Question

假设NOTICE: UNDEFINED VARIABLE: MENU数组的维和形状相同，如何基于a数组的唯一值来求和b数组的总和？

换句话说，对于数组b的每个值，我希望有一个包含数组a的总和的输出。（在下面的示例中：值1的总和= xxx，值2的总和= yyy ...值11的总和= zzz）

a = [[ 5  1 10 11  6]
     [ 5  3  8 10  9]
     [ 2  1 10  8  7]
     [ 7 10  7  8 11]
     [10 10  3  0 11]]
b = [[508 220 316 557 737]
    [625 419 161 736 426]
    [389 608 760 885 232] 
    [396 309 522 204 842]
    [403 831 225 549 797]]

Answer 1

您可以使用numpy：

import numpy as np

a = np.array(
    [[ 5,  1, 10, 11,  6],
     [ 5,  3,  8, 10,  9],
     [ 2,  1, 10,  8,  7],
     [ 7, 10,  7,  8, 11],
     [10, 10,  3,  0, 11]])
b = np.array(
    [[508, 220, 316, 557, 737],
    [625, 419, 161, 736, 426],
    [389, 608, 760, 885, 232],
    [396, 309, 522, 204, 842],
    [403, 831, 225, 549, 797]])

values = np.unique(a)
# will be [ 0  1  2  3  5  6  7  8  9 10 11]

out = {}
for value in values:
    out[value] = sum(b[np.where(a==value)])

print(out)
# {0: 549, 1: 828, 2: 389, 3: 644, 5: 1133, 6: 737, 7: 1150, 8: 1250, 9: 426, 10: 3355, 11: 2196}

或具有dict理解，全部一行：

out = {value: sum(b[np.where(a==value)]) for value in np.unique(a)}

Answer 2

或手动：

from itertools import chain
from collections import defaultdict

a = [[ 5,  1, 10, 11,  6],
     [ 5,  3,  8, 10,  9],
     [ 2,  1, 10,  8,  7],
     [ 7, 10,  7,  8, 11],
     [10, 10,  3,  0, 11]]
b = [[508, 220, 316, 557, 737],
    [625, 419, 161, 736, 426],
    [389, 608, 760, 885, 232],
    [396, 309, 522, 204, 842],
    [403, 831, 225, 549, 797]]

result = defaultdict(int)

for aa, bb in zip(chain(*a), chain(*b)):
    result[aa] += bb

print(result)

#defaultdict(<class 'int'>, {5: 1133, 1: 828, 10: 3355, 11: 2196, 6: 737, 3: 644, 8: 1250, 9: 426, 2: 389, 7: 1150, 0: 549})

Answer 3

熊猫是直接有效的方法：

df=pd.DataFrame(data=b.ravel(),index=a.ravel()) 
sums=df.groupby(level=0).sum()

#        0
# 0    549
# 1    828
# 2    389
# 3    644
# 5   1133
# 6    737
# 7   1150
# 8   1250
# 9    426
# 10  3355
# 11  2196

基准：

a=np.random.randint(0,10**4,size=10**5)
b=np.random.randint(0,10**6,size=10**5)

In [19]: %timeit pd.DataFrame(b,a).groupby(level=0).sum()
58.7 ms ± 12.6 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [20]: %timeit for aa, bb in zip(a,b):result[aa] += bb
223 ms ± 36.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [21]: %timeit for value in np.unique(a): out[value] = np.sum(b[np.where(a==value)])
5.67 s ± 933 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

如何基于另一个数组的引用值对一个数组的值求和

3 个答案: