matplotlib-users Mailing List for matplotlib

Sadly no :( I want the day of the month that I'm processing *OR* the 
last day. The worst case for this is obviously the 31st of each month. 
 The rrule code I've given provides exactly that. When transferred to 
mpl that doesn't work.
On 04/04/2013 17:31, Phil Elson wrote:
> Hi Mark,
>
> Thanks for persevering :-)
>
> What is it you want to achieve? Is it that you just want the last day of
> each month as the located value?
>
> Changing your locator to:
>
> ax.xaxis.set_major_locator(MonthLocator(bymonthday = -1))
>
> Seems to do the trick for me (I've never looked at the mpl date magic,
> so I can give no guarantees).
>
> HTH,
>
>
> On 4 April 2013 17:18, Mark Lawrence
> <bre...@ya...
> <mailto:bre...@ya...>> wrote:
>
> On 01/04/2013 14:48, Mark Lawrence wrote:
> > On 29/03/2013 15:49, Mark Lawrence wrote:
> >> Hi all,
> >>
> >> From http://labix.org/python-dateutil
> >>
> >> "To generate a rrule for the use case of "a date on the
> specified day of
> >> the month, unless it is beyond the end of month, in which case
> it will
> >> be the last day of the month" use the following:
> >>
> >> rrule(MONTHLY, bymonthday=(some_day, -1), bysetpos=1)
> >>
> >> This will generate a value for every calendar month regardless
> of the
> >> day of the month it is started from."
> >>
> >> Using bymonthday with MonthLocator gives ticks on the day given
> and the
> >> last day of the month, which looks extremely ugly. Code below
> demonstrates.
> >>
> >> from dateutil.rrule import *
> >> import datetime
> >> import matplotlib.pyplot as plt
> >> from matplotlib.ticker import FormatStrFormatter, MultipleLocator
> >> from matplotlib.dates import DateFormatter, MonthLocator, DayLocator
> >>
> >> start = datetime.date(2013, 3, 29)
> >> until = datetime.date(2014, 3, 29)
> >> dates = rrule(MONTHLY, bymonthday=(29, -1), bysetpos=1, until=until)
> >> for d in dates:print(d)
> >>
> >> dates = [start, until]
> >> values = [0, 1]
> >> plt.ylabel('Balance')
> >> plt.grid()
> >> ax = plt.subplot(111)
> >> plt.plot_date(dates, values, fmt = 'rx-')
> >> ax.xaxis.set_major_locator(MonthLocator(bymonthday =
> (dates[0].day, -1)))
> >> ax.xaxis.set_major_formatter(DateFormatter('%d/%m/%y'))
> >> ax.yaxis.set_major_formatter(FormatStrFormatter('£%0.2f'))
> >> ax.yaxis.set_minor_locator(MultipleLocator(5))
> >> plt.axis(xmin=dates[0], xmax=dates[-1])
> >> plt.setp(plt.gca().get_xticklabels(), rotation = 45, fontsize = 10)
> >> plt.setp(plt.gca().get_yticklabels(), fontsize = 10)
> >> plt.show()
> >>
> >
> > Seems an apt date to realise that I didn't say much :(
> >
> > Assuming that I'm correct would you like an issue raised on the bug
> > tracker? If not please correct the mistake I've made, presumably in
> > reading the docs, which I think are excellent by the way.
> >
>
> Anybody?
>
> --
> If you're using GoogleCrapTM please read this
> http://wiki.python.org/moin/GoogleGroupsPython.
>
-- 
If you're using GoogleCrapTM please read this 
http://wiki.python.org/moin/GoogleGroupsPython.
Mark Lawrence

I've been seeing some discrepancy between display and saved figures. See
attached. One is the result of the save, the other is the screen capture
of the display. I'm using 1.2.0 on a mac.
On Thu, Apr 4, 2013 at 9:46 AM, Derek Thomas <der...@gm...> wrote:
> I get a different result.
>
>
> On Thu, Apr 4, 2013 at 9:25 AM, Phil Elson <pel...@gm...> wrote:
>
>> Hi Derek,
>>
>> What are we looking at here?
>>
>> The following code:
>>
>> import matplotlib.pyplot as plt
>> import matplotlib.transforms as mtrans
>> import numpy as np
>>
>> plt.figure()
>> ax = plt.subplot(111)
>> base_trans = ax.transData
>> mtx = np.array([[1,1,0],
>> [0,1,0],
>> [0,0,1]])
>> tr = mtrans.Affine2D(matrix=mtx) + base_trans
>>
>> plt.plot([1,2,3], [1,2,3], 'gray', transform=tr)
>> plt.scatter([1,2,3], [1,2,3], c='k', marker='D', transform=tr)
>> plt.show()
>>
>> produces the following plot on v1.2.0:
>> [image: Inline images 1]
>> Is this unexpected or are you getting a different result to me?
>>
>> Regards,
>>
>>
>>
>> On 4 April 2013 17:06, Derek Thomas <der...@gm...> wrote:
>>
>>> I posted a related question on stackoverflow
>>> (
>>> http://stackoverflow.com/questions/15815862/apply-affine-transform-to-quiver-in-python-matplotlib
>>> )
>>> but I've produced a simple enough example with strange results that I
>>> think it merits attention here. I'm trying to apply affine transforms
>>> to quiver and scatter plots. In all cases that I've considered, the
>>> scatter and quiver plots transform opposite the regular plot. Here's
>>> a minimal case:
>>>
>>> import matplotlib as mpl
>>> from pylab import figure, subplot, plot, scatter, show, axis
>>>
>>> figure()
>>> ax = subplot(111)
>>> base_trans = ax.transData
>>> tr = mpl.transforms.Affine2D(matrix =
>>> array([[1,1,0],[0,1,0],[0,0,1]])) + base_trans
>>>
>>>
>>> plot( [1,2,3], [1,2,3], 'k.', transform = tr )
>>> scatter( [1,2,3], [1,2,3], c = 'k', marker = 'D', transform = tr )
>>> axis([0,7,0,7])
>>> show()
>>>
>>> Thanks,
>>>
>>> Derek
>>>
>>>
>>> ------------------------------------------------------------------------------
>>> Minimize network downtime and maximize team effectiveness.
>>> Reduce network management and security costs.Learn how to hire
>>> the most talented Cisco Certified professionals. Visit the
>>> Employer Resources Portal
>>> http://www.cisco.com/web/learning/employer_resources/index.html
>>> _______________________________________________
>>> Matplotlib-users mailing list
>>> Mat...@li...
>>> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>>>
>>
>>
>

I get a different result.
On Thu, Apr 4, 2013 at 9:25 AM, Phil Elson <pel...@gm...> wrote:
> Hi Derek,
>
> What are we looking at here?
>
> The following code:
>
> import matplotlib.pyplot as plt
> import matplotlib.transforms as mtrans
> import numpy as np
>
> plt.figure()
> ax = plt.subplot(111)
> base_trans = ax.transData
> mtx = np.array([[1,1,0],
> [0,1,0],
> [0,0,1]])
> tr = mtrans.Affine2D(matrix=mtx) + base_trans
>
> plt.plot([1,2,3], [1,2,3], 'gray', transform=tr)
> plt.scatter([1,2,3], [1,2,3], c='k', marker='D', transform=tr)
> plt.show()
>
> produces the following plot on v1.2.0:
> [image: Inline images 1]
> Is this unexpected or are you getting a different result to me?
>
> Regards,
>
>
>
> On 4 April 2013 17:06, Derek Thomas <der...@gm...> wrote:
>
>> I posted a related question on stackoverflow
>> (
>> http://stackoverflow.com/questions/15815862/apply-affine-transform-to-quiver-in-python-matplotlib
>> )
>> but I've produced a simple enough example with strange results that I
>> think it merits attention here. I'm trying to apply affine transforms
>> to quiver and scatter plots. In all cases that I've considered, the
>> scatter and quiver plots transform opposite the regular plot. Here's
>> a minimal case:
>>
>> import matplotlib as mpl
>> from pylab import figure, subplot, plot, scatter, show, axis
>>
>> figure()
>> ax = subplot(111)
>> base_trans = ax.transData
>> tr = mpl.transforms.Affine2D(matrix =
>> array([[1,1,0],[0,1,0],[0,0,1]])) + base_trans
>>
>>
>> plot( [1,2,3], [1,2,3], 'k.', transform = tr )
>> scatter( [1,2,3], [1,2,3], c = 'k', marker = 'D', transform = tr )
>> axis([0,7,0,7])
>> show()
>>
>> Thanks,
>>
>> Derek
>>
>>
>> ------------------------------------------------------------------------------
>> Minimize network downtime and maximize team effectiveness.
>> Reduce network management and security costs.Learn how to hire
>> the most talented Cisco Certified professionals. Visit the
>> Employer Resources Portal
>> http://www.cisco.com/web/learning/employer_resources/index.html
>> _______________________________________________
>> Matplotlib-users mailing list
>> Mat...@li...
>> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>>
>
>

Hi Mark,
Thanks for persevering :-)
What is it you want to achieve? Is it that you just want the last day of
each month as the located value?
Changing your locator to:
ax.xaxis.set_major_locator(MonthLocator(bymonthday = -1))
Seems to do the trick for me (I've never looked at the mpl date magic, so I
can give no guarantees).
HTH,
On 4 April 2013 17:18, Mark Lawrence <bre...@ya...> wrote:
> On 01/04/2013 14:48, Mark Lawrence wrote:
> > On 29/03/2013 15:49, Mark Lawrence wrote:
> >> Hi all,
> >>
> >> From http://labix.org/python-dateutil
> >>
> >> "To generate a rrule for the use case of "a date on the specified day of
> >> the month, unless it is beyond the end of month, in which case it will
> >> be the last day of the month" use the following:
> >>
> >> rrule(MONTHLY, bymonthday=(some_day, -1), bysetpos=1)
> >>
> >> This will generate a value for every calendar month regardless of the
> >> day of the month it is started from."
> >>
> >> Using bymonthday with MonthLocator gives ticks on the day given and the
> >> last day of the month, which looks extremely ugly. Code below
> demonstrates.
> >>
> >> from dateutil.rrule import *
> >> import datetime
> >> import matplotlib.pyplot as plt
> >> from matplotlib.ticker import FormatStrFormatter, MultipleLocator
> >> from matplotlib.dates import DateFormatter, MonthLocator, DayLocator
> >>
> >> start = datetime.date(2013, 3, 29)
> >> until = datetime.date(2014, 3, 29)
> >> dates = rrule(MONTHLY, bymonthday=(29, -1), bysetpos=1, until=until)
> >> for d in dates:print(d)
> >>
> >> dates = [start, until]
> >> values = [0, 1]
> >> plt.ylabel('Balance')
> >> plt.grid()
> >> ax = plt.subplot(111)
> >> plt.plot_date(dates, values, fmt = 'rx-')
> >> ax.xaxis.set_major_locator(MonthLocator(bymonthday = (dates[0].day,
> -1)))
> >> ax.xaxis.set_major_formatter(DateFormatter('%d/%m/%y'))
> >> ax.yaxis.set_major_formatter(FormatStrFormatter('£%0.2f'))
> >> ax.yaxis.set_minor_locator(MultipleLocator(5))
> >> plt.axis(xmin=dates[0], xmax=dates[-1])
> >> plt.setp(plt.gca().get_xticklabels(), rotation = 45, fontsize = 10)
> >> plt.setp(plt.gca().get_yticklabels(), fontsize = 10)
> >> plt.show()
> >>
> >
> > Seems an apt date to realise that I didn't say much :(
> >
> > Assuming that I'm correct would you like an issue raised on the bug
> > tracker? If not please correct the mistake I've made, presumably in
> > reading the docs, which I think are excellent by the way.
> >
>
> Anybody?
>
> --
> If you're using GoogleCrapTM please read this
> http://wiki.python.org/moin/GoogleGroupsPython.
>
> Mark Lawrence
>
>
>
> ------------------------------------------------------------------------------
> Minimize network downtime and maximize team effectiveness.
> Reduce network management and security costs.Learn how to hire
> the most talented Cisco Certified professionals. Visit the
> Employer Resources Portal
> http://www.cisco.com/web/learning/employer_resources/index.html
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>

Hi Derek,
What are we looking at here?
The following code:
import matplotlib.pyplot as plt
import matplotlib.transforms as mtrans
import numpy as np
plt.figure()
ax = plt.subplot(111)
base_trans = ax.transData
mtx = np.array([[1,1,0],
 [0,1,0],
 [0,0,1]])
tr = mtrans.Affine2D(matrix=mtx) + base_trans
plt.plot([1,2,3], [1,2,3], 'gray', transform=tr)
plt.scatter([1,2,3], [1,2,3], c='k', marker='D', transform=tr)
plt.show()
produces the following plot on v1.2.0:
[image: Inline images 1]
Is this unexpected or are you getting a different result to me?
Regards,
On 4 April 2013 17:06, Derek Thomas <der...@gm...> wrote:
> I posted a related question on stackoverflow
> (
> http://stackoverflow.com/questions/15815862/apply-affine-transform-to-quiver-in-python-matplotlib
> )
> but I've produced a simple enough example with strange results that I
> think it merits attention here. I'm trying to apply affine transforms
> to quiver and scatter plots. In all cases that I've considered, the
> scatter and quiver plots transform opposite the regular plot. Here's
> a minimal case:
>
> import matplotlib as mpl
> from pylab import figure, subplot, plot, scatter, show, axis
>
> figure()
> ax = subplot(111)
> base_trans = ax.transData
> tr = mpl.transforms.Affine2D(matrix =
> array([[1,1,0],[0,1,0],[0,0,1]])) + base_trans
>
>
> plot( [1,2,3], [1,2,3], 'k.', transform = tr )
> scatter( [1,2,3], [1,2,3], c = 'k', marker = 'D', transform = tr )
> axis([0,7,0,7])
> show()
>
> Thanks,
>
> Derek
>
>
> ------------------------------------------------------------------------------
> Minimize network downtime and maximize team effectiveness.
> Reduce network management and security costs.Learn how to hire
> the most talented Cisco Certified professionals. Visit the
> Employer Resources Portal
> http://www.cisco.com/web/learning/employer_resources/index.html
> _______________________________________________
> Matplotlib-users mailing list
> Mat...@li...
> https://lists.sourceforge.net/lists/listinfo/matplotlib-users
>

On 01/04/2013 14:48, Mark Lawrence wrote:
> On 29/03/2013 15:49, Mark Lawrence wrote:
>> Hi all,
>>
>> From http://labix.org/python-dateutil
>>
>> "To generate a rrule for the use case of "a date on the specified day of
>> the month, unless it is beyond the end of month, in which case it will
>> be the last day of the month" use the following:
>>
>> rrule(MONTHLY, bymonthday=(some_day, -1), bysetpos=1)
>>
>> This will generate a value for every calendar month regardless of the
>> day of the month it is started from."
>>
>> Using bymonthday with MonthLocator gives ticks on the day given and the
>> last day of the month, which looks extremely ugly. Code below demonstrates.
>>
>> from dateutil.rrule import *
>> import datetime
>> import matplotlib.pyplot as plt
>> from matplotlib.ticker import FormatStrFormatter, MultipleLocator
>> from matplotlib.dates import DateFormatter, MonthLocator, DayLocator
>>
>> start = datetime.date(2013, 3, 29)
>> until = datetime.date(2014, 3, 29)
>> dates = rrule(MONTHLY, bymonthday=(29, -1), bysetpos=1, until=until)
>> for d in dates:print(d)
>>
>> dates = [start, until]
>> values = [0, 1]
>> plt.ylabel('Balance')
>> plt.grid()
>> ax = plt.subplot(111)
>> plt.plot_date(dates, values, fmt = 'rx-')
>> ax.xaxis.set_major_locator(MonthLocator(bymonthday = (dates[0].day, -1)))
>> ax.xaxis.set_major_formatter(DateFormatter('%d/%m/%y'))
>> ax.yaxis.set_major_formatter(FormatStrFormatter('£%0.2f'))
>> ax.yaxis.set_minor_locator(MultipleLocator(5))
>> plt.axis(xmin=dates[0], xmax=dates[-1])
>> plt.setp(plt.gca().get_xticklabels(), rotation = 45, fontsize = 10)
>> plt.setp(plt.gca().get_yticklabels(), fontsize = 10)
>> plt.show()
>>
>
> Seems an apt date to realise that I didn't say much :(
>
> Assuming that I'm correct would you like an issue raised on the bug
> tracker? If not please correct the mistake I've made, presumably in
> reading the docs, which I think are excellent by the way.
>
Anybody?
-- 
If you're using GoogleCrapTM please read this 
http://wiki.python.org/moin/GoogleGroupsPython.
Mark Lawrence

I posted a related question on stackoverflow
(http://stackoverflow.com/questions/15815862/apply-affine-transform-to-quiver-in-python-matplotlib)
but I've produced a simple enough example with strange results that I
think it merits attention here. I'm trying to apply affine transforms
to quiver and scatter plots. In all cases that I've considered, the
scatter and quiver plots transform opposite the regular plot. Here's
a minimal case:
import matplotlib as mpl
from pylab import figure, subplot, plot, scatter, show, axis
figure()
ax = subplot(111)
base_trans = ax.transData
tr = mpl.transforms.Affine2D(matrix =
array([[1,1,0],[0,1,0],[0,0,1]])) + base_trans
plot( [1,2,3], [1,2,3], 'k.', transform = tr )
scatter( [1,2,3], [1,2,3], c = 'k', marker = 'D', transform = tr )
axis([0,7,0,7])
show()
Thanks,
Derek

On 4 April 2013 06:45, Sudheer Joseph <sud...@ya...> wrote:
> Below is a sample script I got from windrose pack. I would like to place 2 windroses side by side
...
>
> from windrose import WindroseAxes
> from matplotlib import pyplot as plt
...
> def new_axes():
> fig = plt.figure(figsize=(8, 8), dpi=80, facecolor='w', edgecolor='w')
> rect = [0.1, 0.1, 0.8, 0.8]
> ax = WindroseAxes(fig, rect, axisbg='w')
> fig.add_axes(ax)
> return ax
I'm not familiar with the windrose package, but it looks like the rect
parameter to WindroseAxes specifies the size of the generated axes in
figure co-ordinates (see
http://matplotlib.org/api/figure_api.html?highlight=add_axes#matplotlib.figure.Figure.add_axes).
You should be able to pass in a different list of co-ordinates for
each WindroseAxes to get side-by-side axes on the same figure...
Cheers,
Scott

Dear users,
     Below is a sample script I got from windrose pack. I would like to place 2 windroses side by side so that a comparison can be made. For example I have created additional variables ws1 wd1, and I would like that to be placed in the same row as a 1 row 2 column way. Any help in this regard will be great. (I tried subplot(221) subplot(222) but it do not work as the windrose uses new axis each time.)
from windrose import WindroseAxes
from matplotlib import pyplot as plt
import matplotlib.cm as cm
from numpy.random import random
from numpy import arange
#Create wind speed and direction variables
ws = random(500)*6
wd = random(500)*360
ws1 = random(500)*6
wd1 = random(500)*360
def new_axes():
  fig = plt.figure(figsize=(8, 8), dpi=80, facecolor='w', edgecolor='w')
  rect = [0.1, 0.1, 0.8, 0.8]
  ax = WindroseAxes(fig, rect, axisbg='w')
  fig.add_axes(ax)
  return ax
def set_legend(ax):
  l = ax.legend(axespad=-0.10)
  plt.setp(l.get_texts(), fontsize=8)
#windrose like a stacked histogram with normed (displayed in percent) results
ax = new_axes()
ax.bar(wd, ws, normed=True, opening=0.8, edgecolor='white')
set_legend(ax)
#Another stacked histogram representation, not normed, with bins limits
##print ax._info
plt.show()
 
***************************************************************
Sudheer Joseph 
Indian National Centre for Ocean Information Services
Ministry of Earth Sciences, Govt. of India
POST BOX NO: 21, IDA Jeedeemetla P.O.
Via Pragathi Nagar,Kukatpally, Hyderabad; Pin:5000 55
Tel:+91-40-23886047(O),Fax:+91-40-23895011(O),
Tel:+91-40-23044600(R),Tel:+91-40-9440832534(Mobile)
E-mail:sjo...@gm...;sud...@ya...
Web- http://oppamthadathil.tripod.com
***************************************************************