删除矩阵子视图中的第一个元素 [英] delete the first element in subview of a matrix

问题描述

我有一个像这样的数据集:

I have a dataset like this:

[[0,1],
 [0,2],
 [0,3],
 [0,4],
 [1,5],
 [1,6],
 [1,7],
 [2,8],
 [2,9]]

我需要删除第一列所定义的数据每个子视图的第一元素.因此，首先获取第一列中所有具有0的元素，然后删除第一行:[0,1].然后，我在第一列中获得带有1的元素，并删除第一行[1,5]，下一步我删除[2,8]，依此类推.最后，我想要一个像这样的数据集:

I need to delete the first elements of each subview of the data as defined by the first column. So first I get all elements that have 0 in the first column, and delete the first row: [0,1]. Then I get the elements with 1 in the first column and delete the first row [1,5], next step I delete [2,8] and so on and so forth. In the end, I would like to have a dataset like this:

[[0,2],
 [0,3],
 [0,4],
 [1,6],
 [1,7],
 [2,9]]

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这可以在numpy中完成吗?我的数据集非常大，因此所有元素上的for循环至少需要4分钟才能完成.

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Can this be done in numpy? My dataset is very large so for loops on all elements take at least 4 minutes to complete.

推荐答案

根据要求，提供numpy解决方案:

As requested, a numpy solution:

import numpy as np
a = np.array([[0,1], [0,2], [0,3], [0,4], [1,5], [1,6], [1,7], [2,8], [2,9]])
_,i = np.unique(a[:,0], return_index=True)

b = np.delete(a, i, axis=0)

---

(以上内容经过编辑后加入了@Jaime的解决方案，这是我为后代着想的原始遮罩解决方案)

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(above is edited to incorporate @Jaime's solution, here is my original masking solution for posterity's sake)

m = np.ones(len(a), dtype=bool)
m[i] = False
b = a[m]

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有趣的是，面具似乎更快:

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Interestingly, the mask seems to be faster:

In [225]: def rem_del(a):
   .....:     _,i = np.unique(a[:,0], return_index=True)
   .....:     return np.delete(a, i, axis = 0)
   .....: 

In [226]: def rem_mask(a):
   .....:     _,i = np.unique(a[:,0], return_index=True)
   .....:     m = np.ones(len(a), dtype=bool)
   .....:     m[i] = False
   .....:     return a[m]
   .....: 

In [227]: timeit rem_del(a)
10000 loops, best of 3: 181 us per loop

In [228]: timeit rem_mask(a)
10000 loops, best of 3: 59 us per loop

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