本文主要是介绍【Python】numpy保留数组中不同元素，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
numpy.unique(ar, return_index=False, return_inverse=False, return_counts=False)
去除重复的，然后按从大到小排列
return_counts，对应的元素在原array中出现的次数
Parameters:
ar : array_like
Input array. This will be flattened if it is not already 1-D.
return_index : bool, optional
If True, also return the indices of ar that result in the unique array.
return_inverse : bool, optional
If True, also return the indices of the unique array that can be used to reconstruct ar.
return_counts : bool, optional
If True, also return the number of times each unique value comes up in ar.
New in version 1.9.0.
Returns:
unique : ndarray
The sorted unique values.
unique_indices : ndarray, optional
The indices of the first occurrences of the unique values in the (flattened) original array. Only provided if return_index is True.
unique_inverse : ndarray, optional
The indices to reconstruct the (flattened) original array from the unique array. Only provided if return_inverse is True.
unique_counts : ndarray, optional
The number of times each of the unique values comes up in the original array. Only provided if return_counts is True.
New in version 1.9.0.
>>> np.unique([1, 1, 2, 2, 3, 3])
array([1, 2, 3])
>>> a = np.array([[1, 1], [2, 3]])
>>> np.unique(a)
array([1, 2, 3])
>>> a = np.array(['a', 'b', 'b', 'c', 'a'])
>>> u, indices = np.unique(a, return_index=True)
>>> u
array(['a', 'b', 'c'],dtype='|S1')
>>> indices #表示新不重复数组对应原来的索引值
array([0, 1, 3])
>>> a[indices]
array(['a', 'b', 'c'],dtype='|S1')
>>> a = np.array([1, 2, 6, 4, 2, 3, 2])
>>> u, indices = np.unique(a, return_inverse=True)
>>> u
array([1, 2, 3, 4, 6])
>>> indices #表示原数组使用新数组的重构需要的新数组的索引值
array([0, 1, 4, 3, 1, 2, 1])
>>> u[indices]
array([1, 2, 6, 4, 2, 3, 2])
这篇关于【Python】numpy保留数组中不同元素的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
