matplotlib-users Mailing List for matplotlib

On Wed, Jul 20, 2011 at 5:41 PM, C M <cmp...@gm...> wrote:

> On Wed, Jul 20, 2011 at 7:24 PM, Buchholz, Greg
> <gbu...@in...> wrote:
> >>-----Original Message-----
> >>From: C M [mailto:cmp...@gm...]
> >>
> >>Sorry, this is super-simple, but I'm lost in the whole
> >>locator/formatter part of the docs.
> >>
> >>How can I make a locator that just places a tick at every multiple of
> >>0.5 around the data?  So the y axis would look like:
> >>
> >>3.5 --
> >>3.0 --
> >>2.5 --
> >>2.0 --
> >>1.5 --
> >>1.0 --
> >
> > Do you want something like:
> >
> >    ylim(1.0,3.5)
> >    yticks(arrange(1.0,4.0,0.5))
>
> I'm not sure, because I can't try it out--I'm using the OO matplotlib,
> not Pyplot.  What's the equivalent of this in the OO API?
>


    ax.axis((xmin, xmax, ymin, ymax))
    ax.yaxis.set_ticks(np.arange(1.0, 4.0, 0.5))


>
> Thanks,
> Che
>
>
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-- 
Gökhan

On Wed, Jul 20, 2011 at 7:24 PM, Buchholz, Greg
<gbu...@in...> wrote:
>>-----Original Message-----
>>From: C M [mailto:cmp...@gm...]
>>
>>Sorry, this is super-simple, but I'm lost in the whole
>>locator/formatter part of the docs.
>>
>>How can I make a locator that just places a tick at every multiple of
>>0.5 around the data?  So the y axis would look like:
>>
>>3.5 --
>>3.0 --
>>2.5 --
>>2.0 --
>>1.5 --
>>1.0 --
>
> Do you want something like:
>
>    ylim(1.0,3.5)
>    yticks(arrange(1.0,4.0,0.5))

I'm not sure, because I can't try it out--I'm using the OO matplotlib,
not Pyplot.  What's the equivalent of this in the OO API?

Thanks,
Che

>-----Original Message-----
>From: C M [mailto:cmp...@gm...] 
>
>Sorry, this is super-simple, but I'm lost in the whole
>locator/formatter part of the docs.
>
>How can I make a locator that just places a tick at every multiple of
>0.5 around the data?  So the y axis would look like:
>
>3.5 --
>3.0 --
>2.5 --
>2.0 --
>1.5 --
>1.0 --

Do you want something like:

    ylim(1.0,3.5)
    yticks(arrange(1.0,4.0,0.5))

Sorry, this is super-simple, but I'm lost in the whole
locator/formatter part of the docs.

How can I make a locator that just places a tick at every multiple of
0.5 around the data?  So the y axis would look like:

3.5 --
3.0 --
2.5 --
2.0 --
1.5 --
1.0 --

etc.

Thanks,
Che

hi there,

I would like to draw a a set of lines on top of an image.
Somehow I do not get the result I want

these are the points ((267, 140), (380, 773), (267, 958))

one of my divers atempts is:

pic = plt.imread('../hlwd/effizienz_balken_01.jpg')
pic = np.fliplr(np.rot90(pic, k=2))
plt.imshow(pic)

frame1 = plt.gca()

lx = []
ly = []
for pt in ((267, 140), (380, 773), (267, 958)):
     lx.append(pt[0])
     ly.append(pt[1])
x,y = np.array([lx, ly])
line = mlines.Line2D(x, y, lw=5., alpha=0.4)

frame1.add_line(line)

plt.show()

which produces on line instad of two.

thanks for any pointers
robert

On Wed, Jul 20, 2011 at 12:15 PM, Pete Shepard <pet...@gm...>wrote:

> Hello List,
>
> I am trying to use the pylab.contourf(X,Y,Z,100) function but I would like
> to mplot a 1 dimensional heatmap instead of a 2 dimensional heatmap, Perhaps
> "contourf" is not the solution but I do like these plots.
>
> Any suggestions?
>
> TIA
>
>
An example image of what you would like to see might be useful.

Ben Root

Hello List,

I am trying to use the pylab.contourf(X,Y,Z,100) function but I would like
to mplot a 1 dimensional heatmap instead of a 2 dimensional heatmap, Perhaps
"contourf" is not the solution but I do like these plots.

Any suggestions?

TIA

There could be a few ways. What I recommend as a matter of fact is to
use other method to draw gridlines. On the other hand, given that you
have a working example, it could be better to have different axis to
draw ticklabels in locations where you want. Here is a diff.

Regards,

-JJ

*** qqwee2.py	2011-07-20 22:02:37.973960916 +0900
--- qqwee.py	2011-07-20 22:04:46.063960883 +0900
***************
*** 36,41 ****
--- 36,60 ----
     ax1 = floating_axes.FloatingSubplot(fig, rect, grid_helper=grid_helper)
     fig.add_subplot(ax1)

+    grid_locator1 = FixedLocator([j-0.5*pi for j in theta[::10]])
+    grid_locator2 = FixedLocator([i for i in rad[::5]])
+
+    grid_helper2 = floating_axes.GridHelperCurveLinear(tr,
+                                        extremes=(.5*pi-0.17,
-.5*pi+0.17, 120, 1),
+                                        grid_locator1=grid_locator1,
+                                        grid_locator2=grid_locator2,
+                                        tick_formatter1=None,
+                                        tick_formatter2=None,
+                                        )
+
+    ax1.axis["left2"] = grid_helper2.new_fixed_axis("left", axes=ax1)
+    ax1.axis["left2"].line.set_visible(False)
+    ax1.axis["left2"].toggle(ticks=False)
+
+    ax1.axis["bottom2"] = grid_helper2.new_fixed_axis("bottom", axes=ax1)
+    ax1.axis["bottom2"].line.set_visible(False)
+    ax1.axis["bottom2"].toggle(ticks=False)
+
     # create a parasite axes whose transData in RA, cz
     aux_ax = ax1.get_aux_axes(tr)




On Tue, Jul 12, 2011 at 2:12 AM, bhargav vaidya <coo...@gm...> wrote:
> Hello all,
>
> I need help to have customize tick labels using axis artist.
> My problem is to create a grid representation in polar co-ordinates (see fig) with r = logspace(log(1),log(120),128)
> and theta = linspace(0.17,2.97,64) # in radians
>
> I managed to to create the proper gridlines. But unfortunately I have to remove the tick labels as the way I describe the grid locator they crowd the axis.
> I would like to choose only certain points on the axis and label them as the plot would look nice.
>
>
> Regards
> Bhargav Vaidya.
>
> Here is my code modified to my need from already existing code found in the web :
>
> from matplotlib.transforms import Affine2D
> import mpl_toolkits.axisartist.floating_axes as floating_axes
>
> import numpy as np
> import  mpl_toolkits.axisartist.angle_helper as angle_helper
> from matplotlib.projections import PolarAxes
> from mpl_toolkits.axisartist.grid_finder import FixedLocator, MaxNLocator, \
>     DictFormatter
>
> def setup_axes2(fig, rect):
>
>    tr = PolarAxes.PolarTransform()
>
>    pi = np.pi
>    rad = np.logspace(np.log(1.),np.log(120.),128) # Radial
>    theta = np.linspace(0.17,2.97,64) # Theta Co-ordinates
>    angle_ticks = [(0, r"$\frac{1}{2}\pi$"),
>                   (.25*pi, r"$\frac{1}{4}\pi$"),
>                   (.5*pi, r"$0$"),
>                   (-.25*pi, r"$\frac{3}{4}\pi$"),
>                   (-0.5*pi, r"$\pi$")]
>    grid_locator1 = FixedLocator([j-0.5*pi for j in theta])
>    #grid_locator1 = FixedLocator([v for v, s in angle_ticks])
>
>
>    grid_locator2 = FixedLocator([i for i in rad])
>
>    grid_helper = floating_axes.GridHelperCurveLinear(tr,
>                                        extremes=(.5*pi-0.17, -.5*pi+0.17, 120, 1),
>                                        grid_locator1=grid_locator1,
>                                        grid_locator2=grid_locator2,
>                                        tick_formatter1=None,
>                                        tick_formatter2=None,
>                                        )
>
>    ax1 = floating_axes.FloatingSubplot(fig, rect, grid_helper=grid_helper)
>    fig.add_subplot(ax1)
>
>    # create a parasite axes whose transData in RA, cz
>    aux_ax = ax1.get_aux_axes(tr)
>
>    aux_ax.patch = ax1.patch # for aux_ax to have a clip path as in ax
>    ax1.patch.zorder=0.9 # but this has a side effect that the patch is
>                        # drawn twice, and possibly over some other
>                        # artists. So, we decrease the zorder a bit to
>                        # prevent this.
>
>    return ax1, aux_ax
> if 1:
>    import matplotlib.ticker as mpltick
>    import matplotlib.pyplot as plt
>    fig = plt.figure(1, figsize=(10, 10))
>
>
>    ax2, aux_ax2 = setup_axes2(fig, 111)
>
>
>    ax2.axis["left"].major_ticklabels.set_visible(False)
>    ax2.axis["bottom"].major_ticklabels.set_visible(False)
>
>    ax2.grid(color='k',linestyle='-',linewidth=0.5)
>
>    plt.show()
>
> Here is the eps file
>
>
>
>
>
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>

Have a closer look at the example I gave.

The currently released version of matplotlib doesn't support PySide at 
all.  So I cheated and simply drew to the generic Agg backend and then 
copied the whole figure (gcf = get current figure) into a PySide QImage 
object at the end.  The QImage can then be displayed however you want 
inside your Qt application.  I used a QGraphicsScene but there are other 
options.

If you really wanted to I guess you could use FigureCanvasAgg as an 
intermediary - but the process is fundamentally different.  You can't 
just drop that it into your PySide app as a widget like you can with 
FigureCanvasQTAgg.

As mentioned earlier, if you'd like to use the same code simply wait for 
the next release of matplotlib which will support PySide or you can get 
a copy of the source from github master today that also support PySide.

Gerald.

On 20/07/2011 3:59 PM, lionel chiron wrote:
> Hi Gerald,
>
> I found yesterday interesting informations on a forum where you 
> answered about Matplotlib and pyside .. but some details are missing 
> to make what I want.
> Few days ago I developped stuff in PyQt I 'd like to recuperate in 
> Pyside.. the central difficulty is to import Matplotlib in Pyside.
> In PyQt I was using FigureCanvasQTAgg but in Pyside I couldn't find 
> something equivalent allowing to link Mpl and pyside..
> It seems you're able to make drawings (with add.patch) but how to do 
> for inserting a figure?
>
> Thanks
>
> Best
>
> Lionel

On Tue, Jul 19, 2011 at 10:22 PM, gary ruben <gr...@bi...> wrote:

> Thanks Ben, that works nicely. Good work :) (except that inkscape is
> not nearly as good as matplotlib itself at optimising the resulting
> vector-based pdf to keep the file size down - not mpl's fault though).
> I just remembered, while trying this out, that there are two of every
> object forming the axis parts - two of every patch, grid line, tick
> line and label. It was this way before the latest changes also, but is
> there a reason, or is it a bug? It doesn't impact visually though.
>
> thanks for the great work on this,
> Gary
>
>
Glad that helped.  I haven't noticed the doubling issue, but I have my
suspicions.  It is likely an inadvertent side-effect of the current design.
I will look out for that, but I am hopeful that this will resolve itself as
I continue to refactor mplot3d.

If the problem hasn't resolved itself before the next+1 release (i.e.,
whatever will come after v1.1.0), then ping me again for a reminder.

Cheers,
Ben Root

Thanks Ben, that works nicely. Good work :) (except that inkscape is
not nearly as good as matplotlib itself at optimising the resulting
vector-based pdf to keep the file size down - not mpl's fault though).
I just remembered, while trying this out, that there are two of every
object forming the axis parts - two of every patch, grid line, tick
line and label. It was this way before the latest changes also, but is
there a reason, or is it a bug? It doesn't impact visually though.

thanks for the great work on this,
Gary

On Wed, Jul 20, 2011 at 12:41 PM, Benjamin Root <ben...@ou...> wrote:
> On Tue, Jul 19, 2011 at 9:25 PM, gary ruben <gr...@bi...> wrote:
>>
>> I haven't had a chance to look properly at the new mplot3d
>> improvements that Ben Root has been working on, but I wonder whether
>> it is easy now to set the axis properties so that the patches that
>> form the axes no longer have an alpha value of 0.5? I really want them
>> to be solid. The use case is that I often save images in a vector
>> format for editing within inkscape, do some fiddling, then re-export
>> as eps or pdf. If there are any semi-transparent objects, inkscape
>> will rasterize the whole image, so it becomes necessary to first go
>> through and manually set the alphas of all these patches to 1.0 before
>> saving.
>> A cursory look at the new code makes me hopeful that this is now
>> possible since the setting from _AXINFO has been moved to the Axis
>> constructor. Does that mean I'll be able to do something like
>> ax._axinfo['x']['color']=(0.3,0.3,0.3,1) with the new version?
>>
>> Gary
>>
>
> Gary,
>
> Glad to hear that you are kicking the tires.  To make it clear, the _axinfo
> dictionary is in the Axis3D object (of which there are 3 in a Axes3D
> object).  So, it would be something like:
>
> ax.xaxis._axinfo['color'] = (0.3, 0.3, 0.3, 1)
>
> At least, in theory.  Part of the reason why I did not want to make this
> dictionary official is because the above would not actually work as
> expected.  Although something similar for tick line colors might, for
> example.  Because of the inconsistencies and because I did not want to paint
> myself into a corner, I have made this dictionary explicitly "users beware".
>
> However, there is hope for your problem!  Use ax.xaxis.set_pane_color((0.3,
> 0.3, 0.3, 1)) instead!
>
> Let me know if you encounter any other problems.
> Ben Root
>
>

On Tue, Jul 19, 2011 at 9:25 PM, gary ruben <gr...@bi...> wrote:

> I haven't had a chance to look properly at the new mplot3d
> improvements that Ben Root has been working on, but I wonder whether
> it is easy now to set the axis properties so that the patches that
> form the axes no longer have an alpha value of 0.5? I really want them
> to be solid. The use case is that I often save images in a vector
> format for editing within inkscape, do some fiddling, then re-export
> as eps or pdf. If there are any semi-transparent objects, inkscape
> will rasterize the whole image, so it becomes necessary to first go
> through and manually set the alphas of all these patches to 1.0 before
> saving.
> A cursory look at the new code makes me hopeful that this is now
> possible since the setting from _AXINFO has been moved to the Axis
> constructor. Does that mean I'll be able to do something like
> ax._axinfo['x']['color']=(0.3,0.3,0.3,1) with the new version?
>
> Gary
>
>
Gary,

Glad to hear that you are kicking the tires.  To make it clear, the _axinfo
dictionary is in the Axis3D object (of which there are 3 in a Axes3D
object).  So, it would be something like:

ax.xaxis._axinfo['color'] = (0.3, 0.3, 0.3, 1)

At least, in theory.  Part of the reason why I did not want to make this
dictionary official is because the above would not actually work as
expected.  Although something similar for tick line colors might, for
example.  Because of the inconsistencies and because I did not want to paint
myself into a corner, I have made this dictionary explicitly "users beware".

However, there is hope for your problem!  Use ax.xaxis.set_pane_color((0.3,
0.3, 0.3, 1)) instead!

Let me know if you encounter any other problems.
Ben Root

I haven't had a chance to look properly at the new mplot3d
improvements that Ben Root has been working on, but I wonder whether
it is easy now to set the axis properties so that the patches that
form the axes no longer have an alpha value of 0.5? I really want them
to be solid. The use case is that I often save images in a vector
format for editing within inkscape, do some fiddling, then re-export
as eps or pdf. If there are any semi-transparent objects, inkscape
will rasterize the whole image, so it becomes necessary to first go
through and manually set the alphas of all these patches to 1.0 before
saving.
A cursory look at the new code makes me hopeful that this is now
possible since the setting from _AXINFO has been moved to the Axis
constructor. Does that mean I'll be able to do something like
ax._axinfo['x']['color']=(0.3,0.3,0.3,1) with the new version?

Gary