在 Numpy 中连接空数组

Question

在 Matlab 中我这样做：

>> 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，但我有问题，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

如果您事先知道列数：

>>> 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.]])

如果没有：

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

Answer 2

如果你想这样做只是因为你不能在循环中连接一个数组和一个初始化的空数组，那么只需使用条件语句， 例如

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

Answer 3

在 Python 中，如果可能使用单独的向量，您应该使用 list.append() 追加添加

>>> 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])]

然后在所有追加操作完成后，就这样返回到np.array

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

Answer 4

我为解决此类问题而构建的东西。它还处理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 5

E = np.array([
    
]).reshape(0, 5)
print("E: \n{}\nShape {}\n".format(E, E.shape))

A = np.vstack([
    [1, 2, 3, 4, 5], 
    [10, 20, 30, 40, 50]]
)
print("A:\n{}\nShape {}\n".format(A, A.shape))

C = np.r_[
    E, 
    A
].astype(np.int32)

print("C:\n{}\nShape {}\n".format(C, C.shape))

E: 
[]
Shape (0, 5)

A:
[[ 1  2  3  4  5]
 [10 20 30 40 50]]
Shape (2, 5)

C:
[[ 1  2  3  4  5]
 [10 20 30 40 50]]
Shape (2, 5)

Answer 6

一种解决方案是使用None 对象和np.concatenate、np.hstack 或np.vstack。

>>> 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 7

np.concatenate、np.hstack 和 np.vstack 会做你想做的事。但是请注意，NumPy 数组不适合用作动态数组。为此，请改用 Python 列表。