pandas.MultiIndex.set_codes — pandas 2.2.3 documentation (original) (raw)

MultiIndex.set_codes(codes, *, level=None, verify_integrity=True)[source]#

Set new codes on MultiIndex. Defaults to returning new index.

Parameters:

codessequence or list of sequence

New codes to apply.

levelint, level name, or sequence of int/level names (default None)

Level(s) to set (None for all levels).

verify_integritybool, default True

If True, checks that levels and codes are compatible.

Returns:

new index (of same type and class…etc) or None

The same type as the caller or None if inplace=True.

Examples

idx = pd.MultiIndex.from_tuples( ... [(1, "one"), (1, "two"), (2, "one"), (2, "two")], names=["foo", "bar"] ... ) idx MultiIndex([(1, 'one'), (1, 'two'), (2, 'one'), (2, 'two')], names=['foo', 'bar'])

idx.set_codes([[1, 0, 1, 0], [0, 0, 1, 1]]) MultiIndex([(2, 'one'), (1, 'one'), (2, 'two'), (1, 'two')], names=['foo', 'bar']) idx.set_codes([1, 0, 1, 0], level=0) MultiIndex([(2, 'one'), (1, 'two'), (2, 'one'), (1, 'two')], names=['foo', 'bar']) idx.set_codes([0, 0, 1, 1], level='bar') MultiIndex([(1, 'one'), (1, 'one'), (2, 'two'), (2, 'two')], names=['foo', 'bar']) idx.set_codes([[1, 0, 1, 0], [0, 0, 1, 1]], level=[0, 1]) MultiIndex([(2, 'one'), (1, 'one'), (2, 'two'), (1, 'two')], names=['foo', 'bar'])