Python - 将数组列表转换为二维数组
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Python - Conversion of list of arrays to 2D array
提问by user3182080
I have a dataset that is formatted like this:
我有一个格式如下的数据集：
A=[(Num1,Num2,Num3), (Num4,Num5,Num6), (Num7,Num8,Num9)]
with
和
A.shape = (3,)
and I would like to convert this into a 2D numpy array:
我想将其转换为 2D numpy 数组：
A=[[Num1,Num2,Num3],[Num4,Num5,Num6],[Num7,Num8,Num9]]
with
和
A.shape = (3,3)
How do I do this, preferably without loops? Thanks.
我该怎么做，最好没有循环？谢谢。
回答by devalentino
I think like:
我认为：
A = map(lambda t: list(t), A)
回答by YS-L
Not sure if I understood the question correctly, but does this work for you?
不确定我是否正确理解了这个问题，但这对你有用吗？
import numpy as np
A = [[1,2,3],[4,5,6],[7,8,9]]
A = np.array(A)
If Ais a list of numpy array, how about this:
如果A是一个 numpy 数组列表，那么这个怎么样：
Ah = np.vstack(A)
Av = np.hstack(A)
回答by M4rtini
If what you have is a array of tupples. A (3,) array with dtype=object.
There is no way, that i am aware of, to elegantly unpack them into a (3,3) array through broadcasting. Converting back to a list, and then creating a new array seems the easiest.
如果您拥有的是一组元组。dtype=object 的 (3,) 数组。
据我所知，没有办法通过广播将它们优雅地解包到 (3,3) 数组中。转换回列表，然后创建一个新数组似乎是最简单的。
In [314]: data
Out[314]: array([(1, 2, 3), (4, 5, 6), (7, 8, 9)], dtype=object)
In [315]: data.shape
Out[315]: (3L,)
data2 = np.empty((3,3), dtype=int)
#Neither of these work.
data2[:] = data[:]
data2[:] = data[:, None]
#This will work, but requires looping
data2[0,:] = data[0]
data2[1,:] = data[1]
data2[2,:] = data[2]
#This is the easies way i could find
data2 = np.array(data.tolist())
回答by Sergey Antopolskiy
If I understood correctly what you're asking, you have a case where numpy did not convert array of arrays into 2d array. This can happen when your arrays are not of the same size. Example:
如果我正确理解您的要求，您会遇到 numpy 未将数组数组转换为二维数组的情况。当您的数组大小不同时，可能会发生这种情况。例子：
Automatic conversion to 2d array:
自动转换为二维数组：
import numpy as np
a = np.array([np.array([1,2,3]),np.array([2,3,4]),np.array([6,7,8])])
print a
Output:
输出：
>>>[[1 2 3]
[2 3 4]
[6 7 8]]
No automatic conversion (look for the change in the second subarray):
没有自动转换（寻找第二个子数组的变化）：
import numpy as np
b = np.array([np.array([1,2,3]),np.array([2,3,4,5]),np.array([6,7,8])])
print b
Output:
输出：
>>>[array([1, 2, 3]) array([2, 3, 4, 5]) array([6, 7, 8])]
I found a couple of ways of converting an array of arrays to 2d array. In any case you need to get rid of subarrays which have different size. So you will need a mask to select only "good" subarrays. Then you can use this mask with list comprehensions to recreate array, like this:
我找到了几种将数组数组转换为二维数组的方法。在任何情况下，您都需要摆脱具有不同大小的子数组。所以你需要一个掩码来只选择“好”的子数组。然后你可以使用这个带有列表推导式的掩码来重新创建数组，如下所示：
import numpy as np
a = np.array([np.array([1,2,3]),np.array([2,3,4,5]),np.array([6,7,8])])
mask = np.array([True, False, True])
c = np.array([element for (i,element) in enumerate(a) if mask[i]])
print a
print c
Output:
输出：
>>>>[array([1, 2, 3]) array([2, 3, 4, 5]) array([6, 7, 8])]
>>>>[[1 2 3]
[6 7 8]]
Or you can delete "bad" subarrays and use vstack(), like this:
或者您可以删除“坏”子数组并使用 vstack()，如下所示：
import numpy as np
a = np.array([np.array([1,2,3]),np.array([2,3,4,5]),np.array([6,7,8])])
mask = np.array([True, False, True])
d = np.delete(a,np.where(mask==False))
e = np.vstack(d)
print a
print e
Output:
输出：
>>>>[array([1, 2, 3]) array([2, 3, 4, 5]) array([6, 7, 8])]
>>>>[[1 2 3]
[6 7 8]]
I believe second method would be faster for large arrays, but I haven't tested the timing.
我相信第二种方法对于大型阵列会更快，但我还没有测试时间。
