如何在列表传输单个元素是列表

Question

我

[["1","2","3"],["7","6","5"]] 

预期输出：

[[[1],[2],[3]],[[7],[6],[5]]] 

Answer 1

您可以使用列表解析来得到这样的结果

data = [["1","2","3"],["7","6","5"]] 
print [[[int(j)] for j in i] for i in data ] 

输出

[[[1], [2], [3]], [[7], [6], [5]]] 

Answer 2

替代@ thefourtheye的方法：

>>> import itertools 
>>> l = [["1","2","3"],["7","6","5"]] 
>>> result = [[int(i)] for i in itertools.chain.from_iterable(l)] 
>>> result 
[[1], [2], [3], [7], [6], [5]] 

其分解，这里是用正常的循环方式：

result = [] 
for i in itertools.chain.from_iterable(l): 
    result.append([int(i)]) 

Answer 3

递归函数不会使输入任何假设：

def list2list(l): 
    for i, e in enumerate(l): 
     if type(e) == list: list2list(e) 
     else: l[i] = [int(e)] 
    return l 

l = [["1","2","3"],["7","6","5"]] 
list2list(l) 
print l 

产品：

[[[1], [2], [3]], [[7], [6], [5]]]