在Numpy中连接空数组

Question

在Matlab中，我是这样做的：

>> E = [];
>> A = [1 2 3 4 5; 10 20 30 40 50];
>> E = [E ; A]

E =

     1     2     3     4     5
    10    20    30    40    50

现在我想在Numpy中做同样的事情，但遇到了一些问题，看看这个：

>>> E = array([],dtype=int)
>>> E
array([], dtype=int64)
>>> A = array([[1,2,3,4,5],[10,20,30,40,50]])

>>> E = vstack((E,A))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/Extras/lib/python/numpy/core/shape_base.py", line 226, in vstack
    return _nx.concatenate(map(atleast_2d,tup),0)
ValueError: array dimensions must agree except for d_0

当我这样做时，我也遇到了类似的情况：

>>> E = concatenate((E,A),axis=0)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: arrays must have same number of dimensions

或者：

>>> E = append([E],[A],axis=0)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/Extras/lib/python/numpy/lib/function_base.py", line 3577, in append
    return concatenate((arr, values), axis=axis)
ValueError: arrays must have same number of dimensions

Answer 1

一种解决方法是使用 None 对象，以及 np.concatenate、np.hstack 或 np.vstack 这些函数。

>>> arr=None
>>> p=np.array([0,1,2,3])

>>> for i in range(0,2):
>>>     arr = (np.vstack((arr, p)) if (arr is not None) else p)

array([[ 0, 1, 2, 3],
      [[ 0, 1, 2, 3]])

Answer 2

我做了一个东西来解决这类问题。它还可以处理 list 输入，而不是 np.array：

import numpy as np


def cat(tupleOfArrays, axis=0):
    # deals with problems of concating empty arrays
    # also gives better error massages

    # first check that the input is correct
    assert isinstance(tupleOfArrays, tuple), 'first var should be tuple of arrays'

    firstFlag = True
    res = np.array([])

    # run over each element in tuple
    for i in range(len(tupleOfArrays)):
        x = tupleOfArrays[i]
        if len(x) > 0:  # if an empty array\list - skip
            if isinstance(x, list):  # all should be ndarray
                x = np.array(x)
            if x.ndim == 1:  # easier to concat 2d arrays
                x = x.reshape((1, -1))
            if firstFlag:  # for the first non empty array, just swich the empty res array with it
                res = x
                firstFlag = False
            else:  # actual concatination

                # first check that concat dims are good
                if axis == 0:
                    assert res.shape[1] == x.shape[1], "Error concating vertically element index " + str(i) + \
                                                       " with prior elements: given mat shapes are " + \
                                                       str(res.shape) + " & " + str(x.shape)
                else:  # axis == 1:
                    assert res.shape[0] == x.shape[0], "Error concating horizontally element index " + str(i) + \
                                                       " with prior elements: given mat shapes are " + \
                                                       str(res.shape) + " & " + str(x.shape)

                res = np.concatenate((res, x), axis=axis)
    return res


if __name__ == "__main__":
    print(cat((np.array([]), [])))
    print(cat((np.array([1, 2, 3]), np.array([]), [1, 3, 54+1j]), axis=0))
    print(cat((np.array([[1, 2, 3]]).T, np.array([]), np.array([[1, 3, 54+1j]]).T), axis=1))
    print(cat((np.array([[1, 2, 3]]).T, np.array([]), np.array([[3, 54]]).T), axis=1))  # a bad one

Answer 3

如果你想这么做，仅仅是因为你在循环中不能把一个数组和一个已经初始化的空数组连接起来，那你可以使用一个条件语句来解决这个问题，比如：

if (i == 0): 
   do the first assignment
else:  
   start your contactenate 

Answer 4

在Python中，如果你想单独处理每个向量，添加新元素时应该使用列表的.append()方法。

>>> E = []
>>> B = np.array([1,2,3,4,5])
>>> C = np.array([10,20,30,40,50])
>>> E = E.append(B)
>>> E = E.append(C)
[array([1, 2, 3, 4, 5]), array([10, 20, 30, 40, 50])]

然后在所有添加操作完成后，再把它们转换成np.array，这样做。

>>> E = np.array(E)
array([[ 1,  2,  3,  4,  5],
   [10, 20, 30, 40, 50]])

Answer 5

如果你事先知道列的数量：

>>> xs = np.array([[1,2,3,4,5],[10,20,30,40,50]])
>>> ys = np.array([], dtype=np.int64).reshape(0,5)
>>> ys
array([], shape=(0, 5), dtype=int64)
>>> np.vstack([ys, xs])
array([[  1.,   2.,   3.,   4.,   5.],
       [ 10.,  20.,  30.,  40.,  50.]])

如果不知道的话：

>>> ys = np.array([])
>>> ys = np.vstack([ys, xs]) if ys.size else xs
array([[ 1,  2,  3,  4,  5],
       [10, 20, 30, 40, 50]])