Python -- 如何对当前元素和下一个元素进行操作？

Question

我想对列表中的当前元素和下一个元素执行一个函数，比如打印。请问怎么做才能在到达列表的最后一个元素时不出错呢？

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"

我尝试过的：

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

Answer 1

给你看看：

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

输出结果：

curr_item :  one
next_item :  two

curr_item :  two
next_item :  three

curr_item :  three

Answer 2

你可以使用itertools里的成对配方，像这样：

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 3

我们可以通过列表的索引来遍历到倒数第二个元素。列表的索引可以通过计算列表的长度和使用范围函数来得到，方法如下：

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

输出结果：

one two
two three

Answer 4

你可以使用 itertools.izip_longest 来实现这个功能：

>>> 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 5

你可以把这个列表压缩成一个更小的形式：

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

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

或者如果你想要获取end这个值的话：

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