如何迭代 python 中的多维任意列表

Question

我有一个表达式 Tree，它生成树列表。列表因用于计算数字的表达式而异，例如，

T = ['-', ['+', ['-', 75, ['-', 10, 3]], ['-', 100, 50]], 3]

或

T = ['+', ['*', 5, 4] ,['-', 100, ['/', 20, 2] ]]

我想遍历树的每一个元素，并将运算符的索引、所有运算符的列表、数字的索引和数字列表存储在一个数组中。为此，我试图找出一种可以循环到列表中的方法，然后只检查类型是否为字符串，这意味着它是一个运算符，并通过附加将该索引和值存储在相应的数组中，否则如果类型是数字，然后将其与所有数字一起存储在数组中。

我尝试了下面的代码

T = T =  ['-', ['+', ['-', 75, ['-', 10, 3]], ['-', 100, 50]], 3]
for i in range(0,len(T)):
    if type(T[i]) != int:
        for j in range(0,len(T[i])):
            print(T[i][j])
    else:
        print(T[i])

运行代码给出输出：

-
+
['-', 75, ['-', 10, 3]]
['-', 100, 50]
3

我们基本上想要实现的是进入列表，直到我们可以根据列表大小。关于这如何可能的任何想法？

我的回答基本上是：

address list of the operators is  [[0], [1, 0], [1, 1, 0], [1, 1, 2, 0], [1, 2, 0]] 
    
list of the operators is ['-', '+', '-', '-', '-'] 
    
address of the numbers is [[1, 1, 1], [1, 1, 2, 1], [1, 1, 2, 2], [1, 2, 1], [1, 2, 2], [2]] 
    
list of the numbers is [75, 10, 3, 100, 50, 3]

Answer 1

您可以使用递归函数。可以改进以下内容，例如同时处理运算符和数字。

T = ['-', ['+', ['-', 75, ['-', 10, 3]], ['-', 100, 50]], 3]

def evaluate(arg): # not requested, but why not?
    if type(arg) is list:
        return eval(f'{evaluate(arg[1])} {arg[0]} {evaluate(arg[2])}')
    else:
        return arg

def get_ops(arg): # returns (list of operators, list of their indices)
    ops = [arg[0]] # first elem is an operator, so add it
    idx = [[0]] # add this index 0
    for i in (1, 2): # for each position 1 and 2,
        if type(arg[i]) is list: # if NOT scalar
            ops_sub, idx_sub = get_ops(arg[i]) # recurse!
            ops += ops_sub # add the list of ops from the sublist
            idx += [[i] + x for x in idx_sub] # add idx list from the sublist,
                                              # while the position of the sublist
                                              # being PREPENDED to each idx
    return ops, idx

def get_nums(arg): # basically the same goes here
    nums = []
    idx = []
    for i in (1, 2):
        if type(arg[i]) is list:
            nums_sub, idx_sub = get_nums(arg[i])
            nums += nums_sub
            idx += [[i] + x for x in idx_sub]
        else: # if scalar, this IS a number, so add it to the output
            nums.append(arg[i])
            idx.append([i])
    return nums, idx

print(get_ops(T))
print(get_nums(T))
print(evaluate(T))

结果：

(['-', '+', '-', '-', '-'], [[0], [1, 0], [1, 1, 0], [1, 1, 2, 0], [1, 2, 0]])
([75, 10, 3, 100, 50, 3], [[1, 1, 1], [1, 1, 2, 1], [1, 1, 2, 2], [1, 2, 1], [1, 2, 2], [2]])
115

Answer 2

您可以将单个递归生成器函数与 collections.defaultdict 一起使用：

from collections import defaultdict
def get_ops(t, p = []):
   if isinstance(t, int):
      yield ('val', t, p)
   else:
      yield ('op', t[0], p+[0])
      yield from get_ops(t[1], p+[1])
      yield from get_ops(t[2], p+[2])

T, d = ['-', ['+', ['-', 75, ['-', 10, 3]], ['-', 100, 50]], 3], defaultdict(list)
for a, b, c in get_ops(T):
    d[f'{a}_index'].append(c)
    d[f'{a}s'].append(b)

print(dict(d))

输出：

{
  'op_index': [[0], [1, 0], [1, 1, 0], [1, 1, 2, 0], [1, 2, 0]], 
  'ops': ['-', '+', '-', '-', '-'], 
  'val_index': [[1, 1, 1], [1, 1, 2, 1], [1, 1, 2, 2], [1, 2, 1], [1, 2, 2], [2]], 
  'vals': [75, 10, 3, 100, 50, 3]
}

如何迭代 python 中的多维任意列表

How to iterate over an multidimensional arbitrary list in python

python

iterator

for-loop

list

python-3.x