python – NumPy – 迭代2D列表和打印(行,列)索引

使用NumPy和/或Pandas处理2D列表时遇到困难：

>获取所有元素的唯一组合的总和,而无需再次从同一行中选择(下面的数组应该是81种组合).
>打印组合中每个元素的行和列.

例如：

arr = [[1, 2, 4], [10, 3, 8], [16, 12, 13], [14, 4, 20]]

(1,3,12,20), Sum = 36 and (row, col) =  [(0,0),(1,1),(2,1),(3,2)]

(4,10,16,20), Sum = 50 and (row, col) =[(0,2),(1,0),(2,0),(3,2)]

解决方法:

通过创建所有这样的组合和求和的方法：这是使用itertools.product和数组索引的矢量化方法 –

from itertools import product

a = np.asarray(arr)  # Convert to array for ease of use and indexing
m,n = a.shape
combs = np.array(list(product(range(n), repeat=m)))
out = a[np.arange(m)[:,None],combs.T].sum(0)

样品运行 –

In [296]: arr = [[1, 2, 4], [10, 3, 8], [16, 12, 13], [14, 4, 20]]

In [297]: a = np.asarray(arr)
     ...: m,n = a.shape
     ...: combs = np.array(list(product(range(n), repeat=m)))
     ...: out = a[np.arange(m)[:,None],combs.T].sum(0)
     ...: 

In [298]: out
Out[298]: 
array([41, 31, 47, 37, 27, 43, 38, 28, 44, 34, 24, 40, 30, 20, 36, 31, 21,
       37, 39, 29, 45, 35, 25, 41, 36, 26, 42, 42, 32, 48, 38, 28, 44, 39,
       29, 45, 35, 25, 41, 31, 21, 37, 32, 22, 38, 40, 30, 46, 36, 26, 42,
       37, 27, 43, 44, 34, 50, 40, 30, 46, 41, 31, 47, 37, 27, 43, 33, 23,
       39, 34, 24, 40, 42, 32, 48, 38, 28, 44, 39, 29, 45])

记忆效率方法：这是一种不创造所有这些组合的方法,而是使用即时broadcasted总结,其理念深受this other post的启发 –

a = np.asarray(arr)
m,n = a.shape
out = a[0]
for i in range(1,m):
    out = out[...,None]  + a[i]
out.shape = out.size # Flatten