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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def _check_symbol_mappings(df, exchanges, asset_exchange):
 """Check that there are no cases where multiple symbols resolve to the same asset at the same time in the same country. Parameters df : pd.DataFrame The equity symbol mappings table. exchanges : pd.DataFrame The exchanges table. asset_exchange : pd.Series A series that maps sids to the exchange the asset is in. Raises ------ ValueError Raised when there are ambiguous symbol mappings. """

    mappings = df.set_index('sid')[list(mapping_columns)].copy()
    mappings['country_code'] = exchanges['country_code'][
        asset_exchange.loc[df['sid']]
    ].values
    ambigious = {}

    def check_intersections(persymbol):
        intersections = list(intersecting_ranges(map(
            from_tuple,
            zip(persymbol.start_date, persymbol.end_date),
        )))
        if intersections:
            data = persymbol[
                ['start_date', 'end_date']
            ].astype('datetime64[ns]')
            # indent the dataframe string, also compute this early because
            # ``persymbol`` is a view and ``astype`` doesn't copy the index
            # correctly in pandas 0.22
            msg_component = '\n  '.join(str(data).splitlines())
            ambigious[persymbol.name] = intersections, msg_component

    mappings.groupby(['symbol', 'country_code']).apply(check_intersections)

    if ambigious:
        raise ValueError(
            'Ambiguous ownership for %d symbol%s, multiple assets held the'
            ' following symbols:\n%s' % (
                len(ambigious),
                '' if len(ambigious) == 1 else 's',
                '\n'.join(
                    '%s (%s):\n  intersections: %s\n  %s' % (
                        symbol,
                        country_code,
                        tuple(map(_format_range, intersections)),
                        cs,
                    )
                    for (symbol, country_code), (intersections, cs) in sorted(
                        ambigious.items(),
                        key=first,
                    ),
                ),
            )
        )