Try this.
def get_indx(irow, icol):
return irow*4+icol
ax = plt.subplot(gs[get_indx(0,3):get_indx(3,3)])
With 1d slicing, the axes will occupy the rectangle defined by the
start and stop location.
For example,
gs[i:j]
will occupy the rectangular area between
gs[i] and gs[j-1].
Let me know if this does not work (this is only tested w/ the svn
version and may not work with v1.0).
Regards,
-JJ
On Tue, Oct 26, 2010 at 12:30 AM, Nikolaus Rath <nikol...@rath.org> wrote:
> On 10/25/2010 11:18 AM, Jae-Joon Lee wrote:
>> On Mon, Oct 25, 2010 at 10:45 PM, Nikolaus Rath <nikol...@rath.org> wrote:
>>> So I have to instantiate GridSpec with a (rows, column), but when I
>>> index the grid I have to use (column, row).
>>>
>>> Is there any reason for this counterintuitive behaviour?
>>>
>>
>> This is not an intended behavior but a bug which affects a grid of
>> non-square shape.
>> This has been fixed in the svn version.
>>
>> Meanwhile, you may use 1-d indexing. e.g.,
>>
>> import matplotlib.pyplot as plt
>> from matplotlib.gridspec import GridSpec
>>
>> gs = GridSpec(3, 4)
>>
>> for irow in range(3):
>> for icol in range(4):
>> ax = plt.subplot(gs[irow*4+icol])
>
> I see, thanks. Is there also a way to use this workaround for slices? I
> want a subplot in column 4 that spans all rows...
>
>
> Best,
>
> -Nikolaus
>
> --
> »Time flies like an arrow, fruit flies like a Banana.«
>
> PGP fingerprint: 5B93 61F8 4EA2 E279 ABF6 02CF A9AD B7F8 AE4E 425C
>
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