10.5 5.736883145
> 46.5 10.75 6.057519881
> 46.5 11 5.147381739
> 46.5 11.25 3.520306012
> 46.5 11.5 3.856783573
> 46.5 11.75 4.148941623
> 46.5 12 4.71833512
> 46.5 12.25 4.71833512
> 46.5 12.5 3.701277511
> 46.5 12.75 3.889851131
> 46.5 13 3.32326373
> 46.75 7 1.859766825
> 46.75 7.25 2.198852355
> 46.75 7.5 2.345277833
> 46.75 7.75 2.517818813
> 46.75 8 3.856783573
> 46.75 8.25 3.856783573
> 46.75 8.5 5.06399168
> 46.75 8.75 4.184077131
> 46.75 9 5.829636932
> 46.75 9.25 3.644313708
> 46.75 9.5 3.765429652
> 46.75 9.75 5.309043916
> 46.75 10 6.009422688
> 46.75 10.25 5.147381739
> 46.75 10.5 5.609155594
> 46.75 10.75 5.783100444
> 46.75 11 5.147381739
> 46.75 11.25 3.581868928
> 46.75 11.5 4.908706364
> 46.75 11.75 3.465643983
> 46.75 12 3.465643983
> 46.75 12.25 4.148941623
> 46.75 12.5 3.98369323
> 46.75 12.75 3.581868928
> 46.75 13 3.644313708
>
>
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--
Jeffrey S. Whitaker Phone : (303)497-6313
Meteorologist FAX : (303)497-6449
NOAA/OAR/PSD R/PSD1 Email : Jef...@no...
325 Broadway Office : Skaggs Research Cntr 1D-113
Boulder, CO, USA 80303-3328 Web : http://tinyurl.com/5telg
|
|
From: Tim M. <tim...@gm...> - 2008-07-14 19:23:19
|
Hello Jeff, > Timme: Here's one way to do it many thanks so far. I still have to inspect and improve my script. But at least your code lead me to some contourd surface. I will come back and tell if it worked. Unfortunately I cannot disclose the data nor the results because of copyright issues. So this mail is just to thank you for your responsiveness to my questions. Timmie |
|
From: Tim M. <tim...@gm...> - 2008-07-29 21:55:45
|
Hello Jeff, I just wanna give feedback on what got me going here: #### data preparation ### data is loaded from a CSV file ### lats = y # data[:,0] ## lon => x lons = x # data[:,1] ## values => z values = z #data[:,2] ### lat_uniq = list(set(lats.tolist())) nlats = len(lat_uniq) lon_uniq = list(set(lons.tolist())) nlons = len(lon_uniq) color_map = plt.cm.spectral print lats.shape, nlats, nlons yre = lats.reshape(nlats,nlons) xre = lons.reshape(nlats,nlons) zre = values.reshape(nlats,nlons) #### later in the defined map CT = m.contourf(xre, yre, zre, cmap=color_map) Of course, this can be simplified. But, really, in the end the solution was just simple and I hadn't thought of it... If you want I can try to create a short example for the distribution. Thanks for your help again, Timmie |
