对 numpy 数组中的非零元素求和

Question

我想对数组（1-d）的非零元素求和，但对正整数和负整数分别进行求和（它们只能是一和二），并在它们所在的位置显示零。

数组的示例：

array = np.array([0, 0, 0, -1, -1, 0, 1, 2, 1, 1, 0, -1, 0, 1, 1, -1, -2])

输出：

array([0, 0, 0, -2, 0, 5, 0, -1, 0, 2, -3])

我认为我的问题是我不知道如何分离数组中正值和负值的序列。

Answer 1

这是一种方法 -def sum_by_signs(a):&nbsp; &nbsp; m1 = a<0&nbsp; &nbsp; m2 = a>0&nbsp; &nbsp; m0 = a==0 # or ~m1 & ~m2&nbsp; &nbsp; p = np.flatnonzero(m0[:-1] | np.diff(m1) | np.diff(m2))+1&nbsp; &nbsp; return np.add.reduceat(a, np.r_[0,p])或者将该np.r_[0部分带入布尔构造部分 -def sum_by_signs_v2(a):&nbsp; &nbsp; m1 = a<0&nbsp; &nbsp; m2 = a>0&nbsp; &nbsp; m0 = a==0 # or ~m1 & ~m2&nbsp; &nbsp; p = np.flatnonzero(np.r_[True, m0[:-1] | np.diff(m1) | np.diff(m2)])&nbsp; &nbsp; return np.add.reduceat(a, p)解释我们的想法是根据符号变化或当我们遇到一系列 0 时将数组拆分为“岛”，在这种情况下，我们希望将每个元素拆分为单独的元素。通过拆分，可以将其视为列表的列表，如果这样更容易理解的话。现在，游戏是我们如何获得这些分裂。我们需要表示这些岛屿的开始、停止索引的索引。如前所述，存在三种情况，符号从 变为+或-相反，或者 0 序列。因此，布尔掩码的结构用于为这些索引提供一次性切片，以检测从+到 的符号变化-，反之亦然np.diff(m1) | np.diff(m2)。最后一个m0[:-1]是针对 的序列0s。然后将这些指数输入np.add.reduceat以获得间隔求和。样本运行 -In [208]: aOut[208]: array([ 0,&nbsp; 0,&nbsp; 0, -1, -1,&nbsp; 0,&nbsp; 1,&nbsp; 2,&nbsp; 1,&nbsp; 1,&nbsp; 0, -1,&nbsp; 0,&nbsp; 1,&nbsp; 1, -1, -2])In [209]: sum_by_signs(a)Out[209]: array([ 0,&nbsp; 0,&nbsp; 0, -2,&nbsp; 0,&nbsp; 5,&nbsp; 0, -1,&nbsp; 0,&nbsp; 2, -3])In [211]: aOut[211]: array([ 1,&nbsp; 2,&nbsp; 0, -1, -1,&nbsp; 0,&nbsp; 1,&nbsp; 2,&nbsp; 1,&nbsp; 1,&nbsp; 0, -1,&nbsp; 0,&nbsp; 1,&nbsp; 1, -1, -2])In [212]: sum_by_signs(a)Out[212]: array([ 3,&nbsp; 0, -2,&nbsp; 0,&nbsp; 5,&nbsp; 0, -1,&nbsp; 0,&nbsp; 2, -3])In [214]: aOut[214]:&nbsp;array([ 1,&nbsp; 2,&nbsp; 0, -1, -1,&nbsp; 0,&nbsp; 1,&nbsp; 2,&nbsp; 1,&nbsp; 1,&nbsp; 0, -1,&nbsp; 0,&nbsp; 1,&nbsp; 1, -1, -2,&nbsp; &nbsp; &nbsp; &nbsp; 0])In [215]: sum_by_signs(a)Out[215]: array([ 3,&nbsp; 0, -2,&nbsp; 0,&nbsp; 5,&nbsp; 0, -1,&nbsp; 0,&nbsp; 2, -3,&nbsp; 0])

Answer 2

这解决了问题，但肯定有更聪明的方法来做到这一点array = [0, 0, 0, -1, -1, 0, 1, 2, 1, 1, 0, -1, 0, 1, 1, -1, -2]switch = 0while switch == 0:&nbsp; &nbsp; for i in range(len(array)):&nbsp; &nbsp; &nbsp; &nbsp; try:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; array[i+1]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if array[i] > 0 and array[i+1] > 0:&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; array[i] += array[i + 1]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; array.pop(i + 1)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; break&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; elif array[i] < 0 and array[i+1] < 0:&nbsp; &nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; array[i] += array[i + 1]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; array.pop(i + 1)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; break&nbsp; &nbsp; &nbsp; &nbsp; except:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; switch = 1&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; break最后，数组的值为 [0, 0, 0, -2, 0, 5, 0, -1, 0, 2, -3]