numpy库数组拼接np.concatenate

在实践过程中，会经常遇到数组拼接的问题，基于numpy库的concatenate是一个非常好用的数组操作函数。

1、concatenate((a1, a2, …), axis=0)官方文档详解

concatenate(...)
concatenate((a1, a2, ...), axis=0)

Join a sequence of arrays along an existing axis.

Parameters
----------
a1, a2, ... : sequence of array_like
The arrays must have the same shape, except in the dimension
corresponding to `axis` (the first, by default).
axis : int, optional
The axis along which the arrays will be joined.  Default is 0.

Returns
-------
res : ndarray
The concatenated array.

See Also
--------
ma.concatenate : Concatenate function that preserves input masks.
array_split : Split an array into multiple sub-arrays of equal or
near-equal size.
split : Split array into a list of multiple sub-arrays of equal size.
hsplit : Split array into multiple sub-arrays horizontally (column wise)
vsplit : Split array into multiple sub-arrays vertically (row wise)
dsplit : Split array into multiple sub-arrays along the 3rd axis (depth).
stack : Stack a sequence of arrays along a new axis.
hstack : Stack arrays in sequence horizontally (column wise)
vstack : Stack arrays in sequence vertically (row wise)
dstack : Stack arrays in sequence depth wise (along third dimension)

2、Parameters参数

传入的参数必须是一个多个数组的元组或者列表

另外需要指定拼接的方向，默认是 axis = 0，也就是说对0轴的数组对象进行纵向的拼接（纵向的拼接沿着axis= 1方向）；注：一般axis = 0，就是对该轴向的数组进行操作，操作方向是另外一个轴，即axis=1。

In [23]: a = np.array([[1, 2], [3, 4]])

In [24]: b = np.array([[5, 6]])

In [25]: np.concatenate((a, b), axis=0)
Out[25]:
array([[1, 2],
[3, 4],
[5, 6]])

传入的数组必须具有相同的形状，这里的相同的形状可以满足在拼接方向axis轴上数组间的形状一致即可
如果对数组对象进行 axis= 1 轴的拼接，方向是横向0轴，a是一个22维数组，axis= 0轴为2，b是一个12维数组，axis= 0 是1，两者的形状不等，这时会报错

In [27]: np.concatenate((a,b),axis = 1)
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-27-aa1228decc36> in <module>()
----> 1 np.concatenate((a,b),axis = 1)

ValueError: all the input array dimensions except for the concatenation axis must match exactly

将b进行转置，得到b为2*1维数组：

In [28]: np.concatenate((a,b.T),axis = 1)
Out[28]:
array([[1, 2, 5],
[3, 4, 6]])

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