Creating a new file with input file name

Question

I have a question. How can I rename the output file using the input filename?

For example, my input filename is:

Field_52_combined_final_roughcal.fits

I would like to obtain an output filename like:

Field_52_traitement_1.fits

I know that I could write :

hdu.writeto('Field_52_traitement_1.fits')

But, I have an other script which loop on 200 files, and I would like the output filename being automatically generated by the input filename.

My script looks like this (for a single input file):

#!/usr/bin/python
# coding: utf-8

from astropy.io import fits
from astropy.table import Table
import numpy as np

                ###################################
                # Fichier contenant le champ brut #
                ###################################

filename = 'E:/Fields/Field52_combined_final_roughcal.fits'

# Ouverture du fichier à l'aide d'astropy  
field = fits.open(filename)          

# Lecture des données fits
tbdata = field[1].data               

                #######################################################
                # Application du tri en fonction de divers paramètres #
                #######################################################

Several conditions / sort

        ###################################################
        # Ecriture du résultat dans nouveau fichier .fits #
        ###################################################

hdu = fits.BinTableHDU(data=tbdata_final)
hdu.writeto('{}_{}'.format(filename,'traitement_1'))

But, with this kind of script, I get:

Field_52_combined_final_roughcal.fits_traitement_1

Tell me, if you have any ideas, websites or something else :) Thank you for your answers!

You could use the os.path functions to separate the filename from the extension or, since you seem to consistently use underscores, simply do a filename.split("_") to get a list of all the parts of the filename. You can then piece it back together in a different order via the list indices. index 0 would be "Field", index 1 would be "52" and so on. You can then "_".join()` afterwards to create a new filename — jDo
– jDo, Commented Mar 24, 2016 at 12:54
Thank you to your answer. I used the split method and it's work : — Essex
– Essex, Commented Mar 24, 2016 at 13:00

doron · Accepted Answer · 2016-03-24 13:05:22Z

2

You can use a simple string replace method and create a variable for the output filename.

filename = 'E:/Fields/Field52_combined_final_roughcal.fits'
outname = filename.replace('combined_final_roughcal', 'traitement_1')

Now just write to the file named 'outname', which is now:

E:/Fields/Field52_traitement_1.fits

edited Mar 24, 2016 at 13:05

answered Mar 24, 2016 at 13:00

doron

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3 Comments

Essex Over a year ago

Perfect ! Both solutions are very interesting ! This one is particulary smart !

Tony Babarino Over a year ago

@Astrolabe1993 This will work only if every input file has 'combined_final_roughcal' at the end though. If that's Your case use this solution because it should be a little bit faster than mine.

Essex Over a year ago

@Tony Babarino At the moment, all my input files have the same name. Only the field number is different about each file.

Tony Babarino · Accepted Answer · 2016-03-24 12:57:20Z

1

>>> filename = 'Field_52_combined_final_roughcal.fits'
>>> filename.split('_')
['Field', '52', 'combined', 'final', 'roughcal.fits']
>>> filename.split('_')[:2]
['Field', '52']
>>> '_'.join(filename.split('_')[:2])
'Field_52'

So applying that to Your code, use:

hdu.writeto('{}_{}'.format('_'.join(filename.split('_')[:2]),'traitement_1'))

instead of:

hdu.writeto('{}_{}'.format(filename,'traitement_1'))

answered Mar 24, 2016 at 12:57

Tony Babarino

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1 Comment

Essex Over a year ago

Thank you so much ! I had the same beginning, but I was blocking on the hdu.writeto. So thank you !

jDo · Accepted Answer · 2016-03-24 13:15:27Z

1

Here's one way of doing it. The "best" way depends on how dynamic you want the filename to be. E.g. whether you want to increment "traitement" or not.

def create_new_filename(old_filename, traitement):
    pieces = old_filename.split("_")
    return "_".join([pieces[0], pieces[1], "traitement", str(traitement)]) +  ".fits"

In the interpreter:

>>> print create_new_filename("Field_52_combined_final_roughcal.fits", 1)
Field_52_traitement_1.fits

To use it in your case, you'd pass in the old filename and the traitement number you want:

hdu.writeto(create_new_filename("Field_52_combined_final_roughcal.fits", 1))

edited Mar 24, 2016 at 13:15

answered Mar 24, 2016 at 13:08

jDo

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2 Comments

Essex Over a year ago

Your script is useful. But, I don't need to get an increment "traitement" because this string represent the result of my first script. After that, I will take the ouput files and I will execute my second script. I will obtain (for example) : Field_52_traitement_2.fits etc .. ;)

jDo Over a year ago

Cool cool. I just always try to make it a little generic. It might not be necessary for you, but that way you can have everything in one script and run all tests automatically in a for loop if you want.

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