我在解决遇到的问题时遇到了麻烦.
I am having some trouble with solving a problem I encountered.
我有一个包含价格的数组:
I have an array with prices:
>>> x = np.random.randint(10, size=10) array([6, 1, 7, 6, 9, 0, 8, 2, 1, 8])以及(随机)生成的Poisson分布式到达数组:
And a (randomly) generated array of Poisson distributed arrivals:
>>> arrivals = np.random.poisson(1, size=10) array([4, 0, 1, 1, 3, 2, 1, 3, 2, 1])每个单次到达都应与相同指数的价格相关联.因此，在上述情况下，第一个元素(x [0])应该被选择4次(y [0]).第二个元素(x [1])应该被选择0次(y [1])...因此结果应该是:
Each single arrival should be associated with the price at the same index. So in the case above, the first element ( x[0] ) should be selected 4 times ( y[0] ). The second element ( x[1] ) should be selected 0 times ( y[1] )... The result thus should be:
array([6, 6, 6, 6, 7, 6, 9, 9, 9, 0, 0, 8, 2, 2, 2, 1, 1, 8])是否有任何(快速)方法可以完成此任务，而无需遍历数组?任何帮助将不胜感激.
Is there any (fast) way to accomplish this, without iterating over the arrays? Any help would be greatly appreciated.
推荐答案您可以使用 np.repeat :
In [43]: x = np.array([6, 1, 7, 6, 9, 0, 8, 2, 1, 8]) In [44]: arrivals = np.array([4, 0, 1, 1, 3, 2, 1, 3, 2, 1]) In [45]: np.repeat(x, arrivals) Out[45]: array([6, 6, 6, 6, 7, 6, 9, 9, 9, 0, 0, 8, 2, 2, 2, 1, 1, 8])但请注意，对于某些计算，可能有必要避免形成此中间数组.例如，请参见 scipy.stats.binned_statistic .
but note that for certain calculations, it might be possible to avoid having to form this intermediate array. See for example, scipy.stats.binned_statistic.