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Méthode pandas.Index.asof

Signature de la méthode asof

def asof(self, label) 

Description

asof.__doc__

        Return the label from the index, or, if not present, the previous one.

        Assuming that the index is sorted, return the passed index label if it
        is in the index, or return the previous index label if the passed one
        is not in the index.

        Parameters
        ----------
        label : object
            The label up to which the method returns the latest index label.

        Returns
        -------
        object
            The passed label if it is in the index. The previous label if the
            passed label is not in the sorted index or `NaN` if there is no
            such label.

        See Also
        --------
        Series.asof : Return the latest value in a Series up to the
            passed index.
        merge_asof : Perform an asof merge (similar to left join but it
            matches on nearest key rather than equal key).
        Index.get_loc : An `asof` is a thin wrapper around `get_loc`
            with method='pad'.

        Examples
        --------
        `Index.asof` returns the latest index label up to the passed label.

        >>> idx = pd.Index(['2013-12-31', '2014-01-02', '2014-01-03'])
        >>> idx.asof('2014-01-01')
        '2013-12-31'

        If the label is in the index, the method returns the passed label.

        >>> idx.asof('2014-01-02')
        '2014-01-02'

        If all of the labels in the index are later than the passed label,
        NaN is returned.

        >>> idx.asof('1999-01-02')
        nan

        If the index is not sorted, an error is raised.

        >>> idx_not_sorted = pd.Index(['2013-12-31', '2015-01-02',
        ...                            '2014-01-03'])
        >>> idx_not_sorted.asof('2013-12-31')
        Traceback (most recent call last):
        ValueError: index must be monotonic increasing or decreasing