轉換嵌套列表嵌套元組

Question

請我如何能遍歷嵌套列表得到的元組的嵌套列表出來的，例如循環盆栽得到的rslt

pot = [[1,2,3,4],[5,6,7,8]] 

我試圖

b = [] 

for i in pot: 
    for items in i: 
     b = zip(pot[0][0:],pot[0][1:]) 

但沒有得到所期望的輸出由於

希望的結果=

rslt = [[(1,2),(3,4)],[(5,6),(7,8)]] 

Answer 1

基礎上grouperrecipe in the itertools documentation，你可以嘗試這樣的事情（假設你的子表的長度已指示）：

>>> def grouper(iterable, n): 
    args = [iter(iterable)] * n # creates a list of n references to the same iterator object (which is exhausted after one iteration) 
    return zip(*args) 

現在你可以測試一下：

>>> pot = [[1,2,3,4],[5,6,7,8]] 
>>> rslt = [] 
>>> for sublist in pot: 
    rslt.append(grouper(sublist, 2)) 
>>> rslt 
[[(1, 2), (3, 4)], [(5, 6), (7, 8)]] 

Answer 2

可以也使用列表理解：

[[(a, b), (c, d)] for a, b, c, d in l]