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，我有类似的情况：

>>> 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.]])

如果不是：

>>> 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

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 3

如果您只是因为无法在循环中将数组与已初始化的空数组连接而想要这样做，则只需使用条件语句，例如

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

Answer 4

在Python中，如果可以使用单个向量，则应使用list .append()

>>> 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])]

，然后在完成所有附加操作之后，立即返回np.array

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

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

np.concatenate，np.hstack和np.vstack会做你想要的。但请注意，NumPy数组不适合用作动态数组。为此目的使用Python列表。

在Numpy中连接空数组

6 个答案: