Python 对 numpy 数组中的每个第 n 个条目进行子采样
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subsampling every nth entry in a numpy array
提问by Rich Williams
I am a beginner with numpy, and I am trying to extract some data from a long numpy array. What I need to do is start from a defined position in my array, and then subsample every nth data point from that position, until the end of my array.
我是 numpy 的初学者，我正在尝试从长 numpy 数组中提取一些数据。我需要做的是从数组中定义的位置开始，然后从该位置对每个第 n 个数据点进行子采样，直到数组结束。
basically if I had
基本上如果我有
a = [1,2,3,4,1,2,3,4,1,2,3,4....]
I want to subsample this to start at a[1]and then sample every fourth point from there, to produce something like
我想对它进行二次采样a[1]，然后从那里开始每四个点采样一次，以产生类似的东西
b = [2,2,2.....]
回答by behzad.nouri
You can use numpy's slicing, simply start:stop:step.
您可以简单地使用numpy 的切片start:stop:step。
>>> xs
array([1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4])
>>> xs[1::4]
array([2, 2, 2])
This creates a viewof the the original data, so it's constant time. It'll also reflect changes to the original array and keep the whole original array in memory:
这将创建原始数据的视图，因此它是常数 time。它还将反映对原始数组的更改并将整个原始数组保留在内存中：
>>> a
array([1, 2, 3, 4, 5])
>>> b = a[::2] # O(1), constant time
>>> b[:] = 0 # modifying the view changes original array
>>> a # original array is modified
array([0, 2, 0, 4, 0])
so if either of the above things are a problem, you can make a copy explicitly:
因此，如果上述任何一项都有问题，您可以明确复制：
>>> a
array([1, 2, 3, 4, 5])
>>> b = a[::2].copy() # explicit copy, O(n)
>>> b[:] = 0 # modifying the copy
>>> a # original is intact
array([1, 2, 3, 4, 5])
This isn't constant time, but the result isn't tied to the original array. The copy also contiguous in memory, which can make some operations on it faster.
这不是常数时间，但结果与原始数组无关。副本在内存中也是连续的，这可以使对其进行一些操作更快。
