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From: William C. <wcc...@lb...> - 2009-11-02 22:30:27
|
Hello Mitra, If you haven't solved your problem yet, I would highly recommend cleaning up previous versions and using the make.osx file supplied with the matplotlib download. Edit the PREFIX to replace the example target directory with your own target directory. Make sure the MACOSX_DEPLOYMENT_TARGET and the MacOSX versions in the CFLAGS and LDFLAGS are all set to the version you are using, presumably 10.4. The make.osx script will check all the required dependencies and go get the ones it needs for you. I had a lot of trouble installing matplotlib on 10.6 until I followed this procedure. Good luck, Bill On 11/2/09 12:11 PM, "R. Mitra" <rm...@uc...> wrote: > Hello Again > Since I got no replies for my last question and figuring that it > must be a problem with numpy (as Eric pointed out) I now have running > python 2.6.4 and numpy 1.3.0 on OSX 10.4.11. Is there a need to > uninstall previous matplotlib and reinstall it again? This page > sounds pretty complicated for me: http://matplotlib.sourceforge.net/ > users/installing.html#build-osx? > > There are multiple versions still existing on my computer. How can I > clean up and install the latest version? Thank you all so much > > Mitra > > ------------------------------------------------------------------------------ > Come build with us! The BlackBerry(R) Developer Conference in SF, CA > is the only developer event you need to attend this year. Jumpstart your > developing skills, take BlackBerry mobile applications to market and stay > ahead of the curve. Join us from November 9 - 12, 2009. Register now! > http://p.sf.net/sfu/devconference > _______________________________________________ > Matplotlib-users mailing list > Mat...@li... > https://lists.sourceforge.net/lists/listinfo/matplotlib-users |
From: Pierre de B. <pd...@ul...> - 2009-11-02 21:34:08
|
From memory, you just need to make a length one list contour(z, [i]) Pierre Le 2 nov. 09 à 22:19, Brendan Arnold a écrit : > Hi there, > > I can draw a single contour line in MATLAB using > > contour(z, [i i]) > > however, > > contour(z, [i, i]) > > using matplotlib gives an error. In fact any plot that plots a single > line (i.e. contour(z, 1)) also gives an error as follows, > > TypeError: unhashable type: 'numpy.ndarray' > > How do I draw a single contour line using matplotlib? > > regards, > > Brendan > > ---------------------------------------------------------------------- > -------- > Come build with us! The BlackBerry(R) Developer Conference in SF, CA > is the only developer event you need to attend this year. Jumpstart > your > developing skills, take BlackBerry mobile applications to market > and stay > ahead of the curve. Join us from November 9 - 12, 2009. Register now! > http://p.sf.net/sfu/devconference > _______________________________________________ > Matplotlib-users mailing list > Mat...@li... > https://lists.sourceforge.net/lists/listinfo/matplotlib-users |
From: John H. <jd...@gm...> - 2009-11-02 21:28:10
|
On Mon, Nov 2, 2009 at 3:19 PM, Brendan Arnold <bre...@gm...> wrote: > Hi there, > > I can draw a single contour line in MATLAB using > > contour(z, [i i]) > > however, > > contour(z, [i, i]) > > using matplotlib gives an error. In fact any plot that plots a single > line (i.e. contour(z, 1)) also gives an error as follows, > > TypeError: unhashable type: 'numpy.ndarray' > > How do I draw a single contour line using matplotlib? I've used: ax.contour(R, F, dR, levels=[0]) |
From: Brendan A. <bre...@gm...> - 2009-11-02 21:19:46
|
Hi there, I can draw a single contour line in MATLAB using contour(z, [i i]) however, contour(z, [i, i]) using matplotlib gives an error. In fact any plot that plots a single line (i.e. contour(z, 1)) also gives an error as follows, TypeError: unhashable type: 'numpy.ndarray' How do I draw a single contour line using matplotlib? regards, Brendan |
From: Pierre de B. <pd...@ul...> - 2009-11-02 20:43:05
|
Hello, > There are multiple versions still existing on my computer. How can I > clean up and install the latest version? Thank you all so much You need to check the different paths accessed by Python. >>> import sys >>> print sys.path and remove the possible matplotib files in these paths. Typically, this includes /usr/lib/python2.6/site-packages, /usr/local/lib/..., / opt/lib/..., /sw/lib/... (on mac) Then, download the .tar.gz file from matplotlib website, uncompress and in the folder run $ python setup.py build That command will check dependencies before building, indicating what is needed. If some dependencies are not met, I suggest to use a package manager to install them. I use Fink on OS 10.4.11 with no problem, but I am still at Python 2.5 running matplotlib 1.0.svn . Pierre Le 2 nov. 09 à 20:11, R. Mitra a écrit : > Hello Again > Since I got no replies for my last question and figuring that it > must be a problem with numpy (as Eric pointed out) I now have running > python 2.6.4 and numpy 1.3.0 on OSX 10.4.11. Is there a need to > uninstall previous matplotlib and reinstall it again? This page > sounds pretty complicated for me: http://matplotlib.sourceforge.net/ > users/installing.html#build-osx? > Mitra > > ---------------------------------------------------------------------- > -------- > Come build with us! The BlackBerry(R) Developer Conference in SF, CA > is the only developer event you need to attend this year. Jumpstart > your > developing skills, take BlackBerry mobile applications to market > and stay > ahead of the curve. Join us from November 9 - 12, 2009. Register now! > http://p.sf.net/sfu/devconference > _______________________________________________ > Matplotlib-users mailing list > Mat...@li... > https://lists.sourceforge.net/lists/listinfo/matplotlib-users |
From: R. M. <rm...@uc...> - 2009-11-02 19:16:34
|
Hello Again Since I got no replies for my last question and figuring that it must be a problem with numpy (as Eric pointed out) I now have running python 2.6.4 and numpy 1.3.0 on OSX 10.4.11. Is there a need to uninstall previous matplotlib and reinstall it again? This page sounds pretty complicated for me: http://matplotlib.sourceforge.net/ users/installing.html#build-osx? There are multiple versions still existing on my computer. How can I clean up and install the latest version? Thank you all so much Mitra |
From: Eric F. <ef...@ha...> - 2009-11-02 17:56:11
|
R. Mitra wrote: > Hello All > Thanks a lot Eric. I think I have some module missing but I cannot > figure out what it is. I get the following. Any idea. This started > hapening after I reinstalled matplotlib. This is a numpy installation or version problem--notice that at the bottom of the traceback it is something from numpy that is not being found. Installation of numpy and mpl on OSX seems to be subject to all sorts of complications, and I don't use OSX, so I will have to leave this for someone else on the list. Eric > > File "/Users/Kennel/Pythoncodes/coolingmodel.py", line 6, in <module> > import matplotlib > File > "/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/site-packages/matplotlib/__init__.py", > line 129, in <module> > from rcsetup import defaultParams, validate_backend, validate_toolbar > File > "/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/site-packages/matplotlib/rcsetup.py", > line 19, in <module> > from matplotlib.colors import is_color_like > File > "/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/site-packages/matplotlib/colors.py", > line 54, in <module> > import matplotlib.cbook as cbook > File > "/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/site-packages/matplotlib/cbook.py", > line 10, in <module> > import numpy.ma as ma > File > "/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/site-packages/numpy/ma/__init__.py", > line 47, in <module> > import extras > File > "/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/site-packages/numpy/ma/extras.py", > line 45, in <module> > from numpy.lib.index_tricks import AxisConcatenator > ImportError: cannot import name AxisConcatenator > > > The first few lines of my codes are: > > from math import * > import exceptions > import pmag > import matplotlib > matplotlib.use("TkAgg") > import numpy as np > import matplotlib.pyplot as plt > > Thanks a lot for all your help. > > Mitra > > > On Nov 1, 2009, at 11:32 PM, Eric Firing wrote: > >> R. Mitra wrote: >>> Hi Eric >>> from math import * >>> import matplotlib >>> matplotlib.use("TkAgg") >>> from pylab import * >>> This is very basic I know. The above are the modules I am using. Is >>> plt a separate module? Thanks. >> >> With your method above, you don't need the "plt." part. "from pylab >> import *" is discouraged, however, except for quick interactive use. >> For scripts, it is generally considered better practice to use >> >> import matplotlib >> matplotlib.use("tkagg") >> import numpy as np >> import matplotlib.pyplot as plt >> >> pyplot is the plot-related part of the pylab namespace; or to put it >> another way, pylab is pyplot plus numpy. >> >> See http://matplotlib.sourceforge.net/faq/usage_faq.html >> >> Eric >> >> >> >>> RM >>> On Nov 1, 2009, at 7:26 PM, Eric Firing wrote: >>>> R. Mitra wrote: >>>>> Hi >>>>> I am having problems with contourf(). Suppose I have two 10X30 >>>>> arrays X,Y and a corresponding Z value array. How do I make the >>>>> upper left to be the origin? I cannot use contour >>>>> (Z,origin='upper') because the axis values gets messed up. >>>> >>>> It sounds like maybe you want to reverse the y-axis. Try something >>>> like plt.gca().invert_yaxis() after your call to contourf. >>>> >>>> Eric >> > |
From: R. M. <rm...@uc...> - 2009-11-02 17:48:25
|
Hello All Thanks a lot Eric. I think I have some module missing but I cannot figure out what it is. I get the following. Any idea. This started hapening after I reinstalled matplotlib. File "/Users/Kennel/Pythoncodes/coolingmodel.py", line 6, in <module> import matplotlib File "/Library/Frameworks/Python.framework/Versions/2.5/lib/ python2.5/site-packages/matplotlib/__init__.py", line 129, in <module> from rcsetup import defaultParams, validate_backend, validate_toolbar File "/Library/Frameworks/Python.framework/Versions/2.5/lib/ python2.5/site-packages/matplotlib/rcsetup.py", line 19, in <module> from matplotlib.colors import is_color_like File "/Library/Frameworks/Python.framework/Versions/2.5/lib/ python2.5/site-packages/matplotlib/colors.py", line 54, in <module> import matplotlib.cbook as cbook File "/Library/Frameworks/Python.framework/Versions/2.5/lib/ python2.5/site-packages/matplotlib/cbook.py", line 10, in <module> import numpy.ma as ma File "/Library/Frameworks/Python.framework/Versions/2.5/lib/ python2.5/site-packages/numpy/ma/__init__.py", line 47, in <module> import extras File "/Library/Frameworks/Python.framework/Versions/2.5/lib/ python2.5/site-packages/numpy/ma/extras.py", line 45, in <module> from numpy.lib.index_tricks import AxisConcatenator ImportError: cannot import name AxisConcatenator The first few lines of my codes are: from math import * import exceptions import pmag import matplotlib matplotlib.use("TkAgg") import numpy as np import matplotlib.pyplot as plt Thanks a lot for all your help. Mitra On Nov 1, 2009, at 11:32 PM, Eric Firing wrote: > R. Mitra wrote: >> Hi Eric >> from math import * >> import matplotlib >> matplotlib.use("TkAgg") >> from pylab import * >> This is very basic I know. The above are the modules I am using. >> Is plt a separate module? Thanks. > > With your method above, you don't need the "plt." part. "from > pylab import *" is discouraged, however, except for quick > interactive use. For scripts, it is generally considered better > practice to use > > import matplotlib > matplotlib.use("tkagg") > import numpy as np > import matplotlib.pyplot as plt > > pyplot is the plot-related part of the pylab namespace; or to put > it another way, pylab is pyplot plus numpy. > > See http://matplotlib.sourceforge.net/faq/usage_faq.html > > Eric > > > >> RM >> On Nov 1, 2009, at 7:26 PM, Eric Firing wrote: >>> R. Mitra wrote: >>>> Hi >>>> I am having problems with contourf(). Suppose I have two >>>> 10X30 arrays X,Y and a corresponding Z value array. How do I >>>> make the upper left to be the origin? I cannot use contour >>>> (Z,origin='upper') because the axis values gets messed up. >>> >>> It sounds like maybe you want to reverse the y-axis. Try >>> something like plt.gca().invert_yaxis() after your call to >>> contourf. >>> >>> Eric > |
From: Jeff W. <js...@fa...> - 2009-11-02 15:53:00
|
Stephane Raynaud wrote: > Ross, > > > one way is to mask (or remove) ocean points using the _geoslib module > provided with basemap. > When you create a Basemap instance, you can retrieve all its polygons > land (continents and islands) with "mymap.coastpolygons". > Thay are stored as numpy arrays, and you can convert them to > _geoslib.Polygon objects : > > poly = _geoslib.Polygon(N.asarray(coastalpoly).T) > > Then you loop over all Polygons and all (x,y) points and test : > > good_point = _geoslib.Point((x,y)).within(poly) > > Thanks to this method, you can choose you optimal resolution. > You can even compute the intersection of you hexagons with coastal > polygons using .intersection() and .area (instead of simply checking > if the center is inside) and then reject points depending the fraction > of the cell covered by land (or ocean). Following Stephane's excellent suggestion, here's a prototype Basemap method that checks to see if a point is on land or over water. Ross - if you find it useful I'll include it in the next release. Note that it will be slow for lots of points or large map regions. -Jeff > > On Mon, Nov 2, 2009 at 8:07 AM, <Ros...@ga... > <mailto:Ros...@ga...>> wrote: > > Listers, > > I'm using basemap to plot randomly sampled values (x,y,z) through > hexbin. This produces a very nice result. Some sample code is: > ---------- > import numpy as np > from numpy.random import seed > import matplotlib.pyplot as plt > from mpl_toolkits.basemap import Basemap > from matplotlib.mlab import griddata > > ll_lat = -38.39477 # extent of area of interest > ll_lon = 144.54767 > ur_lat = -37.51642 > ur_lon = 145.67144 > > num_points = 100 # sample points > > # create random sampling over the area of interest > seed(0) > data = np.ones((3, num_points)) > data[0,:] *= ll_lon + np.random.random((num_points))*(ur_lon-ll_lon) > data[1,:] *= ll_lat + np.random.random((num_points))*(ur_lat-ll_lat) > data[2,:] *= np.random.random((num_points))*10000 > > # plot the data > fig = plt.figure() > ax = fig.add_subplot(111) > m = Basemap(projection='cyl', llcrnrlat=ll_lat, urcrnrlat=ur_lat, > llcrnrlon=ll_lon, urcrnrlon=ur_lon, resolution='f', > suppress_ticks=False, area_thresh=0.5) > plt.hexbin(data[0,:], data[1,:], data[2,:], zorder=3) > m.fillcontinents(color=(0.8,0.8,0.8,0), zorder=1) > m.drawcoastlines(linewidth=0.25, color='k', zorder=2) > plt.show() > ---------- > > This contrived example shows a sparse set of hexagons on both land > and ocean. I would like the hexagons over the ocean to be hidden. > I can make the ones on land disappear by changing the 'zorder' > parameter of .hexbin() to 0. However I have found no way of doing > the inverse and hiding hexagons over the ocean. > > Using drawlsmask() is too crude at a 5-minute resolution. > > Any ideas? > > Thanks, > Ross > > ------------------------------------------------------------------------------ > Come build with us! The BlackBerry(R) Developer Conference in SF, CA > is the only developer event you need to attend this year. > Jumpstart your > developing skills, take BlackBerry mobile applications to market > and stay > ahead of the curve. Join us from November 9 - 12, 2009. Register now! > http://p.sf.net/sfu/devconference > _______________________________________________ > Matplotlib-users mailing list > Mat...@li... > <mailto:Mat...@li...> > https://lists.sourceforge.net/lists/listinfo/matplotlib-users > > > > > -- > Stephane Raynaud > ------------------------------------------------------------------------ > > ------------------------------------------------------------------------------ > Come build with us! The BlackBerry(R) Developer Conference in SF, CA > is the only developer event you need to attend this year. Jumpstart your > developing skills, take BlackBerry mobile applications to market and stay > ahead of the curve. Join us from November 9 - 12, 2009. Register now! > http://p.sf.net/sfu/devconference > ------------------------------------------------------------------------ > > _______________________________________________ > Matplotlib-users mailing list > Mat...@li... > https://lists.sourceforge.net/lists/listinfo/matplotlib-users > -- 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: Stephane R. <ste...@gm...> - 2009-11-02 09:00:32
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Ross, one way is to mask (or remove) ocean points using the _geoslib module provided with basemap. When you create a Basemap instance, you can retrieve all its polygons land (continents and islands) with "mymap.coastpolygons". Thay are stored as numpy arrays, and you can convert them to _geoslib.Polygon objects : poly = _geoslib.Polygon(N.asarray(coastalpoly).T) Then you loop over all Polygons and all (x,y) points and test : good_point = _geoslib.Point((x,y)).within(poly) Thanks to this method, you can choose you optimal resolution. You can even compute the intersection of you hexagons with coastal polygons using .intersection() and .area (instead of simply checking if the center is inside) and then reject points depending the fraction of the cell covered by land (or ocean). On Mon, Nov 2, 2009 at 8:07 AM, <Ros...@ga...> wrote: > Listers, > > I'm using basemap to plot randomly sampled values (x,y,z) through hexbin. > This produces a very nice result. Some sample code is: > ---------- > import numpy as np > from numpy.random import seed > import matplotlib.pyplot as plt > from mpl_toolkits.basemap import Basemap > from matplotlib.mlab import griddata > > ll_lat = -38.39477 # extent of area of interest > ll_lon = 144.54767 > ur_lat = -37.51642 > ur_lon = 145.67144 > > num_points = 100 # sample points > > # create random sampling over the area of interest > seed(0) > data = np.ones((3, num_points)) > data[0,:] *= ll_lon + np.random.random((num_points))*(ur_lon-ll_lon) > data[1,:] *= ll_lat + np.random.random((num_points))*(ur_lat-ll_lat) > data[2,:] *= np.random.random((num_points))*10000 > > # plot the data > fig = plt.figure() > ax = fig.add_subplot(111) > m = Basemap(projection='cyl', llcrnrlat=ll_lat, urcrnrlat=ur_lat, > llcrnrlon=ll_lon, urcrnrlon=ur_lon, resolution='f', > suppress_ticks=False, area_thresh=0.5) > plt.hexbin(data[0,:], data[1,:], data[2,:], zorder=3) > m.fillcontinents(color=(0.8,0.8,0.8,0), zorder=1) > m.drawcoastlines(linewidth=0.25, color='k', zorder=2) > plt.show() > ---------- > > This contrived example shows a sparse set of hexagons on both land and > ocean. I would like the hexagons over the ocean to be hidden. I can make > the ones on land disappear by changing the 'zorder' parameter of .hexbin() > to 0. However I have found no way of doing the inverse and hiding hexagons > over the ocean. > > Using drawlsmask() is too crude at a 5-minute resolution. > > Any ideas? > > Thanks, > Ross > > > ------------------------------------------------------------------------------ > Come build with us! The BlackBerry(R) Developer Conference in SF, CA > is the only developer event you need to attend this year. Jumpstart your > developing skills, take BlackBerry mobile applications to market and stay > ahead of the curve. Join us from November 9 - 12, 2009. Register now! > http://p.sf.net/sfu/devconference > _______________________________________________ > Matplotlib-users mailing list > Mat...@li... > https://lists.sourceforge.net/lists/listinfo/matplotlib-users > -- Stephane Raynaud |
From: <Ros...@ga...> - 2009-11-02 07:08:21
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Listers, I'm using basemap to plot randomly sampled values (x,y,z) through hexbin. This produces a very nice result. Some sample code is: ---------- import numpy as np from numpy.random import seed import matplotlib.pyplot as plt from mpl_toolkits.basemap import Basemap from matplotlib.mlab import griddata ll_lat = -38.39477 # extent of area of interest ll_lon = 144.54767 ur_lat = -37.51642 ur_lon = 145.67144 num_points = 100 # sample points # create random sampling over the area of interest seed(0) data = np.ones((3, num_points)) data[0,:] *= ll_lon + np.random.random((num_points))*(ur_lon-ll_lon) data[1,:] *= ll_lat + np.random.random((num_points))*(ur_lat-ll_lat) data[2,:] *= np.random.random((num_points))*10000 # plot the data fig = plt.figure() ax = fig.add_subplot(111) m = Basemap(projection='cyl', llcrnrlat=ll_lat, urcrnrlat=ur_lat, llcrnrlon=ll_lon, urcrnrlon=ur_lon, resolution='f', suppress_ticks=False, area_thresh=0.5) plt.hexbin(data[0,:], data[1,:], data[2,:], zorder=3) m.fillcontinents(color=(0.8,0.8,0.8,0), zorder=1) m.drawcoastlines(linewidth=0.25, color='k', zorder=2) plt.show() ---------- This contrived example shows a sparse set of hexagons on both land and ocean. I would like the hexagons over the ocean to be hidden. I can make the ones on land disappear by changing the 'zorder' parameter of .hexbin() to 0. However I have found no way of doing the inverse and hiding hexagons over the ocean. Using drawlsmask() is too crude at a 5-minute resolution. Any ideas? Thanks, Ross |
From: Eric F. <ef...@ha...> - 2009-11-02 03:26:56
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R. Mitra wrote: > Hi > I am having problems with contourf(). Suppose I have two 10X30 > arrays X,Y and a corresponding Z value array. How do I make the > upper left to be the origin? I cannot use contour (Z,origin='upper') > because the axis values gets messed up. It sounds like maybe you want to reverse the y-axis. Try something like plt.gca().invert_yaxis() after your call to contourf. Eric |