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df[['A','B','C']]

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pd.set_option('display.max_columns', 0)

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pd.reset_option('all')

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df1[~df1.package_req_id.str.get(0).isin(['Y'])]

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cols = list(df1)
cols.insert(0, cols.pop(cols.index('return_datetime'))) #3rd
cols.insert(0, cols.pop(cols.index('order_datetime')))  #2nd
cols.insert(0, cols.pop(cols.index('package_req_id')))  #1st 
df1 = df1.ix[:, cols]

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pd.set_option('display.max_columns', 0)

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pd.reset_option('all')

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####Temporary solution 

from IPython.core.display import display, HTML
display(HTML("<style>.container { width:100% !important; }</style>"))

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def flattenHierarchicalCol(col,sep = '_'):
    if not type(col) is tuple:
        return col
    else:
        new_col = ''
        for leveli,level in enumerate(col):
            if not level == '':
                if not leveli == 0:
                    new_col += sep
                new_col += level

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jupyter nbconvert --to notebook --execute testipynb.ipynb --output testipynb.ipynb

nbconvert ~/shared/projects/CallCenterHiringFC/code/rates_calculations.ipynb --to script