如何将新行添加到空的numpy数组

问题描述

使用标准的Python数组，我可以执行以下操作:

Using standard Python arrays, I can do the following:

arr = []
arr.append([1,2,3])
arr.append([4,5,6])
# arr is now [[1,2,3],[4,5,6]]

但是，我不能在numpy中做同样的事情.例如:

However, I cannot do the same thing in numpy. For example:

arr = np.array([])
arr = np.append(arr, np.array([1,2,3]))
arr = np.append(arr, np.array([4,5,6]))
# arr is now [1,2,3,4,5,6]

我也查看了vstack，但是当我在空数组上使用vstack时，我得到了:

I also looked into vstack, but when I use vstack on an empty array, I get:

ValueError: all the input array dimensions except for the concatenation axis must match exactly

那我该如何将新行追加到numpy中的空数组?

So how do I do append a new row to an empty array in numpy?

推荐答案

启动"所需数组的方法是:

The way to "start" the array that you want is:

arr = np.empty((0,3), int)

这是一个空数组，但具有适当的维数.

Which is an empty array but it has the proper dimensionality.

>>> arr
array([], shape=(0, 3), dtype=int64)

然后确保沿轴0附加:

arr = np.append(arr, np.array([[1,2,3]]), axis=0)
arr = np.append(arr, np.array([[4,5,6]]), axis=0)

但是，@ jonrsharpe是正确的.实际上，如果要循环添加，则像第一个示例中那样将其添加到列表中会更快得多，然后最后转换为numpy数组，因为您实际上并没有将numpy用作打算在循环中使用:

But, @jonrsharpe is right. In fact, if you're going to be appending in a loop, it would be much faster to append to a list as in your first example, then convert to a numpy array at the end, since you're really not using numpy as intended during the loop:

In [210]: %%timeit
   .....: l = []
   .....: for i in xrange(1000):
   .....:     l.append([3*i+1,3*i+2,3*i+3])
   .....: l = np.asarray(l)
   .....:
1000 loops, best of 3: 1.18 ms per loop

In [211]: %%timeit
   .....: a = np.empty((0,3), int)
   .....: for i in xrange(1000):
   .....:     a = np.append(a, 3*i+np.array([[1,2,3]]), 0)
   .....:
100 loops, best of 3: 18.5 ms per loop

In [214]: np.allclose(a, l)
Out[214]: True

执行numpythonic的方式取决于您的应用程序，但更像是:

The numpythonic way to do it depends on your application, but it would be more like:

In [220]: timeit n = np.arange(1,3001).reshape(1000,3)
100000 loops, best of 3: 5.93 µs per loop

In [221]: np.allclose(a, n)
Out[221]: True

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