matplotlib-users Mailing List for matplotlib

All,
Several days ago I tested the waters and asked you guys, the
community, how useful you thought a command-line front-end to
matplotlib would be. The overwhelmingly positive feedback was enough
for me to sit down and figure out how to do this well, and in a way
that would mimic existing tools to achieve such tasks. One example
being the `graph` utility, which is a part of GNU plotutils. Though
there are subtle differences between mpl_binutils and GNU plotutils
they, in my opinion, improve the user experience and reduce the
ambiguity regarding the parsing of command-line options.
I am announcing that mpl_binutils is in a state ready to be tested by
you guys. Hopefully you'll find it useful. You can check out the
source code here: https://github.com/dmcdougall/mpl_binutils
Without getting into details, I ran into some serious limitations with
argparse. At the end of the day, nothing is perfect, but some tools
are better than others. One such tool, docopt, was shown to me by Mark
Lawrence. docopt will change the way I do any python from the
command-line in the future. docopt is a light-weight command-line
parsing library written in python with no dependencies.
mpl_binutils has two dependencies: docopt and matplotlib. Most of you
should already have one of these! For the other, a simple `pip install
docopt` should work but I had no problems installing it from source
(python setup.py install) on OS X. mpl_binutils is currently a single
script (a python script), called mpl-graph. There is example usage on
the github readme if you'd like to take a look.
Currently, mpl-graph doesn't fail gracefully. It should, but I wanted
to get something working first. Command-line option validation is next
on my todo list and since there are only a handful of command line
options implemented (albeit the most useful ones, in my opinion), this
shouldn't be too big of a job.
Go forth and fork!
-- 
Damon McDougall
http://www.damon-is-a-geek.com
B2.39
Mathematics Institute
University of Warwick
Coventry
West Midlands
CV4 7AL
United Kingdom

On Sat, Oct 20, 2012 at 11:37 PM, Benjamin Root <ben...@ou...> wrote:
>
>
> On Saturday, October 20, 2012, Damon McDougall wrote:
>>
>> On Sat, Oct 20, 2012 at 10:25 PM, Steven Boada <bo...@ph...>
>> wrote:
>> > It'd be cool if we could do something like
>> >
>> > bins = [(0.0,0.05,0.1),(0.05,0.1,0.15)...]
>> >
>> > Where I have specified the left edge, center and right edge of each
>> > bin. Yeah, that'd be pretty slick.
>> >
>> > S
>> >
>> > On Sat Oct 20 16:21:41 2012, Steven Boada wrote:
>> >> Let's say I generate a bunch of random numbers from 0-1. Then, I'd
>> >> like to make a histogram of it. But here's the clincher. I'd like my
>> >> bins to overlap a bit. For example, if the first bin is from 0 - 0.1,
>> >> centered on 0.05, I'd like the next (second) bin to be centered on 0.1
>> >> and range from 0.05 - 0.15.
>> >>
>> >> So basically, I want the width of each bin to be greater than the
>> >> spacing.
>> >>
>> >> Is this something that could be done with the histogram function? I
>> >> did a couple of google searches and couldn't come up with anything
>> >> meaningful. Apparently, 'rwidth' in the hist function just makes the
>> >> displayed bars bigger or smaller.
>> >>
>> >> Any thoughts?
>> >>
>> >
>> > --
>> >
>> > Steven Boada
>> >
>> > Doctoral Student
>> > Dept of Physics and Astronomy
>> > Texas A&M University
>> > bo...@ph...
>>
>> My thoughts are that this goes against everything a histogram is set
>> out to do; attempt to provide a 'discretised' probability distribution
>> function given a set of discrete samples. Lets say a sample lies in
>> the region where two bins overlap. How do you define which bin the
>> sample lies in? Both? If both, how do you define the value of the
>> approximated probability distribution on a bin? You could just take
>> the height of the bin, but some of the bin's mass lies in each of the
>> neighbouring bins.
>>
>> If you don't want to apply mass to the neighbouring bins for a sample
>> that lies in the region where two bins overlap, you could just pick
>> one. You then have the problem of non-uniqueness. If you'd picked the
>> other bin you'd have a different probability distribution function.
>> This a bad property to have.
>>
>> If you don't want to pick a neighbouring bin to apply more mass, and
>> just increase the width of the each bin's matplotlib.patches.Patch
>> object, then that is more sensible. Except now you have the problem of
>> displaying the histogram. Which bin gets displayed over its left
>> neighbour? And its right neighbour?
>>
>> I dread to think what this would imply if you also wanted to stack
>> such histograms. A potential can of worms.
>>
>
> The closest I could think of as something reasonable is to apply a
> convolution of some sort to the discrete pdf to produce an approximation of
> a continuous PDF.
>
> Cheers!
> Ben Root
Yes. That's possible. The issue here, though, is getting the discrete
case to start with. There are multiple ways to do it depending on your
choice of bin, and the result is not independent of this choice.
-- 
Damon McDougall
http://www.damon-is-a-geek.com
B2.39
Mathematics Institute
University of Warwick
Coventry
West Midlands
CV4 7AL
United Kingdom

On Saturday, October 20, 2012, Damon McDougall wrote:
> On Sat, Oct 20, 2012 at 10:25 PM, Steven Boada <bo...@ph...<javascript:;>>
> wrote:
> > It'd be cool if we could do something like
> >
> > bins = [(0.0,0.05,0.1),(0.05,0.1,0.15)...]
> >
> > Where I have specified the left edge, center and right edge of each
> > bin. Yeah, that'd be pretty slick.
> >
> > S
> >
> > On Sat Oct 20 16:21:41 2012, Steven Boada wrote:
> >> Let's say I generate a bunch of random numbers from 0-1. Then, I'd
> >> like to make a histogram of it. But here's the clincher. I'd like my
> >> bins to overlap a bit. For example, if the first bin is from 0 - 0.1,
> >> centered on 0.05, I'd like the next (second) bin to be centered on 0.1
> >> and range from 0.05 - 0.15.
> >>
> >> So basically, I want the width of each bin to be greater than the
> >> spacing.
> >>
> >> Is this something that could be done with the histogram function? I
> >> did a couple of google searches and couldn't come up with anything
> >> meaningful. Apparently, 'rwidth' in the hist function just makes the
> >> displayed bars bigger or smaller.
> >>
> >> Any thoughts?
> >>
> >
> > --
> >
> > Steven Boada
> >
> > Doctoral Student
> > Dept of Physics and Astronomy
> > Texas A&M University
> > bo...@ph... <javascript:;>
>
> My thoughts are that this goes against everything a histogram is set
> out to do; attempt to provide a 'discretised' probability distribution
> function given a set of discrete samples. Lets say a sample lies in
> the region where two bins overlap. How do you define which bin the
> sample lies in? Both? If both, how do you define the value of the
> approximated probability distribution on a bin? You could just take
> the height of the bin, but some of the bin's mass lies in each of the
> neighbouring bins.
>
> If you don't want to apply mass to the neighbouring bins for a sample
> that lies in the region where two bins overlap, you could just pick
> one. You then have the problem of non-uniqueness. If you'd picked the
> other bin you'd have a different probability distribution function.
> This a bad property to have.
>
> If you don't want to pick a neighbouring bin to apply more mass, and
> just increase the width of the each bin's matplotlib.patches.Patch
> object, then that is more sensible. Except now you have the problem of
> displaying the histogram. Which bin gets displayed over its left
> neighbour? And its right neighbour?
>
> I dread to think what this would imply if you also wanted to stack
> such histograms. A potential can of worms.
>
>
The closest I could think of as something reasonable is to apply a
convolution of some sort to the discrete pdf to produce an approximation of
a continuous PDF.
Cheers!
Ben Root

On Sat, Oct 20, 2012 at 10:25 PM, Steven Boada <bo...@ph...> wrote:
> It'd be cool if we could do something like
>
> bins = [(0.0,0.05,0.1),(0.05,0.1,0.15)...]
>
> Where I have specified the left edge, center and right edge of each
> bin. Yeah, that'd be pretty slick.
>
> S
>
> On Sat Oct 20 16:21:41 2012, Steven Boada wrote:
>> Let's say I generate a bunch of random numbers from 0-1. Then, I'd
>> like to make a histogram of it. But here's the clincher. I'd like my
>> bins to overlap a bit. For example, if the first bin is from 0 - 0.1,
>> centered on 0.05, I'd like the next (second) bin to be centered on 0.1
>> and range from 0.05 - 0.15.
>>
>> So basically, I want the width of each bin to be greater than the
>> spacing.
>>
>> Is this something that could be done with the histogram function? I
>> did a couple of google searches and couldn't come up with anything
>> meaningful. Apparently, 'rwidth' in the hist function just makes the
>> displayed bars bigger or smaller.
>>
>> Any thoughts?
>>
>
> --
>
> Steven Boada
>
> Doctoral Student
> Dept of Physics and Astronomy
> Texas A&M University
> bo...@ph...
My thoughts are that this goes against everything a histogram is set
out to do; attempt to provide a 'discretised' probability distribution
function given a set of discrete samples. Lets say a sample lies in
the region where two bins overlap. How do you define which bin the
sample lies in? Both? If both, how do you define the value of the
approximated probability distribution on a bin? You could just take
the height of the bin, but some of the bin's mass lies in each of the
neighbouring bins.
If you don't want to apply mass to the neighbouring bins for a sample
that lies in the region where two bins overlap, you could just pick
one. You then have the problem of non-uniqueness. If you'd picked the
other bin you'd have a different probability distribution function.
This a bad property to have.
If you don't want to pick a neighbouring bin to apply more mass, and
just increase the width of the each bin's matplotlib.patches.Patch
object, then that is more sensible. Except now you have the problem of
displaying the histogram. Which bin gets displayed over its left
neighbour? And its right neighbour?
I dread to think what this would imply if you also wanted to stack
such histograms. A potential can of worms.
-- 
Damon McDougall
http://www.damon-is-a-geek.com
B2.39
Mathematics Institute
University of Warwick
Coventry
West Midlands
CV4 7AL
United Kingdom

It'd be cool if we could do something like
bins = [(0.0,0.05,0.1),(0.05,0.1,0.15)...]
Where I have specified the left edge, center and right edge of each 
bin. Yeah, that'd be pretty slick.
S
On Sat Oct 20 16:21:41 2012, Steven Boada wrote:
> Let's say I generate a bunch of random numbers from 0-1. Then, I'd
> like to make a histogram of it. But here's the clincher. I'd like my
> bins to overlap a bit. For example, if the first bin is from 0 - 0.1,
> centered on 0.05, I'd like the next (second) bin to be centered on 0.1
> and range from 0.05 - 0.15.
>
> So basically, I want the width of each bin to be greater than the
> spacing.
>
> Is this something that could be done with the histogram function? I
> did a couple of google searches and couldn't come up with anything
> meaningful. Apparently, 'rwidth' in the hist function just makes the
> displayed bars bigger or smaller.
>
> Any thoughts?
>
--
Steven Boada
Doctoral Student
Dept of Physics and Astronomy
Texas A&M University
bo...@ph...

Let's say I generate a bunch of random numbers from 0-1. Then, I'd like 
to make a histogram of it. But here's the clincher. I'd like my bins to 
overlap a bit. For example, if the first bin is from 0 - 0.1, centered 
on 0.05, I'd like the next (second) bin to be centered on 0.1 and range 
from 0.05 - 0.15.
So basically, I want the width of each bin to be greater than the spacing.
Is this something that could be done with the histogram function? I did 
a couple of google searches and couldn't come up with anything 
meaningful. Apparently, 'rwidth' in the hist function just makes the 
displayed bars bigger or smaller.
Any thoughts?
-- 
Steven Boada
Doctoral Student
Dept of Physics and Astronomy
Texas A&M University
bo...@ph...