通过带有np.where函数的数组添加Numpy Python [英] Adding through an array with np.where function Numpy Python

问题描述

我正在尝试编写一个函数，它一次添加一个 Sequence ，直到达到 Number .因此，由于 Sequence 数组的总和为21，而不是45，因此会将 Sequence 3的第一个元素添加到 Sequence 变成 [3，7，11，3] .由于 [3、7、11、3] 等于24且小于45，因此要在Sequence中添加下一个元素，即7，因此 Sequence 会更新为 [3、7、11、3、7] ，依此类推，直到 Number>np.sum([Sequence] is is not return True .下面的 np.where 函数有问题，我该如何解决?

I am trying to write a function where it adds the Sequence one at a time until the Number is reached. So since the the sum of the Sequence array is 21 and not 45 it will add the first element of the Sequence 3 to the Sequence so it becomes [3, 7, 11, 3]. Since [3, 7, 11, 3] is equal to 24 and is lower than 45 the next element in the Sequence is going to be added which is 7, the Sequence is updated to be [3, 7, 11, 3, 7] and so on until Number> np.sum([Sequence] is does not return True. The np.where function below is faulty how can I fix it?

代码:

Number = 45
Number2= 46
Sequence = np.array([3, 7, 11])
p = 0
np.where(Number> np.sum([Sequence]),np.append(Sequence,Sequence[p], p+= 1))
np.where(Number2> np.sum([Sequence]),np.append(Sequence,Sequence[p], p+= 1))

预期产量

Number = [3,7,11,3,7,11,3]
Number2 = [3,7,11,3,7,11,3,7]

推荐答案

我不认为这可以作为一个整体来完成. np.where 的第二个和第三个参数应该是序列.作为循环很简单:

I'm not convinced this can be done as a one-liner. The second and third parameters to np.where are supposed to be sequences. It's easy as a loop:

seq45 = []
for n in itertools.cycle(Sequence):
    seq45.append(n)
    if sum(seq45) >= 45: break

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