Python numpy函数hstack() vstack() stack() dstack() vsplit() concatenate()

Python numpy函数hstack()，vstack()，stack()，dstack()，vsplit()，concatenate()

在上述的几个函数中，numpy.hstack()和numpy.column_stack()函数略有相似，numpy.vstack()与numpy.row_stack()函数也是挺像的。

stackoverflow上也有类似的讨论 https://stackoverflow.com/questions/16473042/numpy-vstack-vs-column-stack

相关函数的列表：

• stack()：Join a sequence of arrays along a new axis.
• hstack()：Stack arrays in sequence horizontally (column wise).
• dstack()：Stack arrays in sequence depth wise (along third dimension).
• concatenate()：Join a sequence of arrays along an existing axis.
• vsplit ()：Split array into a list of multiple sub-arrays vertically.

numpy.stack()

函数原型：numpy.stack(arrays, axis=0)

程序实例：

>>> arrays = [np.random.randn(3, 4) for _ in range(10)]  >>> np.stack(arrays, axis=0).shape  (10, 3, 4)  >>>  >>> np.stack(arrays, axis=1).shape  (3, 10, 4)  >>>  >>> np.stack(arrays, axis=2).shape  (3, 4, 10)  >>>  >>> a = np.array([1, 2, 3])  >>> b = np.array([2, 3, 4])  >>> np.stack((a, b))  array([[1, 2, 3],         [2, 3, 4]])  >>>  >>> np.stack((a, b), axis=-1)  array([[1, 2],         [2, 3],         [3, 4]])  

numpy.hstack()

函数原型：numpy.hstack(tup)

其中tup是arrays序列，The arrays must have the same shape, except in the dimensioncorresponding to axis (the first, by default).

等价于：np.concatenate(tup, axis=1)

程序实例：

>>> a = np.array((1,2,3))  >>> b = np.array((2,3,4))  >>> np.hstack((a,b))  array([1, 2, 3, 2, 3, 4])  >>> a = np.array([[1],[2],[3]])  >>> b = np.array([[2],[3],[4]])  >>> np.hstack((a,b))  array([[1, 2],         [2, 3],         [3, 4]])  

numpy.vstack()

函数原型：numpy.vstack(tup)

等价于：np.concatenate(tup, axis=0） if tup contains arrays thatare at least 2-dimensional.

程序实例：

>> a = np.array([1, 2, 3])  >>> b = np.array([2, 3, 4])  >>> np.vstack((a,b))  array([[1, 2, 3],         [2, 3, 4]])  >>>  >>> a = np.array([[1], [2], [3]])  >>> b = np.array([[2], [3], [4]])  >>> np.vstack((a,b))  array([[1],         [2],         [3],         [2],         [3],         [4]]) 

numpy.dstack(）

函数原型：numpy.dstack(tup)

等价于：np.concatenate(tup, axis=2)

程序实例：

>>> a = np.array((1,2,3))  >>> b = np.array((2,3,4))  >>> np.dstack((a,b))  array([[[1, 2],          [2, 3],          [3, 4]]])  >>>  >>> a = np.array([[1],[2],[3]])  >>> b = np.array([[2],[3],[4]])  >>> np.dstack((a,b))  array([[[1, 2]],         [[2, 3]],         [[3, 4]]]) 

numpy.concatenate()

函数原型：numpy.concatenate((a1, a2, …), axis=0)

程序实例：

>>> a = np.array([[1, 2], [3, 4]])  >>> b = np.array([[5, 6]])  >>> np.concatenate((a, b), axis=0)  array([[1, 2],         [3, 4],         [5, 6]])  >>> np.concatenate((a, b.T), axis=1)  array([[1, 2, 5],         [3, 4, 6]])  This function will not preserve masking of MaskedArray inputs.  >>>  >>> a = np.ma.arange(3)  >>> a[1] = np.ma.masked  >>> b = np.arange(2, 5)  >>> a  masked_array(data = [0 -- 2],               mask = [False  True False],         fill_value = 999999)  >>> b  array([2, 3, 4])  >>> np.concatenate([a, b])  masked_array(data = [0 1 2 2 3 4],               mask = False,         fill_value = 999999)  >>> np.ma.concatenate([a, b])  masked_array(data = [0 -- 2 2 3 4],               mask = [False  True False False False False],         fill_value = 999999)  

numpy.vsplit()

函数原型：numpy.vsplit(ary, indices_or_sections)

程序实例：

>>> x = np.arange(16.0).reshape(4, 4)  >>> x  array([[  0.,   1.,   2.,   3.],         [  4.,   5.,   6.,   7.],         [  8.,   9.,  10.,  11.],         [ 12.,  13.,  14.,  15.]])  >>> np.vsplit(x, 2)  [array([[ 0.,  1.,  2.,  3.],         [ 4.,  5.,  6.,  7.]]),   array([[  8.,   9.,  10.,  11.],         [ 12.,  13.,  14.,  15.]])]  >>> np.vsplit(x, np.array([3, 6]))  [array([[  0.,   1.,   2.,   3.],         [  4.,   5.,   6.,   7.],         [  8.,   9.,  10.,  11.]]),   array([[ 12.,  13.,  14.,  15.]]),   array([], dtype=float64)]  With a higher dimensional array the split is still along the first axis.  >>>  >>> x = np.arange(8.0).reshape(2, 2, 2)  >>> x  array([[[ 0.,  1.],          [ 2.,  3.]],         [[ 4.,  5.],          [ 6.,  7.]]])  >>> np.vsplit(x, 2)  [array([[[ 0.,  1.],          [ 2.,  3.]]]),   array([[[ 4.,  5.],          [ 6.,  7.]]])]  

Reference

http://www.cnblogs.com/harvey888/p/5967352.html