Python 在Numpy中连接空数组
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Concatenating empty array in Numpy
提问by maxv15
in Matlab I do this:
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
Now I want the same thing in Numpy but I have problems, look at this:
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
I have a similar situation when I do this with:
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
Or:
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
采纳答案by behzad.nouri
if you know the number of columns before hand:
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.]])
if not:
如果不：
>>> ys = np.array([])
>>> ys = np.vstack([ys, xs]) if ys.size else xs
array([[ 1, 2, 3, 4, 5],
[10, 20, 30, 40, 50]])
回答by Sturla Molden
np.concatenate, np.hstackand np.vstackwill do what you want. Note however that NumPy arrays are not suitable for use as dynamic arrays. Use Python lists for that purpose instead.
np.concatenate，np.hstack并且np.vstack会做你想做的。但是请注意，NumPy 数组不适合用作动态数组。为此，请使用 Python 列表。
回答by YoniChechik
Something that I've build to deal with this sort of problem. It's also deals with listinput instead of np.array:
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
回答by Ming
If you wanna do this just because you cannot concatenate an array with an initialized empty array in a loop, then just use a conditional statement, e.g.
如果你想这样做只是因为你不能在循环中将一个数组与一个初始化的空数组连接起来，那么只需使用一个条件语句，例如
if (i == 0):
do the first assignment
else:
start your contactenate
