asof(self, label)
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.
The label up to which the method returns the latest index label.
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.
Return the label from the index, or, if not present, the previous one.
Index.get_loc
An :None:None:`asof` is a thin wrapper around :None:None:`get_loc` with method='pad'.
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.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.
This example is valid syntax, but we were not able to check execution>>> 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.
This example is valid syntax, but we were not able to check execution>>> idx.asof('1999-01-02') nan
If the index is not sorted, an error is raised.
This example is valid syntax, but we were not able to check execution>>> idx_not_sorted = pd.Index(['2013-12-31', '2015-01-02',See :
... '2014-01-03'])
... idx_not_sorted.asof('2013-12-31') Traceback (most recent call last): ValueError: index must be monotonic increasing or decreasing
The following pages refer to to this document either explicitly or contain code examples using this.
pandas.core.indexes.base.Index.asof
pandas.core.indexes.base.Index.asof_locs
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