NumPy - 迭代2D列表和打印（行，列）索引

Question

使用NumPy和/或Pandas處理2D列表時遇到困難：

1. 獲得所有元素的唯一組合的sum ，而無需再次從同一行中選擇（下面的數組應該是81種組合）。

2. 打印組合中每個元素的行和列。

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)]

Answer 1

通過創建所有這些組合和求和的方法：這是使用itertools.product和array-indexing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)

樣品運行 -

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])

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記憶效率方法：這是一種方法，沒有創建所有這些組合，而是使用動態broadcasted總結，哲學很受this other post啟發 -

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

Answer 2

您可以使用itertools product函數：

from itertools import product    
y = [sum(p) for p in product(*arr)]

len(y)
# 81

列表較小的示例：

arr = [[1,2],[3,4],[5,6]]
[sum(p) for p in product(*arr)]
# [9, 10, 10, 11, 10, 11, 11, 12]