Question

我想在列表中的当前和下一个元素上执行一个函数，比如print。当你到达列表中的最后一个元素时，如何在不收到错误的情况下执行此操作？

some_list = ["one", "two", "three"]

期望的输出：

curr_item: "one"
next_item: "two"

curr_item: "two"
next_item: "three"

curr_item: "three"
next_item: "end"

我尝试过：

index = 0
for item in list_name[:-1]:
    print item, list_name[index+1]
    index = index+1

Answer 1

您可以压缩列表：

some_list = ["one", "two", "three"]

for cur, nxt in zip (some_list, some_list [1:] ):
    print (cur, nxt)

或者如果您想要end值：

for cur, nxt in zip (some_list, some_list [1:] + ['end'] ):
    print (cur, nxt)

Answer 2

您可以使用itertools.izip_longest：

>>> from itertools import izip_longest
>>> some_list = ["one", "two", "three"]
>>> for x, y in izip_longest(some_list, some_list[1:], fillvalue='end'):
    print 'cur_item', x
    print 'next_item', y


cur_item one
next_item two
cur_item two
next_item three
cur_item three
next_item end

Answer 3

我们可以使用列表索引迭代到倒数第二个元素。可以使用len和range函数计算列表索引，如下所示：

for i in range(len(some_list)-1):
   print some_list[i], some_list[i+1]

输出：

one two
two three

Answer 4

您可以使用itertools中的成对配方，例如：

from itertools import izip_longest, tee

def pairwise(iterable):
    fst, snd = tee(iterable)
    next(snd, None)
    return izip_longest(fst, snd, fillvalue='end')

for fst, snd in pairwise(['one', 'two', 'three']):
    print fst, snd

#one two
#two three
#three end

Answer 5

单线解决方案

[(cur, nxt) for cur, nxt in zip(some_list, some_list[1:])]

Answer 6

您可以尝试以下内容：

for i in range(len(some_list)):
    print("curr_item: " + some_list[i])
    if i < len(some_list)-1:
        print("next_item: " + some_list[i+1])
    else:
        print('next_item: "end"')

Answer 7

您要去的地方

some_list = ["one", "two", "three"]

for f, r in enumerate(some_list):
try:
    print('curr_item : ',some_list[f])
    print('next_item : ',some_list[f+1])
    print()

except IndexError :
    pass

输出：

curr_item :  one
next_item :  two

curr_item :  two
next_item :  three

curr_item :  three

Answer 8

使用链表来评估列表

class Node:
    def __init__(self, dataval=None):
       self.dataval = dataval
       self.nextval = None
       self.previousval=None

class LinkedList:
    def __init__(self):
       self.headval = None

    def listprint(self):
       node = self.headval
       while node is not None:
           if node.previousval!=None:
                print("Previous Node",node.previousval.dataval)
           else:
                print("Previous Node is None")
           print("CurrentNode", node.dataval)
           print("\n")
           node = node.nextval        
    
list1 = LinkedList()
some_list = ["one", "two", "three"]
for item in some_list:
    if list1.headval==None:
        currentNode=Node(item)
        list1.headval = currentNode
        currentNode.previousval=None
        previousNode=currentNode
    else:
        currentNode=Node(item)
        currentNode.previousval=previousNode
        previousNode.nextval=currentNode
        previousNode=currentNode
    
list1.listprint()

输出：

Previous Node is None
CurrentNode one

Previous Node one
CurrentNode two

Previous Node two
CurrentNode three

Python - 对当前元素和列表中的下一个元素做些什么？

8 个答案: