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如何从两端迭代序列?

iterable iteration python-3.x python

greybeard · 技术社区 · 7 年前

int start = 2, beyond = 7;
for (int forward = start, back = beyond ; forward <= --back ; forward++)
    printf("%d %d\n", forward, back);

我似乎无法决定我最不喜欢以下哪种变体:

seq = tuple(chr(c) for c in range(ord('a'), ord('z')+1))
start, beyond = 2, 7

print("subtract")
for forward in range(start, beyond):
    back = start + beyond - 1 - forward
    if (back < forward):
        break
    print(seq[forward], seq[back])
#    if (back <= forward): XXX ERROR, as YvesgereY pointed out
#        break
print("reversed")
for forward, back in zip(range(start, beyond),
                         reversed(range(start, beyond))):
    if (back < forward):
        break
    print(seq[forward], seq[back])
print("step -1")
for forward, back in zip(range(start, beyond), range(beyond-1, start-1, -1)):
    if (back < forward):
        break
    print(seq[forward], seq[back])
print("naked code")
forward, back = start, beyond-1 # thanks again, YvesgereY
while (forward <= back):
    print(seq[forward], seq[back])
    forward += 1
    back -= 1

有没有一种令人愉快/理智/惯用的方法来处理这个问题?

5 回复 | 直到 7 年前

cs95 7 年前

为什么不呢? zip

mid = (len(lst) + 1) // 2   
for x, y in zip(lst[:mid], lst[::-1]):
     print((x, y))

(1, 5)
(2, 4)
(3, 3)

拉链

YvesgereY 7 年前

我宁愿写:

size = max((beyond-start+1)//2,0)  # 0 if start>beyond
for i in range(0,size):
   print(start+i,beyond-i-1)

substract 和 step -1 版本不正确,因为它们允许 forward>back 在最后一次迭代中。

while 强调循环条件的版本(减少一个错误的可能性):

forward, back = start, beyond-1
while forward <= back:
    print(forward, back)
    forward += 1
    back -= 1

Chris 7 年前

在我看来,最后一种使用while循环的方法是最好的:

forward, back = start, beyond
while (forward < back):
    back -= 1
    print(forward, back)
    forward += 1

它简洁但仍然可读,具有清晰的语义,与其他具有多个级别的函数调用嵌套的方法相比,它似乎是最“Pythonic”的方法。您可以通过将其包装在生成器中进行改进:

def iterate_to_middle(start_front, start_end):
    while start_front < start_end:
        start_end -= 1
        yield start_front, start_end
        start_front += 1

>>> for indexes in iterate_to_middle(2, 7):
    print(indexes)


(2, 6)
(3, 5)
(4, 4)

Moses Koledoye 7 年前

while 回路正常。

负索引 enumerate 要为每个转发项生成负的对应索引:

lst = [1, 2, 3, 4, 5]

for i, x in enumerate(lst, 1):
    print(x, lst[-i])
    if i * 2 >= len(lst): break

(1, 5)
(2, 4)
(3, 3)

Wes Williams 3 年前

可以使用~运算符获取(索引+1)*-1。

arr = [1, 2, 3, 4, 5]
for i in range((len(arr) + 1) // 2):
    print(arr[i], arr[~i])

1 5
2 4
3 3