Learn how easy it is to sync an existing GitHub or Google Code repo to a SourceForge project! See Demo
Close
From: Bryan Fodness <bryan.fodness@gm...>  20071125 22:11:45

I am wondering if there is a way to view my data with respect to the physical size of what my array element is suppose to be. I have an array that is 60 x 4000 where, the first row has a height of 1.4 the next nine has a height of 1 the next forty has a height of 0.5 the next nine has a height of 1 and the last one has a height of 1.4 When viewing this with contourf or pcolor, the image is more narrow than it should be. Is there an easy way to view this properly. Bryan =20 "The game of science can accurately be described as a neverending insult to human intelligence."=09 Jo=E3o Magueijo 
From: Michael Droettboom <mdroe@st...>  20071126 12:52:52

You can provide mesh coordinates to the pcolor command: X and Y, if given, specify the (x,y) coordinates of the colored quadrilaterals; the quadrilateral for C[i,j] has corners at (X[i,j],Y[i,j]), (X[i,j+1],Y[i,j+1]), (X[i+1,j],Y[i+1,j]), (X[i+1,j+1],Y[i+1,j+1]). Ideally the dimensions of X and Y should be one greater than those of C; if the dimensions are the same, then the last row and column of C will be ignored. Actually generating the mesh is up to you (wink), but hopefully that allows for what you need to do. Cheers, Mike Bryan Fodness wrote: > I am wondering if there is a way to view my data with respect to the > physical size of what my array element is suppose to be. > > I have an array that is 60 x 4000 where, > the first row has a height of 1.4 > the next nine has a height of 1 > the next forty has a height of 0.5 > the next nine has a height of 1 > and the last one has a height of 1.4 > > When viewing this with contourf or pcolor, the image is more narrow > than it should be. Is there an easy way to view this properly. > > Bryan >  Michael Droettboom Science Software Branch Operations and Engineering Division Space Telescope Science Institute Operated by AURA for NASA 
From: Bryan Fodness <bryan.fodness@gm...>  20071126 23:21:51

Could someone give me an idea how to get started with this so it coincides with my array of values. I took a look at the quadmesh_demo in the examples and do not see a straightforward way to do this On Nov 26, 2007 7:52 AM, Michael Droettboom <mdroe@...> wrote: > You can provide mesh coordinates to the pcolor command: > > X and Y, if given, specify the (x,y) coordinates of the colored > quadrilaterals; the quadrilateral for C[i,j] has corners at > (X[i,j],Y[i,j]), (X[i,j+1],Y[i,j+1]), (X[i+1,j],Y[i+1,j]), > (X[i+1,j+1],Y[i+1,j+1]). Ideally the dimensions of X and Y > should be one greater than those of C; if the dimensions are the > same, then the last row and column of C will be ignored. > > Actually generating the mesh is up to you (wink), but hopefully that > allows for what you need to do. > > Cheers, > Mike > > > Bryan Fodness wrote: > > I am wondering if there is a way to view my data with respect to the > > physical size of what my array element is suppose to be. > > > > I have an array that is 60 x 4000 where, > > the first row has a height of 1.4 > > the next nine has a height of 1 > > the next forty has a height of 0.5 > > the next nine has a height of 1 > > and the last one has a height of 1.4 > > > > When viewing this with contourf or pcolor, the image is more narrow > > than it should be. Is there an easy way to view this properly. > > > > Bryan > > > >  > Michael Droettboom > Science Software Branch > Operations and Engineering Division > Space Telescope Science Institute > Operated by AURA for NASA > =20 "The game of science can accurately be described as a neverending insult to human intelligence."=09 Jo=E3o Magueijo 
From: Eric Firing <efiring@ha...>  20071127 00:02:51

Bryan Fodness wrote: > Could someone give me an idea how to get started with this so it > coincides with my array of values. I took a look at the quadmesh_demo > in the examples and do not see a straightforward way to do this Maybe the docstrings make it sound more complicated than it really is. In your case you have an array of rectangles, not general quadrilaterals. All you need are two 1D arrays, one each for the x and y grid boundaries. Something like this: Z = numpy.random.rand(60,4000) X = numpy.arange(4001) Y = numpy.array([0, 1.4] + [1.0]*9 + [0.5]*40 + [1]*9 + [1.4]).cumsum() pcolor(X, Y, Z) pcolormesh should work the same, but when I try it now with svn it doesn't; I don't know what is going on with it. If you are using a release version of mpl, I expect it will work. Eric > > On Nov 26, 2007 7:52 AM, Michael Droettboom <mdroe@...> wrote: >> You can provide mesh coordinates to the pcolor command: >> >> X and Y, if given, specify the (x,y) coordinates of the colored >> quadrilaterals; the quadrilateral for C[i,j] has corners at >> (X[i,j],Y[i,j]), (X[i,j+1],Y[i,j+1]), (X[i+1,j],Y[i+1,j]), >> (X[i+1,j+1],Y[i+1,j+1]). Ideally the dimensions of X and Y >> should be one greater than those of C; if the dimensions are the >> same, then the last row and column of C will be ignored. >> >> Actually generating the mesh is up to you (wink), but hopefully that >> allows for what you need to do. >> >> Cheers, >> Mike >> >> >> Bryan Fodness wrote: >>> I am wondering if there is a way to view my data with respect to the >>> physical size of what my array element is suppose to be. >>> >>> I have an array that is 60 x 4000 where, >>> the first row has a height of 1.4 >>> the next nine has a height of 1 >>> the next forty has a height of 0.5 >>> the next nine has a height of 1 >>> and the last one has a height of 1.4 >>> >>> When viewing this with contourf or pcolor, the image is more narrow >>> than it should be. Is there an easy way to view this properly. >>> >>> Bryan >>> >>  >> Michael Droettboom >> Science Software Branch >> Operations and Engineering Division >> Space Telescope Science Institute >> Operated by AURA for NASA >> > > > 
From: Bryan Fodness <bryan.fodness@gm...>  20071127 00:53:20

Thank You! It works great. On Nov 26, 2007 7:02 PM, Eric Firing <efiring@...> wrote: > Bryan Fodness wrote: > > Could someone give me an idea how to get started with this so it > > coincides with my array of values. I took a look at the quadmesh_demo > > in the examples and do not see a straightforward way to do this > > Maybe the docstrings make it sound more complicated than it really is. > In your case you have an array of rectangles, not general > quadrilaterals. All you need are two 1D arrays, one each for the x and > y grid boundaries. Something like this: > > Z =3D numpy.random.rand(60,4000) > X =3D numpy.arange(4001) > Y =3D numpy.array([0, 1.4] + [1.0]*9 + [0.5]*40 + [1]*9 + [1.4]).cumsum() > pcolor(X, Y, Z) > > pcolormesh should work the same, but when I try it now with svn it > doesn't; I don't know what is going on with it. If you are using a > release version of mpl, I expect it will work. > > Eric > > > > > > On Nov 26, 2007 7:52 AM, Michael Droettboom <mdroe@...> wrote: > >> You can provide mesh coordinates to the pcolor command: > >> > >> X and Y, if given, specify the (x,y) coordinates of the colored > >> quadrilaterals; the quadrilateral for C[i,j] has corners at > >> (X[i,j],Y[i,j]), (X[i,j+1],Y[i,j+1]), (X[i+1,j],Y[i+1,j]), > >> (X[i+1,j+1],Y[i+1,j+1]). Ideally the dimensions of X and Y > >> should be one greater than those of C; if the dimensions are the > >> same, then the last row and column of C will be ignored. > >> > >> Actually generating the mesh is up to you (wink), but hopefully that > >> allows for what you need to do. > >> > >> Cheers, > >> Mike > >> > >> > >> Bryan Fodness wrote: > >>> I am wondering if there is a way to view my data with respect to the > >>> physical size of what my array element is suppose to be. > >>> > >>> I have an array that is 60 x 4000 where, > >>> the first row has a height of 1.4 > >>> the next nine has a height of 1 > >>> the next forty has a height of 0.5 > >>> the next nine has a height of 1 > >>> and the last one has a height of 1.4 > >>> > >>> When viewing this with contourf or pcolor, the image is more narrow > >>> than it should be. Is there an easy way to view this properly. > >>> > >>> Bryan > >>> > >>  > >> Michael Droettboom > >> Science Software Branch > >> Operations and Engineering Division > >> Space Telescope Science Institute > >> Operated by AURA for NASA > >> > > > > > > > > =20 "The game of science can accurately be described as a neverending insult to human intelligence."=09 Jo=E3o Magueijo 
From: Bryan Fodness <bryan.fodness@gm...>  20071215 01:06:30
Attachments:
Message as HTML

I would also like to get the area of the mesh element when I fill the corresponding array element. if a[1,0] area =3D 1.0 * 0.01 if a[30,0] area =3D 0.5 * 0.01 Is this possible? On Nov 26, 2007 7:02 PM, Eric Firing <efiring@...> wrote: > Bryan Fodness wrote: > > Could someone give me an idea how to get started with this so it > > coincides with my array of values. I took a look at the quadmesh_demo > > in the examples and do not see a straightforward way to do this > > Maybe the docstrings make it sound more complicated than it really is. > In your case you have an array of rectangles, not general > quadrilaterals. All you need are two 1D arrays, one each for the x and > y grid boundaries. Something like this: > > Z =3D numpy.random.rand(60,4000) > X =3D numpy.arange(4001) > Y =3D numpy.array([0, 1.4] + [1.0]*9 + [0.5]*40 + [1]*9 + [1.4]).cumsum() > pcolor(X, Y, Z) > > pcolormesh should work the same, but when I try it now with svn it > doesn't; I don't know what is going on with it. If you are using a > release version of mpl, I expect it will work. > > Eric > > > > > On Nov 26, 2007 7:52 AM, Michael Droettboom <mdroe@...> wrote: > >> You can provide mesh coordinates to the pcolor command: > >> > >> X and Y, if given, specify the (x,y) coordinates of the colored > >> quadrilaterals; the quadrilateral for C[i,j] has corners at > >> (X[i,j],Y[i,j]), (X[i,j+1],Y[i,j+1]), (X[i+1,j],Y[i+1,j]), > >> (X[i+1,j+1],Y[i+1,j+1]). Ideally the dimensions of X and Y > >> should be one greater than those of C; if the dimensions are the > >> same, then the last row and column of C will be ignored. > >> > >> Actually generating the mesh is up to you (wink), but hopefully that > >> allows for what you need to do. > >> > >> Cheers, > >> Mike > >> > >> > >> Bryan Fodness wrote: > >>> I am wondering if there is a way to view my data with respect to the > >>> physical size of what my array element is suppose to be. > >>> > >>> I have an array that is 60 x 4000 where, > >>> the first row has a height of 1.4 > >>> the next nine has a height of 1 > >>> the next forty has a height of 0.5 > >>> the next nine has a height of 1 > >>> and the last one has a height of 1.4 > >>> > >>> When viewing this with contourf or pcolor, the image is more narrow > >>> than it should be. Is there an easy way to view this properly. > >>> > >>> Bryan > >>> > >>  > >> Michael Droettboom > >> Science Software Branch > >> Operations and Engineering Division > >> Space Telescope Science Institute > >> Operated by AURA for NASA > >> > > > > > > > > =20 "The game of science can accurately be described as a neverending insult t= o human intelligence."  Jo=E3o Magueijo 
From: Eric Firing <efiring@ha...>  20071215 02:09:46

Bryan Fodness wrote: > I would also like to get the area of the mesh element when I fill the > corresponding array element. > > if a[1,0] > area = 1.0 * 0.01 > > if a[30,0] > area = 0.5 * 0.01 > > Is this possible? I'm sorry, but I don't understand what you are asking. Are you asking how to calculate an array of areas corresponding to the grid? You know what the deltaY values are: dy = numpy.array([1.4] + [1.0]*9 + [0.5]*40 + [1]*9 + [1.4]) Are you looking for area = dy * 0.01 ? Eric > On Nov 26, 2007 7:02 PM, Eric Firing <efiring@... > <mailto:efiring@...>> wrote: > > Bryan Fodness wrote: > > Could someone give me an idea how to get started with this so it > > coincides with my array of values. I took a look at the > quadmesh_demo > > in the examples and do not see a straightforward way to do this > > Maybe the docstrings make it sound more complicated than it really is. > In your case you have an array of rectangles, not general > quadrilaterals. All you need are two 1D arrays, one each for the x > and > y grid boundaries. Something like this: > > Z = numpy.random.rand(60,4000) > X = numpy.arange(4001) > Y = numpy.array([0, 1.4] + [1.0]*9 + [0.5]*40 + [1]*9 + [1.4]).cumsum() > pcolor(X, Y, Z) > > pcolormesh should work the same, but when I try it now with svn it > doesn't; I don't know what is going on with it. If you are using a > release version of mpl, I expect it will work. > > Eric > > > > > On Nov 26, 2007 7:52 AM, Michael Droettboom <mdroe@... > <mailto:mdroe@...>> wrote: > >> You can provide mesh coordinates to the pcolor command: > >> > >> X and Y, if given, specify the (x,y) coordinates of the colored > >> quadrilaterals; the quadrilateral for C[i,j] has corners at > >> (X[i,j],Y[i,j]), (X[i,j+1],Y[i,j+1]), (X[i+1,j],Y[i+1,j]), > >> (X[i+1,j+1],Y[i+1,j+1]). Ideally the dimensions of X and Y > >> should be one greater than those of C; if the dimensions are the > >> same, then the last row and column of C will be ignored. > >> > >> Actually generating the mesh is up to you (wink), but hopefully that > >> allows for what you need to do. > >> > >> Cheers, > >> Mike > >> > >> > >> Bryan Fodness wrote: > >>> I am wondering if there is a way to view my data with respect > to the > >>> physical size of what my array element is suppose to be. > >>> > >>> I have an array that is 60 x 4000 where, > >>> the first row has a height of 1.4 > >>> the next nine has a height of 1 > >>> the next forty has a height of 0.5 > >>> the next nine has a height of 1 > >>> and the last one has a height of 1.4 > >>> > >>> When viewing this with contourf or pcolor, the image is more narrow > >>> than it should be. Is there an easy way to view this properly. > >>> > >>> Bryan > >>> > >>  > >> Michael Droettboom > >> Science Software Branch > >> Operations and Engineering Division > >> Space Telescope Science Institute > >> Operated by AURA for NASA > >> > > > > > > > > > > >  > "The game of science can accurately be described as a neverending > insult to human intelligence."  João Magueijo 
From: Bryan Fodness <bryan.fodness@gm...>  20071215 04:13:28
Attachments:
Message as HTML
Figure.png
