如何从numpy数组中删除一些特定的元素?假设我有
import numpy as np
a = np.array([1,2,3,4,5,6,7,8,9])
然后我想从a
中删除3,4,7
。我只知道值的索引(index=[2,3,6]
)。
发布于 2016-04-07 03:33:33
有一个numpy内置函数可以帮助你做到这一点。
import numpy as np
>>> a = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> b = np.array([3,4,7])
>>> c = np.setdiff1d(a,b)
>>> c
array([1, 2, 5, 6, 8, 9])
发布于 2019-04-26 17:43:13
要按值删除:
modified_array = np.delete(original_array, np.where(original_array == value_to_delete))
发布于 2012-06-12 20:06:15
不是一个麻木的人,我尝试了一下:
>>> import numpy as np
>>> import itertools
>>>
>>> a = np.array([1,2,3,4,5,6,7,8,9])
>>> index=[2,3,6]
>>> a = np.array(list(itertools.compress(a, [i not in index for i in range(len(a))])))
>>> a
array([1, 2, 5, 6, 8, 9])
根据我的测试,它的性能优于numpy.delete()
。我不知道为什么会出现这种情况,可能是因为初始数组太小了?
python -m timeit -s "import numpy as np" -s "import itertools" -s "a = np.array([1,2,3,4,5,6,7,8,9])" -s "index=[2,3,6]" "a = np.array(list(itertools.compress(a, [i not in index for i in range(len(a))])))"
100000 loops, best of 3: 12.9 usec per loop
python -m timeit -s "import numpy as np" -s "a = np.array([1,2,3,4,5,6,7,8,9])" -s "index=[2,3,6]" "np.delete(a, index)"
10000 loops, best of 3: 108 usec per loop
这是一个非常显著的差异(与我预期的相反),有人知道为什么会这样吗?
更奇怪的是,向numpy.delete()
传递一个列表比遍历列表并为其提供单个索引的性能更差。
python -m timeit -s "import numpy as np" -s "a = np.array([1,2,3,4,5,6,7,8,9])" -s "index=[2,3,6]" "for i in index:" " np.delete(a, i)"
10000 loops, best of 3: 33.8 usec per loop
编辑:这似乎与数组的大小有关。对于大型数组,numpy.delete()
的速度要快得多。
python -m timeit -s "import numpy as np" -s "import itertools" -s "a = np.array(list(range(10000)))" -s "index=[i for i in range(10000) if i % 2 == 0]" "a = np.array(list(itertools.compress(a, [i not in index for i in range(len(a))])))"
10 loops, best of 3: 200 msec per loop
python -m timeit -s "import numpy as np" -s "a = np.array(list(range(10000)))" -s "index=[i for i in range(10000) if i % 2 == 0]" "np.delete(a, index)"
1000 loops, best of 3: 1.68 msec per loop
显然,这一切都是非常无关紧要的,因为你应该总是追求清晰,避免重复发明轮子,但我发现它有点有趣,所以我想我把它留在这里。
https://stackoverflow.com/questions/10996140
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