Roger Mason <rma...@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
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