Python 如何将数组的前 N 个元素设置为零？
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How to set first N elements of array to zero?
提问by user3059201
I have a long array:
我有一个长数组：
x= ([2, 5, 4, 7, ...])
for which I need to set the first Nelements to 0. So for N = 2, the desired output would be:
为此，我需要将第一个N元素设置为0. 因此，对于N = 2，所需的输出将是：
x = ([0, 0, 4, 7, ...])
Is there an easy way to do this in Python? Some numpyfunction?
有没有一种简单的方法可以在 Python 中做到这一点？一些numpy功能？
采纳答案by syntonym
Pure python:
纯蟒蛇：
x[:n] = [0] * n
with numpy:
与麻木：
y = numpy.array(x)
y[:n] = 0
also note that x[:n] = 0does notwork if xis a python list (instead of a numpy array).
还注意到，x[:n] = 0确实没有，如果工作x是一个Python列表（而不是numpy的阵列）。
It is also a bad idea to use [{some object here}] * nfor anything mutable, because the list will not contain n different objects but n references to the same object:
[{some object here}] * n用于任何可变对象也是一个坏主意，因为列表将不包含 n 个不同的对象，而是 n 个对同一对象的引用：
>>> a = [[],[],[],[]]
>>> a[0:2] = [["a"]] * 2
>>> a
[['a'], ['a'], [], []]
>>> a[0].append("b")
>>> a
[['a', 'b'], ['a', 'b'], [], []]
回答by jhoepken
Just set them explicitly:
只需明确设置它们：
x[0:2] = 0
