Vfrdtyky

I have a Numpy matrix:

M = [[55, 5],

 [56, 3],

 [57, 7],

 [58, 9],

 [59, 3],

 [60, 8],

 [61, 1]] 

I want to aggregate by group_size (for example into 3 groups):

group_size = math.ceil(M.size/groups) # math.ceil(7/3) = 3

Each aggregated row has a left value being the first left value from the group, and the right value being the sum of all right values from the group.

Expected output:

R = [[55, 15], # 55 first left column value of first group, 15 sum of all right values in group 

 [58, 20], # 58 first left column value of second group, 20 sum of all right values in group

 [61, 1]] # Third group consist only of one row, remainder

Is there an efficient way to solve this with Numpy without looping?

Here's one way with NumPy:

n = 3

x = M[::n, 0]

y = np.add.reduceat(M[:, 1], np.arange(0, M.shape[0], n))



R = np.vstack((x, y)).T



print(R)



array([[55, 15],

       [58, 20],

       [61,  1]])

pandas solution should be use agg with first and sum:

group_size = 3

df = pd.DataFrame(M).groupby(np.arange(len(M)) // group_size).agg({0:'first',1:'sum'})

print (df)

    0   1

0  55  15

1  58  20

2  61   1



a = np.array(df.values.tolist())

print(a)

[[55 15]

 [58 20]

 [61  1]]

A solution using Python:

from operator import itemgetter



M = [[55, 5],

     [56, 3],

     [57, 7],

     [58, 9],

     [59, 3],

     [60, 8],

     [61, 1]]

it = (M[e:e+3] for e in range(0, len(M), 3))

print([[e[0][0], sum(map(itemgetter(1), e))] for e in it])

Output

[[55, 15], [58, 20], [61, 1]]

a = np.array([[2, 3],[5, 6],[7, 9]])

b = numpy.zeros(shape=(len(a[0])))

for i in a:

    b=b+i

    print(b)