Python/numpy数组分区 [英] Python/numpy array partitioning

问题描述

我正在使用Python 3.6和numpy.

I'm using Python 3.6 and numpy.

我从hdf5文件中读取了一个表的一列，该表是2D数组.

From an hdf5 file I read a column of a table that is a 2D array.

数组的每一行都包含一个有限元节点的ID.

Each row of the array holds the ID's of the nodes of a finite element.

表的结构使得它在同一张表中同时包含了低阶和高阶元素(这很糟糕，但不是我可以更改的自由度)

The table is structured such that it holds both lower and higher order elements in the same table (which sucks, but is not a degree of freedom I can change)

所以数组看起来像这样(除了它可能有数百万行)

So the array looks something like this (except that it has potentially millions of rows)

[[1,2,3,4,0,0,0,0],           #<- 4 Node quad data packed with zeros
 [3,4,5,6,0,0,0,0],             
 [7,8,9,10,11,12,13,14],      #<- 8 node quad in the same table as 4 node quad
 [15,16,17,18,19,20,21,22]]

我需要将此信息分成两个单独的数组-一个用于4个节点，一个用于8个节点行.

I need to separate this info into two separate arrays - one for the 4 node an done for 8 node rows.

[[1,2,3,4],          
 [3,4,5,6]]

[[7,8,9,10,11,12,13,14], 
 [15,16,17,18,19,20,21,22]]

现在，我在2D数组上进行迭代，检查每行中第5个值的值，并创建两个索引数组-一个标识4个节点行，一个标识8个节点行.

Right now I'm iterating over the 2D array, checking the value of the 5th value in each row and creating two index arrays - one identifying the 4 node rows and one the 8 node rows.

for element in elements:
    if element[5] == 0:
        tet4indices.append(index)
    else:        
        tet10indices.append(index)
    index+=1  

然后我使用索引数组切片来获取两个数组

Then I use index array slicing to get the two arrays

tet4s=elements[tet4indices, 0:5]
tet10s=elements[tet10indices,0:10]

以上方法有效，但看起来有点难看.

The above works, but seems kinda ugly.

如果有人有更好的解决方案，我将不胜感激.....

If anyone has a better solution, I'd be grateful to hear about it.....

预先感谢

道格

推荐答案

在数组中，很容易找到第5个元素为0或不为0的行:

In an array it's easy to find rows where the 5th element is 0, or not 0:

In [75]: arr = np.array(alist)
In [76]: arr
Out[76]: 
array([[ 1,  2,  3,  4,  0,  0,  0,  0],
       [ 3,  4,  5,  6,  0,  0,  0,  0],
       [ 7,  8,  9, 10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19, 20, 21, 22]])
In [77]: arr[:,5]
Out[77]: array([ 0,  0, 12, 20])
In [78]: eights = np.where(arr[:,5])[0]
In [79]: eights
Out[79]: array([2, 3], dtype=int32)
In [80]: arr[eights,:]
Out[80]: 
array([[ 7,  8,  9, 10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19, 20, 21, 22]])
In [81]: fours = np.where(arr[:,5]==0)[0]
In [82]: arr[fours,:]
Out[82]: 
array([[1, 2, 3, 4, 0, 0, 0, 0],
       [3, 4, 5, 6, 0, 0, 0, 0]])

或者带有布尔掩码

In [83]: mask = arr[:,5]>0
In [84]: arr[mask,:]
Out[84]: 
array([[ 7,  8,  9, 10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19, 20, 21, 22]])
In [85]: arr[~mask,:]
Out[85]: 
array([[1, 2, 3, 4, 0, 0, 0, 0],
       [3, 4, 5, 6, 0, 0, 0, 0]])

从某种意义上说，您很幸运有这个清晰的0标记.一些有限元代码重复节点编号以减少数量，例如[1,2,3,3]适用于4节点系统中的3节点元素.但是在那种情况下，即使将2个节点合并为1个，其余的数学工作也很好.

You are lucky, in a sense, to have this clear 0 marker. Some finite element code duplicates node numbers to reduce the number, e.g. [1,2,3,3] for a 3 node element in a 4 node system. But in those cases the rest of the math works fine, even when you merge 2 nodes into one.

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