Re: [O] Output result of source block to a file

[stardiviner]

Roger Mason <address@hidden> 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


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