Re: [Matplotlib-users] Problem with text bounding box
On 09/03/2012 08:57 PM, Eric Firing wrote:
It looks like you can either use the _get_layout() method (which
requires that you specify the renderer), or, if you know the text object
will be among the last 50 for which _get_layout() has been called, you
can use txt1.cached[txt1.get_prop_typ()].
Is there a different way to achieve a fitted text object?
Not that I know of; but someone else may have a suggestion.
Eric
Many thanks for the quick and valuable return. I now use the _get_layout()
method and
it works like a charm :)
Just replaced:
sf1 = 6*72./txt1.cached.items()[0][1][1][0][1][0]
with:
sf1 = 6*72./txt1._get_layout('GTKAgg')[1][0][1][0]
br
Jakob
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Re: [Matplotlib-users] Problem with text bounding box
On 2012/09/03 8:33 PM, Jakob Gager wrote:
On 09/03/2012 08:57 PM, Eric Firing wrote:
It looks like you can either use the _get_layout() method (which
requires that you specify the renderer), or, if you know the text object
will be among the last 50 for which _get_layout() has been called, you
can use txt1.cached[txt1.get_prop_typ()].
Is there a different way to achieve a fitted text object?
Not that I know of; but someone else may have a suggestion.
Eric
Many thanks for the quick and valuable return. I now use the _get_layout()
method and
it works like a charm :)
Just replaced:
sf1 = 6*72./txt1.cached.items()[0][1][1][0][1][0]
with:
sf1 = 6*72./txt1._get_layout('GTKAgg')[1][0][1][0]
Jakob,
What is actually happening here is that the first two lines of
_get_layout() are equivalent to the second method I suggested (that's
where I got the idea for that method), and the renderer argument is
never being used. This is good, because the renderer argument would have
to be a renderer instance, not the string name of a backend.
There is also the caution that the leading underscore means
_get_layout() is intended for internal mpl use only, and subject to
change without notice. So, with some future version of mpl, you may need
to change your code again.
Eric
br
Jakob
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Re: [Matplotlib-users] Problem with text bounding box
On 09/04/2012 09:13 AM, Eric Firing wrote:
On 2012/09/03 8:33 PM, Jakob Gager wrote:
On 09/03/2012 08:57 PM, Eric Firing wrote:
It looks like you can either use the _get_layout() method (which
requires that you specify the renderer), or, if you know the text object
will be among the last 50 for which _get_layout() has been called, you
can use txt1.cached[txt1.get_prop_typ()].
Eric
Many thanks for the quick and valuable return. I now use the _get_layout()
method and
it works like a charm :)
Just replaced:
sf1 = 6*72./txt1.cached.items()[0][1][1][0][1][0]
with:
sf1 = 6*72./txt1._get_layout('GTKAgg')[1][0][1][0]
Jakob,
What is actually happening here is that the first two lines of
_get_layout() are equivalent to the second method I suggested (that's
where I got the idea for that method), and the renderer argument is
never being used. This is good, because the renderer argument would have
to be a renderer instance, not the string name of a backend.
There is also the caution that the leading underscore means
_get_layout() is intended for internal mpl use only, and subject to
change without notice. So, with some future version of mpl, you may need
to change your code again.
Eric
Hi Eric,
does this mean you would suggest to use txt1.cached[txt1.get_prop_tup()] to be
on the save side for
future versions? I'm aware of the leading underscore notation, however, the
txt.cached dictionary has
changed as well (without leading underscore and notice?), therefore, slight
modifications in the scrips
cannot be precluded anyway.
Regarding the renderer argument...
I first thought about passing a renderer instance to the _get_layout method,
but as I had no clue where
to get this instance from. So I simply tried with the backend name and luckly
it worked :).
Thanks again, and have a nice day!
Jakob
--
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threat landscape has changed and how IT managers can respond. Discussions
will include endpoint security, mobile security and the latest in malware
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Re: [Matplotlib-users] Problem with text bounding box
On 2012/09/03 9:36 PM, Jakob Gager wrote:
On 09/04/2012 09:13 AM, Eric Firing wrote:
On 2012/09/03 8:33 PM, Jakob Gager wrote:
On 09/03/2012 08:57 PM, Eric Firing wrote:
It looks like you can either use the _get_layout() method (which
requires that you specify the renderer), or, if you know the text object
will be among the last 50 for which _get_layout() has been called, you
can use txt1.cached[txt1.get_prop_typ()].
Eric
Many thanks for the quick and valuable return. I now use the _get_layout()
method and
it works like a charm :)
Just replaced:
sf1 = 6*72./txt1.cached.items()[0][1][1][0][1][0]
with:
sf1 = 6*72./txt1._get_layout('GTKAgg')[1][0][1][0]
Jakob,
What is actually happening here is that the first two lines of
_get_layout() are equivalent to the second method I suggested (that's
where I got the idea for that method), and the renderer argument is
never being used. This is good, because the renderer argument would have
to be a renderer instance, not the string name of a backend.
There is also the caution that the leading underscore means
_get_layout() is intended for internal mpl use only, and subject to
change without notice. So, with some future version of mpl, you may need
to change your code again.
Eric
Hi Eric,
does this mean you would suggest to use txt1.cached[txt1.get_prop_tup()] to
be on the save side for
future versions? I'm aware of the leading underscore notation, however, the
txt.cached dictionary has
changed as well (without leading underscore and notice?), therefore, slight
modifications in the scrips
cannot be precluded anyway.
You are right; although it lacks the underscore, I doubt the cached
attribute and the get_prop_tup() method were intended for anything but
internal use. There is no safe solution!
Regarding the renderer argument...
I first thought about passing a renderer instance to the _get_layout method,
but as I had no clue where
to get this instance from. So I simply tried with the backend name and luckly
it worked :).
That's the advantage of the txt1.cached[] approach; it doesn't require
you to supply a bogus but unused argument. It's probably what I would pick.
In any case, I'm glad you are back in business.
Eric
Thanks again, and have a nice day!
Jakob
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[Matplotlib-users] Problem with text bounding box
Recently, I switched from matplotlib 0.99 to 1.0.1 and unfortunately discovered that my text scaling script stopped working properly. The script scales the given text to fit into the current figure. A quick example of what I'm doing (works in 0.99, but not in 1.0.1 and 1.1.1rc): import matplotlib.pyplot as plt plt.ion() fig = plt.figure(1, (6,8), dpi=72) fig.add_axes([0,0,1,1]) txt1 = plt.text(0.5,0.5,'foobar',horizontalalignment='center',verticalalignment='center') plt.draw() # compute width of figure in points and divide by txt1 bounding box width sf1 = 6*72./txt1.cached.items()[0][1][1][0][1][0] txt1.set_size(txt1.get_size()*sf1) plt.draw() The problem is related to the fact that starting from mpl 1.0.1, the txt1.cached dictionary contains information of all text objects present rather than only of the specific object. I guess this behavior is intended, however this way it is rather complicated to derive the bounding box of the specific text object without cycling over the whole dictionary. Is there a different way to achieve a fitted text object? Thanks, br Jakob -- Live Security Virtual Conference Exclusive live event will cover all the ways today's security and threat landscape has changed and how IT managers can respond. Discussions will include endpoint security, mobile security and the latest in malware threats. http://www.accelacomm.com/jaw/sfrnl04242012/114/50122263/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] Problem with text bounding box
On 2012/09/03 1:38 AM, Jakob Gager wrote: Recently, I switched from matplotlib 0.99 to 1.0.1 and unfortunately discovered that my text scaling script stopped working properly. The script scales the given text to fit into the current figure. A quick example of what I'm doing (works in 0.99, but not in 1.0.1 and 1.1.1rc): import matplotlib.pyplot as plt plt.ion() fig = plt.figure(1, (6,8), dpi=72) fig.add_axes([0,0,1,1]) txt1 = plt.text(0.5,0.5,'foobar',horizontalalignment='center',verticalalignment='center') plt.draw() # compute width of figure in points and divide by txt1 bounding box width sf1 = 6*72./txt1.cached.items()[0][1][1][0][1][0] txt1.set_size(txt1.get_size()*sf1) plt.draw() The problem is related to the fact that starting from mpl 1.0.1, the txt1.cached dictionary contains information of all text objects present rather than only of the specific object. I guess this behavior is intended, however this way it is rather complicated to derive the bounding box of the specific text object without cycling over the whole dictionary. It looks like you can either use the _get_layout() method (which requires that you specify the renderer), or, if you know the text object will be among the last 50 for which _get_layout() has been called, you can use txt1.cached[txt1.get_prop_typ()]. Is there a different way to achieve a fitted text object? Not that I know of; but someone else may have a suggestion. Eric Thanks, br Jakob -- Live Security Virtual Conference Exclusive live event will cover all the ways today's security and threat landscape has changed and how IT managers can respond. Discussions will include endpoint security, mobile security and the latest in malware threats. http://www.accelacomm.com/jaw/sfrnl04242012/114/50122263/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users -- Live Security Virtual Conference Exclusive live event will cover all the ways today's security and threat landscape has changed and how IT managers can respond. Discussions will include endpoint security, mobile security and the latest in malware threats. http://www.accelacomm.com/jaw/sfrnl04242012/114/50122263/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
