Hello,
I just noticed that in order to get a 3 row by 4 column grid, I have to
do
import matplotlib.pyplot as plt
from matplotlib.gridspec import GridSpec
def make_ticklabels_invisible(fig):
for i, ax in enumerate(fig.axes):
ax.text(0.5, 0.5, "ax%d" % (i+1), va="center", ha="center")
for tl in ax.get_xticklabels() + ax.get_yticklabels():
tl.set_visible(False)
plt.figure()
gs = GridSpec(3, 4)
for i in range(4):
for j in range(3):
plt.subplot(gs[i, j])
plt.suptitle("GridSpec")
make_ticklabels_invisible(plt.gcf())
plt.show()
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?
Best,
-Nikolaus
···
--
»Time flies like an arrow, fruit flies like a Banana.«
PGP fingerprint: 5B93 61F8 4EA2 E279 ABF6 02CF A9AD B7F8 AE4E 425C
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])
Regards,
-JJ
···
On Mon, Oct 25, 2010 at 10:45 PM, Nikolaus Rath <Nikolaus@...3072...> 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?
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
···
On 10/25/2010 11:18 AM, Jae-Joon Lee wrote:
On Mon, Oct 25, 2010 at 10:45 PM, Nikolaus Rath <Nikolaus@...3072...> 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])
--
»Time flies like an arrow, fruit flies like a Banana.«
PGP fingerprint: 5B93 61F8 4EA2 E279 ABF6 02CF A9AD B7F8 AE4E 425C
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 <Nikolaus@...3072...> wrote:
On 10/25/2010 11:18 AM, Jae-Joon Lee wrote:
On Mon, Oct 25, 2010 at 10:45 PM, Nikolaus Rath <Nikolaus@...3072...> 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