xorbits.pandas.Index.searchsorted#

Index.searchsorted(value: NumpyValueArrayLike | ExtensionArray, side: Literal['left', 'right'] = 'left', sorter: NumpySorter | None = None) → npt.NDArray[np.intp] | np.intp#

Find indices where elements should be inserted to maintain order.

Find the indices into a sorted Index self such that, if the corresponding elements in value were inserted before the indices, the order of self would be preserved.

备注

The Index must be monotonically sorted, otherwise wrong locations will likely be returned. Pandas does not check this for you.

参数

  • value (array-like or scalar) – Values to insert into self.

  • side ({'left', 'right'}, optional) – If ‘left’, the index of the first suitable location found is given. If ‘right’, return the last such index. If there is no suitable index, return either 0 or N (where N is the length of self).

  • sorter (1-D array-like, optional) – Optional array of integer indices that sort self into ascending order. They are typically the result of np.argsort.

返回

A scalar or array of insertion points with the same shape as value.

返回类型

int or array of int

参见

sort_values

Sort by the values along either axis.

numpy.searchsorted

Similar method from NumPy.

提示

Binary search is used to find the required insertion points.

实际案例

>>> ser = pd.Series([1, 2, 3])  
>>> ser  
0    1
1    2
2    3
dtype: int64

>>> ser.searchsorted(4)  
3

>>> ser.searchsorted([0, 4])  
array([0, 3])

>>> ser.searchsorted([1, 3], side='left')  
array([0, 2])

>>> ser.searchsorted([1, 3], side='right')  
array([1, 3])

>>> ser = pd.Series(pd.to_datetime(['3/11/2000', '3/12/2000', '3/13/2000']))  
>>> ser  
0   2000-03-11
1   2000-03-12
2   2000-03-13
dtype: datetime64[ns]

>>> ser.searchsorted('3/14/2000')  
3

>>> ser = pd.Categorical(  
...     ['apple', 'bread', 'bread', 'cheese', 'milk'], ordered=True
... )
>>> ser  
['apple', 'bread', 'bread', 'cheese', 'milk']
Categories (4, object): ['apple' < 'bread' < 'cheese' < 'milk']

>>> ser.searchsorted('bread')  
1

>>> ser.searchsorted(['bread'], side='right')  
array([3])

If the values are not monotonically sorted, wrong locations may be returned:

>>> ser = pd.Series([2, 1, 3])  
>>> ser  
0    2
1    1
2    3
dtype: int64

>>> ser.searchsorted(1)  
0  # wrong result, correct would be 1

警告

This method has not been implemented yet. Xorbits will try to execute it with pandas.

This docstring was copied from pandas.core.indexes.base.Index.