Convert xls spreadsheet to yaml file

convertxlstoyaml.py

from pathlib import Path
import ruamel.yaml
import pandas as pd

yaml = ruamel.yaml.YAML()

df = pd.read_excel('../scopus_vexport_with_manualedit_202008.xlsx')
print (df)

# df.columns returns:
# Index(['Authors', 'Authorsv2', 'Authors (short)', 'Author position', 'Title', 'Journal Info', 
# 'Year', 'Article Code', 'IF', 'Cited by',
#      'Quartile and Area', 'ISSN', 'H Index', 'Hyperlink', 'DOI', 'DOI link',
#       'Publisher', 'ISBN', 'CODEN', 'Document Type', 'Publication Stage',
#       'Access Type', 'Source', 'EID', 'Unnamed: 24', 'Journal', 'JournalAbr',
#       'Volume', 'Page start', 'Page end'],
#      dtype='object')

il = 0;
publis = []
for il in range(len(df)):
    aa =  dict(num='('+str(len(df)-il)+')',
                title=df.loc[il,'Title'] ,
                authors=df.loc[il,'Authorsv2'],
                journal='*'+df.loc[il,'Journal']+'*',
                year= str(df.loc[il,'Year']),
                doi=df.loc[il,'DOI'],
                doilink=df.loc[il,'DOI link'],
                )

    publis.append(aa)


with open('publis.yml', 'w') as f:

    data = yaml.dump(publis, f)