Python 的 numpy 中“zip()”的等价物是什么？
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What is the equivalent of "zip()" in Python's numpy?
提问by TimY
I am trying to do the following but with numpy arrays:
我正在尝试使用 numpy 数组执行以下操作：
x = [(0.1, 1.), (0.1, 2.), (0.1, 3.), (0.1, 4.), (0.1, 5.)]
normal_result = zip(*x)
This should give a result of:
这应该给出以下结果：
normal_result = [(0.1, 0.1, 0.1, 0.1, 0.1), (1., 2., 3., 4., 5.)]
But if the input vector is a numpy array:
但是如果输入向量是一个 numpy 数组：
y = np.array(x)
numpy_result = zip(*y)
print type(numpy_result)
It (expectedly) returns a:
它（预期）返回一个：
<type 'list'>
The issue is that I will need to transform the result back into a numpy array after this.
问题是我需要在此之后将结果转换回一个 numpy 数组。
What I would like to know is what is if there is an efficient numpy function that will avoid these back-and-forth transformations?
我想知道的是，是否有一个有效的 numpy 函数可以避免这些来回转换？
采纳答案by Jon Clements
You can just transpose it...
你可以把它转...
>>> a = np.array([(0.1, 1.), (0.1, 2.), (0.1, 3.), (0.1, 4.), (0.1, 5.)])
>>> a
array([[ 0.1, 1. ],
[ 0.1, 2. ],
[ 0.1, 3. ],
[ 0.1, 4. ],
[ 0.1, 5. ]])
>>> a.T
array([[ 0.1, 0.1, 0.1, 0.1, 0.1],
[ 1. , 2. , 3. , 4. , 5. ]])
回答by zenpoy
Try using dstack:
尝试使用dstack：
>>> from numpy import *
>>> a = array([[1,2],[3,4]]) # shapes of a and b can only differ in the 3rd dimension (if present)
>>> b = array([[5,6],[7,8]])
>>> dstack((a,b)) # stack arrays along a third axis (depth wise)
array([[[1, 5],
[2, 6]],
[[3, 7],
[4, 8]]])
so in your case it would be:
所以在你的情况下，它将是：
x = [(0.1, 1.), (0.1, 2.), (0.1, 3.), (0.1, 4.), (0.1, 5.)]
y = np.array(x)
np.dstack(y)
>>> array([[[ 0.1, 0.1, 0.1, 0.1, 0.1],
[ 1. , 2. , 3. , 4. , 5. ]]])
