Python NumPy-迭代二维列表和打印(行、列)索引
我很难使用Python NumPy-迭代二维列表和打印(行、列)索引,python,pandas,numpy,Python,Pandas,Numpy,我很难使用NumPy和/或Pandas来处理2D列表: 获取所有元素的唯一组合的和,无需再次从同一行中进行选择(下面的数组应为81个组合) 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)
NumPy
和/或Pandas
来处理2D
列表:
和
,无需再次从同一行中进行选择(下面的数组应为81个组合)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)]
您可以使用
itertools
中的product
功能:
from itertools import product
y = [sum(p) for p in product(*arr)]
len(y)
# 81
列表较小的示例:
arr = [[1,2],[3,4],[5,6]]
[sum(p) for p in product(*arr)]
# [9, 10, 10, 11, 10, 11, 11, 12]
通过创建所有这些组合并求和来实现此方法:这里是一种使用and
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)
样本运行-
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])
节省内存的方法:这里有一种方法,它不需要创建所有这些组合,而是使用即时总结,其理念深受-
我尝试了常规的python
循环。但我需要使用numpy/panda提供的高性能数据结构。我搜索了很多论坛,但找不到如何循环所有元素的组合。谢谢,它没有打印帮助得出那个总数的元素(行、列)索引。我该怎么做。?
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