Output result of source block to a file

Output result of source block to a file

[Roger Mason]

Hello,

I want to output the result of the evaluation of a (python) source block
to a (graphics) file and have a link to the file inserted in the buffer.

Here is the code:

#+begin_src python :results value file :file scaleplot02.pdf :exports results :var data=test
  import matplotlib.pyplot as plt
  import csv

  scale = []
  lescale = []
  cdscale = []

  energy = []
  leenergy = []
  cdenergy = []

  for row in data:
      scale.append(float(row[0]))
      energy.append(float(row[1]))
      if int(row[2]) != 0:
	 lescale.append(float(row[0]))
	 leenergy.append(float(row[1]))
      if int(row[3]) != 0:
	 cdscale.append(float(row[0]))
	 cdenergy.append(float(row[1]))        

  plt.plot(scale,energy, 'r+')
  plt.plot(lescale,leenergy, 'go')
  plt.plot(cdscale,cdenergy, 'bo')
  plt.xlabel('scale')
  plt.ylabel('energy (Ha)')
  plt.title('Energy vs scale')
  plt.legend()
  plt.savefig(file)
  print ("[[./%s]]" % file)
#+end_src

This is the error:

NameError: name 'file' is not defined

Obviously the syntax on the begin_src line is wrong, but what should it
be instead?

Thanks,
Roger

Re: Output result of source block to a file

[stardiviner]

Roger Mason <rmason@mun.ca> writes:

> Hello,
>
> I want to output the result of the evaluation of a (python) source block
> to a (graphics) file and have a link to the file inserted in the buffer.

For this, you should try updated header arguments like ":results graphics link"
or ":results file link".

>
> Here is the code:
>
> #+begin_src python :results value file :file scaleplot02.pdf :exports results :var data=test
>   import matplotlib.pyplot as plt
>   import csv
>
>   scale = []
>   lescale = []
>   cdscale = []
>
>   energy = []
>   leenergy = []
>   cdenergy = []
>
>   for row in data:
>       scale.append(float(row[0]))
>       energy.append(float(row[1]))
>       if int(row[2]) != 0:
> 	 lescale.append(float(row[0]))
> 	 leenergy.append(float(row[1]))
>       if int(row[3]) != 0:
> 	 cdscale.append(float(row[0]))
> 	 cdenergy.append(float(row[1]))        
>
>   plt.plot(scale,energy, 'r+')
>   plt.plot(lescale,leenergy, 'go')
>   plt.plot(cdscale,cdenergy, 'bo')
>   plt.xlabel('scale')
>   plt.ylabel('energy (Ha)')
>   plt.title('Energy vs scale')
>   plt.legend()
>   plt.savefig(file)
>   print ("[[./%s]]" % file)
> #+end_src
>
> This is the error:
>
> NameError: name 'file' is not defined
>
> Obviously the syntax on the begin_src line is wrong, but what should it
> be instead?
>
> Thanks,
> Roger


-- 
[ stardiviner ]
       I try to make every word tell the meaning what I want to express.

       Blog: https://stardiviner.github.io/
       IRC(freenode): stardiviner, Matrix: stardiviner
       GPG: F09F650D7D674819892591401B5DF1C95AE89AC3
      

Re: Output result of source block to a file

[Ken Mankoff]

On 2019-06-05 at 12:41 +02, Roger Mason <rmason@mun.ca> wrote...
> I want to output the result of the evaluation of a (python) source
> block to a (graphics) file and have a link to the file inserted in the
> buffer.
>
> Here is the code:
>
> #+begin_src python :results value file :file scaleplot02.pdf :exports results :var data=test
>   print ("[[./%s]]" % file)
> #+end_src

What about :var file="scaleplot02.pdf".

  -k.

Re: Output result of source block to a file

[Roger Mason]

Hello,

stardiviner <numbchild@gmail.com> writes:

> Roger Mason <rmason@mun.ca> writes:
>
>> Hello,
>>
>> I want to output the result of the evaluation of a (python) source block
>> to a (graphics) file and have a link to the file inserted in the buffer.

>> #+begin_src python :results value file :file scaleplot02.pdf :exports results :var data=test

...snip...

>>   plt.savefig(file)
>>   print ("[[./%s]]" % file)
>> #+end_src

> For this, you should try updated header arguments like ":results graphics link"
> or ":results file link".

Thanks for your reply.

The current incarnation:

#+begin_src python :results graphics link :exports results :var data=test :var file="scaleplot2.pdf"

The output:
#+RESULTS: scaleplot02
: None

I'm guessing I have the syntax wrong.  Is it documented somewhere?  I'm
on Org mode version 9.1.13 (release_9.1.13-784-ged7d1d @
/home/rmason/.emacs.d/org-git/lisp/).

Cheers,
Roger

Re: Output result of source block to a file

[Roger Mason]

Hello,
Ken Mankoff <mankoff@gmail.com> writes:

>
> What about :var file="scaleplot02.pdf".

That was my starting point, but the resulting link is prefixed by ':'

Perhaps I should upgrade org.

Cheers,
Roger

Re: Output result of source block to a file

[Roger Mason]

Hello again,

stardiviner <numbchild@gmail.com> writes:

> Roger Mason <rmason@mun.ca> writes:
>
>> Hello,
>>
>> I want to output the result of the evaluation of a (python) source block
>> to a (graphics) file and have a link to the file inserted in the buffer.
>
> For this, you should try updated header arguments like ":results graphics link"
> or ":results file link".

I updated to Org mode version 9.2.3 (release_9.2.3-390-gfb5091 @ /home/rmason/.emacs.d/org-git/lisp/)

This works but sets the results to verbatim, hence the exported &
compiled code does not include the graphic.

#+name: scaleplot03
#+begin_src python :results output graphics link :exports results :var data=test :var file="scaleplot03.pdf"
  import matplotlib.pyplot as plt
  import csv

  scale = []
  lescale = []
  cdscale = []

  energy = []
  leenergy = []
  cdenergy = []

  for row in data:
      scale.append(float(row[0]))
      energy.append(float(row[1]))
      if int(row[2]) != 0:
             lescale.append(float(row[0]))
             leenergy.append(float(row[1]))
      if int(row[3]) != 0:
             cdscale.append(float(row[0]))
             cdenergy.append(float(row[1]))        

  plt.plot(scale,energy, 'r+')
  plt.plot(lescale,leenergy, 'go')
  plt.plot(cdscale,cdenergy, 'bo')
  plt.xlabel('scale')
  plt.ylabel('energy (Ha)')
  plt.title('Energy vs scale')
  plt.legend()
  plt.savefig(file)
  print ("./%s" % file)
#+end_src

#+RESULTS: scaleplot03
: ./scaleplot03.pdf

This header does not work either:
#+name: scaleplot04
#+begin_src python :results file link :exports results :var data=test :var file="scaleplot04.pdf"
( same pythone code here)

#+RESULTS: scaleplot04
[[file:None]]
]]
#+end_src

Thank you for your help.

Roger

Re: Output result of source block to a file

[John Kitchin]

[-- Attachment #1: Type: text/plain, Size: 2427 bytes --]

I think you can use something like this:

#+BEGIN_SRC python :results output file :var fname="test.png"
import matplotlib.pyplot as plt
import io

f = io.StringIO()
plt.plot([1, 2, 3, 17])
plt.savefig(fname)
print(fname, end='')
#+END_SRC

It is in Python3.

John

-----------------------------------
Professor John Kitchin
Doherty Hall A207F
Department of Chemical Engineering
Carnegie Mellon University
Pittsburgh, PA 15213
412-268-7803
@johnkitchin
http://kitchingroup.cheme.cmu.edu



On Thu, Jun 6, 2019 at 2:27 PM Roger Mason <rmason@mun.ca> wrote:

> Hello again,
>
> stardiviner <numbchild@gmail.com> writes:
>
> > Roger Mason <rmason@mun.ca> writes:
> >
> >> Hello,
> >>
> >> I want to output the result of the evaluation of a (python) source block
> >> to a (graphics) file and have a link to the file inserted in the buffer.
> >
> > For this, you should try updated header arguments like ":results
> graphics link"
> > or ":results file link".
>
> I updated to Org mode version 9.2.3 (release_9.2.3-390-gfb5091 @
> /home/rmason/.emacs.d/org-git/lisp/)
>
> This works but sets the results to verbatim, hence the exported &
> compiled code does not include the graphic.
>
> #+name: scaleplot03
> #+begin_src python :results output graphics link :exports results :var
> data=test :var file="scaleplot03.pdf"
>   import matplotlib.pyplot as plt
>   import csv
>
>   scale = []
>   lescale = []
>   cdscale = []
>
>   energy = []
>   leenergy = []
>   cdenergy = []
>
>   for row in data:
>       scale.append(float(row[0]))
>       energy.append(float(row[1]))
>       if int(row[2]) != 0:
>              lescale.append(float(row[0]))
>              leenergy.append(float(row[1]))
>       if int(row[3]) != 0:
>              cdscale.append(float(row[0]))
>              cdenergy.append(float(row[1]))
>
>   plt.plot(scale,energy, 'r+')
>   plt.plot(lescale,leenergy, 'go')
>   plt.plot(cdscale,cdenergy, 'bo')
>   plt.xlabel('scale')
>   plt.ylabel('energy (Ha)')
>   plt.title('Energy vs scale')
>   plt.legend()
>   plt.savefig(file)
>   print ("./%s" % file)
> #+end_src
>
> #+RESULTS: scaleplot03
> : ./scaleplot03.pdf
>
> This header does not work either:
> #+name: scaleplot04
> #+begin_src python :results file link :exports results :var data=test :var
> file="scaleplot04.pdf"
> ( same pythone code here)
>
> #+RESULTS: scaleplot04
> [[file:None]]
> ]]
> #+end_src
>
> Thank you for your help.
>
> Roger
>
>
>
>

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Re: Output result of source block to a file

[Roger Mason]

Hello John,

John Kitchin <jkitchin@andrew.cmu.edu> writes:

> I think you can use something like this:
>
> #+BEGIN_SRC python :results output file :var fname="test.png"
> import matplotlib.pyplot as plt
> import io
>
> f = io.StringIO()
> plt.plot([1, 2, 3, 17])
> plt.savefig(fname)
> print(fname, end='')
> #+END_SRC
>
> It is in Python3.

Thank you.  That works.

Best wishes,
Roger

Re: Output result of source block to a file

[stardiviner]

Roger Mason <rmason@mun.ca> writes:

> Hello,
>
> stardiviner <numbchild@gmail.com> writes:
>
>> Roger Mason <rmason@mun.ca> writes:
>>
>>> Hello,
>>>
>>> I want to output the result of the evaluation of a (python) source block
>>> to a (graphics) file and have a link to the file inserted in the buffer.
>
>>> #+begin_src python :results value file :file scaleplot02.pdf :exports results :var data=test
>
> ...snip...
>
>>>   plt.savefig(file)
>>>   print ("[[./%s]]" % file)
>>> #+end_src
>
>> For this, you should try updated header arguments like ":results graphics link"
>> or ":results file link".
>
> Thanks for your reply.
>
> The current incarnation:
>
> #+begin_src python :results graphics link :exports results :var data=test :var file="scaleplot2.pdf"
>
> The output:
> #+RESULTS: scaleplot02
> : None
>

First, check out =org-manual.org= documentation.
And you don't need ~print ("[[./%s]]" % file)~ in Python code if you use ~:results graphics link~.

> I'm guessing I have the syntax wrong.  Is it documented somewhere?  I'm
> on Org mode version 9.1.13 (release_9.1.13-784-ged7d1d @
> /home/rmason/.emacs.d/org-git/lisp/).
>
> Cheers,
> Roger


-- 
[ stardiviner ]
       I try to make every word tell the meaning what I want to express.

       Blog: https://stardiviner.github.io/
       IRC(freenode): stardiviner, Matrix: stardiviner
       GPG: F09F650D7D674819892591401B5DF1C95AE89AC3
      

Re: Output result of source block to a file

[John Kitchin]

you probably figured out the "import io" and "f = io..." line are not
necessary here.

I couldn't figure out a reasonable way to use :results graphics link
that didn't result in repeating the filename more than desired. These
also both work, but seem to both require repeating the filename twice.

#+BEGIN_SRC python :results graphics link :var fname="test.png" :file "test.png"
import matplotlib.pyplot as plt

plt.plot([1, 2, 3, 1])
plt.savefig(fname)
#+END_SRC

#+BEGIN_SRC python :results graphics link :file "test.png"
import matplotlib.pyplot as plt

plt.plot([1, 2, 3])
plt.savefig("test.png")
#+END_SRC

Something like this should work, but there seem to be some extra bytes
getting put in the png file from the decoding, and latin-1 is the only
one I can get to work. If anyone knows how to get this to work, I am
interested in seeing it!

#+BEGIN_SRC python :results value :file "io.png"
import matplotlib.pyplot as plt
import io
buf = io.BytesIO()

plt.plot([1, 2, 3])
plt.savefig(buf, format='png')

s = buf.getvalue()
return s.decode('latin-1')
#+END_SRC


In general though, all of these are much more work than using
ob-ipython, which just puts images in the buffer for you.

Roger Mason <rmason@mun.ca> writes:

> Hello John,
>
> John Kitchin <jkitchin@andrew.cmu.edu> writes:
>
>> I think you can use something like this:
>>
>> #+BEGIN_SRC python :results output file :var fname="test.png"
>> import matplotlib.pyplot as plt
>> import io
>>
>> f = io.StringIO()
>> plt.plot([1, 2, 3, 17])
>> plt.savefig(fname)
>> print(fname, end='')
>> #+END_SRC
>>
>> It is in Python3.
>
> Thank you.  That works.
>
> Best wishes,
> Roger


--
Professor John Kitchin
Doherty Hall A207F
Department of Chemical Engineering
Carnegie Mellon University
Pittsburgh, PA 15213
412-268-7803
@johnkitchin
http://kitchingroup.cheme.cmu.edu

Re: Output result of source block to a file

[Roger Mason]

Hello John,

John Kitchin <jkitchin@andrew.cmu.edu> writes:

> you probably figured out the "import io" and "f = io..." line are not
> necessary here.

Indeed.

> I couldn't figure out a reasonable way to use :results graphics link
> that didn't result in repeating the filename more than desired. These
> also both work, but seem to both require repeating the filename twice.
>
> #+BEGIN_SRC python :results graphics link :var fname="test.png" :file "test.png"
> import matplotlib.pyplot as plt
>
> plt.plot([1, 2, 3, 1])
> plt.savefig(fname)
> #+END_SRC
>
> #+BEGIN_SRC python :results graphics link :file "test.png"
> import matplotlib.pyplot as plt
>
> plt.plot([1, 2, 3])
> plt.savefig("test.png")
> #+END_SRC
>
> Something like this should work, but there seem to be some extra bytes
> getting put in the png file from the decoding, and latin-1 is the only
> one I can get to work. If anyone knows how to get this to work, I am
> interested in seeing it!
>
> #+BEGIN_SRC python :results value :file "io.png"
> import matplotlib.pyplot as plt
> import io
> buf = io.BytesIO()
>
> plt.plot([1, 2, 3])
> plt.savefig(buf, format='png')
>
> s = buf.getvalue()
> return s.decode('latin-1')
> #+END_SRC
>
>
> In general though, all of these are much more work than using
> ob-ipython, which just puts images in the buffer for you.

I will investigate that, thanks for the tip.  I began this bit of work
using gnuplot for making x-y plots, but I find that gnuplot syntax gets
messy for anything but simple data.  I am not a particular fan
of python so I'm also looking into guile & racket for plotting.

Thanks for your help, it is much appreciated.

Roger

Re: Output result of source block to a file

[John Kitchin]

[-- Attachment #1: Type: text/plain, Size: 2394 bytes --]

A similar approach as the last one should work, the problem I was having is
that to print the binary string from python you have to decode it, and
latin-1 seems close to right, but it puts a bunch of extra bytes in it that
lead to a bad png file. I feel like this worked in Python2 with StringIO,
but not in Python3 with BytesIO.



John

-----------------------------------
Professor John Kitchin
Doherty Hall A207F
Department of Chemical Engineering
Carnegie Mellon University
Pittsburgh, PA 15213
412-268-7803
@johnkitchin
http://kitchingroup.cheme.cmu.edu



On Sat, Jun 8, 2019 at 6:52 AM Roger Mason <rmason@mun.ca> wrote:

> Hello John,
>
> John Kitchin <jkitchin@andrew.cmu.edu> writes:
>
> > you probably figured out the "import io" and "f = io..." line are not
> > necessary here.
>
> Indeed.
>
> > I couldn't figure out a reasonable way to use :results graphics link
> > that didn't result in repeating the filename more than desired. These
> > also both work, but seem to both require repeating the filename twice.
> >
> > #+BEGIN_SRC python :results graphics link :var fname="test.png" :file
> "test.png"
> > import matplotlib.pyplot as plt
> >
> > plt.plot([1, 2, 3, 1])
> > plt.savefig(fname)
> > #+END_SRC
> >
> > #+BEGIN_SRC python :results graphics link :file "test.png"
> > import matplotlib.pyplot as plt
> >
> > plt.plot([1, 2, 3])
> > plt.savefig("test.png")
> > #+END_SRC
> >
> > Something like this should work, but there seem to be some extra bytes
> > getting put in the png file from the decoding, and latin-1 is the only
> > one I can get to work. If anyone knows how to get this to work, I am
> > interested in seeing it!
> >
> > #+BEGIN_SRC python :results value :file "io.png"
> > import matplotlib.pyplot as plt
> > import io
> > buf = io.BytesIO()
> >
> > plt.plot([1, 2, 3])
> > plt.savefig(buf, format='png')
> >
> > s = buf.getvalue()
> > return s.decode('latin-1')
> > #+END_SRC
> >
> >
> > In general though, all of these are much more work than using
> > ob-ipython, which just puts images in the buffer for you.
>
> I will investigate that, thanks for the tip.  I began this bit of work
> using gnuplot for making x-y plots, but I find that gnuplot syntax gets
> messy for anything but simple data.  I am not a particular fan
> of python so I'm also looking into guile & racket for plotting.
>
> Thanks for your help, it is much appreciated.
>
> Roger
>

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