如何简化通过索引搜索列表项？

Question

我有一个简单的程序：

pos = [1,2]
searched = [
    [1,3,4,6],
    [2,6,7,8],
    [0,1,2,8],
    [5,6,9,2]
]
print(searched[pos[0]][pos[1]])

现在我想要的是一些摆脱 searched[pos[0]][pos[1]] 的方法，只需输入类似 searched[[pos]].

的内容

有什么办法吗，还是每次都要写出来

I have gotten a lot of suggestions, but what I am searching for is a way to do this in one neat line, simplifying everything.
That means that things like using a function, converting to a specific dictionary or even enumerate don't work for me, so for anyone looking at this post later:
.
I suggest using np.array(variable) while defining said variable so you can use variable[pos]

Answer 1

不知道你是否可以接受，但是一个函数就可以了：

def search_for_list_item_by_index(a_list, row, col):
    return a_list[row][col]

print(search_for_list_item_by_index(searched, 1, 2))

这会按预期打印 7。

Answer 2

您可以将其转换为 numpy 数组并进行搜索：

import numpy as np
pos = [1,2] 
searched = np.array([ 
  [1,3,4,6], 
  [2,6,7,8], 
  [0,1,2,8], 
  [5,6,9,2] 
  ]) 
print(searched[1,2])  
# 7

Answer 3

您可以按照@oppressionslayer 的建议将数组转换为 numpy。另一种方法是创建字典并按如下方式使用它：

pos = [1,2]
searched = [
    [1,3,4,6],
    [2,6,7,8],
    [0,1,2,8],
    [5,6,9,2]
]
m=4 # width of the searched array
n=4 # hight of the searched array

searched = {(i,j):searched[i][j] for j in range(m) for i in range(n)}

print(searched[1,2]) # prints 7
print(searched[tuple(pos)]) # prints 7

希望对您有所帮助！！

Answer 4

您可以使用以下功能：

def func(idx, lst):
    for i in idx:
        lst = lst[i]
    return lst

func(pos, searched)
# 7

Answer 5

我会使用字典方法但有趣的是您可以使用 enumerate:

泛化到非矩形列表

searched = {(row,col):item
              for row,elems in enumerate(searched)
                for col,item in enumerate(elems)}

或使用做同样的事情，但适用于完全任意的嵌套，包括字典。可能不是你想要的方向但值得一提。

data = flatten_data(searched)
pos = [1,2]
print(data[1,2])
print(data[tuple(pos)])

如何简化通过索引搜索列表项？

how do I simplify searching for a list item by an index?

python

list

simplify

python-3.x