NumPy 追加与 Python 追加
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NumPy append vs Python append
提问by leosenko
In Python I can append to an empty array like:
在 Python 中，我可以附加到一个空数组，如：
>>> a = []
>>> a.append([1,2,3])
>>> a.append([1,2,3])
>>> a
[[1, 2, 3], [1, 2, 3]]
How can I do the same in NumPy? np.appendflattens the array, unfortunately (and I need to have an empty array at the beginning).
我如何在 NumPy 中做同样的事情？np.append不幸的是，使数组变平（而且我需要在开头有一个空数组）。
采纳答案by Zero
OP intended to start with empty array. So, here's one approach using NumPy
OP 打算从空数组开始。所以，这是使用 NumPy 的一种方法
In [2]: a = np.empty((0,3), int)
In [3]: a
Out[3]: array([], shape=(0L, 3L), dtype=int32)
In [4]: a = np.append(a, [[1,2,3]], axis=0)
In [5]: a
Out[5]: array([[1, 2, 3]])
In [6]: a = np.append(a, [[1,2,3]], axis=0)
In [7]: a
Out[7]:
array([[1, 2, 3],
[1, 2, 3]])
BUT, if you're appending in a large number of loops. It's faster to append list first and convert to array than appending NumPy arrays.
但是，如果您要附加大量循环。首先附加列表并转换为数组比附加 NumPy 数组更快。
In [8]: %%timeit
...: list_a = []
...: for _ in xrange(10000):
...: list_a.append([1, 2, 3])
...: list_a = np.asarray(list_a)
...:
100 loops, best of 3: 5.95 ms per loop
In [9]: %%timeit
....: arr_a = np.empty((0, 3), int)
....: for _ in xrange(10000):
....: arr_a = np.append(arr_a, np.array([[1,2,3]]), 0)
....:
10 loops, best of 3: 110 ms per loop
回答by ComputerFellow
回答by John1024
Using np.append
使用 np.append
Let's start with an empty 2-D array:
让我们从一个空的二维数组开始：
In [8]: a = np.array([]); a = a.reshape((0, 3)); a
Out[8]: array([], shape=(0, 3), dtype=float64)
Now, let's append some rows:
现在，让我们附加一些行：
In [19]: a = np.append(a, [[1, 2, 3]], axis=0 ); a
Out[19]: array([[ 1., 2., 3.]])
In [20]: a = np.append(a, [[1, 2, 3]], axis=0 ); a
Out[20]:
array([[ 1., 2., 3.],
[ 1., 2., 3.]])
Using np.concatenate:
使用np.concatenate：
Again, let's start with an empty 2-D array:
同样，让我们从一个空的二维数组开始：
In [28]: a = np.array([]); a = a.reshape((0, 3)); a
Out[28]: array([], shape=(0, 3), dtype=float64)
Now, let's concatenate some rows:
现在，让我们连接一些行：
In [29]: a = np.concatenate( (a, [[1, 2, 3]]), axis=0 ); a
Out[29]: array([[ 1., 2., 3.]])
In [30]: a = np.concatenate( (a, [[1, 2, 3]]), axis=0 ); a
Out[30]:
array([[ 1., 2., 3.],
[ 1., 2., 3.]])
