[英]how to move identical elements in numpy array into subarrays
如何有效地将相同元素从已排序的numpy数组移到子数组?
从这里:
import numpy as np
a=np.array([0,0,1,1,1,3,5,5,5])
到这里:
a2=array([[0, 0], [1, 1, 1], [3], [5, 5, 5]], dtype=object)
一种方法是获取移位的位置,在此数字发生变化,并使用这些索引将输入数组拆分为子数组。 为了找到这些索引,您可以使用np.nonzero
差异化阵列上,然后用np.split
一个分裂,就像这样-
np.split(a,np.nonzero(np.diff(a))[0]+1)
样品运行-
In [42]: a
Out[42]: array([2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 6, 6, 6])
In [43]: np.split(a,np.nonzero(np.diff(a))[0]+1)
Out[43]:
[array([2, 2, 2, 2]),
array([3, 3, 3, 3]),
array([4, 4, 4, 4, 4, 4, 4]),
array([5, 5]),
array([6, 6, 6])]
一种方法是使用itertools.groupby
。 范例-
result = np.array([list(g) for _,g in groupby(a)])
这也适用于普通排序列表,而不仅仅是numpy数组。
演示-
In [24]: import numpy as np
In [25]: a=np.array([0,0,1,1,1,3,5,5,5])
In [26]: from itertools import groupby
In [27]: result = np.array([list(g) for _,g in groupby(a)])
In [28]: result
Out[28]: array([[0, 0], [1, 1, 1], [3], [5, 5, 5]], dtype=object)
与其他方法的时间比较-
In [29]: %timeit np.array([list(g) for _,g in groupby(a)])
The slowest run took 6.10 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 9.86 µs per loop
In [30]: %timeit np.split(a,np.where(np.diff(a)>0)[0]+1)
10000 loops, best of 3: 29.2 µs per loop
In [31]: %timeit np.array([list(g) for _,g in groupby(a)])
100000 loops, best of 3: 10.5 µs per loop
In [33]: %timeit np.split(a,np.nonzero(np.diff(a))[0]+1)
The slowest run took 4.32 times longer than the fastest. This could mean that an intermediate result is being cached
10000 loops, best of 3: 25.2 µs per loop
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