本文介绍了NumPy附加vs Python附加的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
在Python中,我可以追加到一个空数组,例如:
In Python I can append to an empty array like:
>>> a = []
>>> a.append([1,2,3])
>>> a.append([1,2,3])
>>> a
[[1, 2, 3], [1, 2, 3]]
我如何在NumPy中做同样的事情? np.append
不幸的是,该数组展平了(我需要在开始时有一个空数组).
How can I do the same in NumPy? np.append
flattens the array, unfortunately (and I need to have an empty array at the beginning).
推荐答案
OP旨在以空数组开头.因此,这是使用NumPy的一种方法
OP intended to start with empty array. So, here's one approach using NumPy
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]])
但(如果要添加大量循环).首先附加列表并转换为数组比附加NumPy数组更快.
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.
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
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