跳过循环中的多次迭代

问题描述：

我有一个循环中的列表，我想在look到达后跳过 3 个元素。在这个答案中提出了一些建议，但我没有很好地利用它们：

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
for sing in song:
    if sing == 'look':
        print sing
        continue
        continue
        continue
        continue
        print 'a' + sing
    print sing

当然，四次continue是无稽之谈，使用四次也next()没有用。

输出应如下所示：

always
look
aside
of
life

解决方案 1：

for用于iter(song)循环；您可以在自己的代码中执行此操作，然后在循环内推进迭代器；iter()再次调用可迭代对象将只返回相同的可迭代对象，因此您可以在循环内推进可迭代对象，并for在下一次迭代中继续进行。

next()使用函数推进迭代器；它在 Python 2 和 3 中都能正常工作，而无需调整语法：

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
    print sing
    if sing == 'look':
        next(song_iter)
        next(song_iter)
        next(song_iter)
        print 'a' + next(song_iter)

通过print sing上移队列，我们​​也可以避免重复。

如果可迭代对象超出值，则使用next()此方式可能会引发异常。StopIteration

next()您可以捕获该异常，但提供第二个参数（即默认值）以忽略异常并返回默认值会更容易：

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
    print sing
    if sing == 'look':
        next(song_iter, None)
        next(song_iter, None)
        next(song_iter, None)
        print 'a' + next(song_iter, '')

我会用它itertools.islice()来跳过 3 个元素；节省重复next()调用：

from itertools import islice

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
song_iter = iter(song)
for sing in song_iter:
    print sing
    if sing == 'look':
        print 'a' + next(islice(song_iter, 3, 4), '')

可迭代对象islice(song_iter, 3, 4)将跳过 3 个元素，然后返回第 4 个元素，然后完成。调用next()该对象即可从中检索第 4 个元素song_iter()。

演示：

>>> from itertools import islice
>>> song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
>>> song_iter = iter(song)
>>> for sing in song_iter:
...     print sing
...     if sing == 'look':
...         print 'a' + next(islice(song_iter, 3, 4), '')
... 
always
look
aside
of
life

解决方案 2：

>>> song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
>>> count = 0
>>> while count < (len(song)):
    if song[count] == "look" :
        print song[count]
        count += 4
        song[count] = 'a' + song[count]
        continue
    print song[count]
    count += 1

Output:

always
look
aside
of
life

解决方案 3：

我认为，使用迭代器就很好了，next这里：

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
it = iter(song)
while True:
    word = next(it, None)
    if not word:
       break
    print word
    if word == 'look':
        for _ in range(4): # skip 3 and take 4th
            word = next(it, None)
        if word:
            print 'a' + word

或者，使用异常处理（正如@Steinar 注意到的，它更短也更强大）：

it = iter(song)
while True:
    try:
        word = next(it)
        print word
        if word == 'look':
            for _ in range(4):
                word = next(it)
            print 'a' + word 
    except StopIteration:
        break

解决方案 4：

您可以在不使用 iter() 的情况下完成此操作，也可以仅使用额外的变量：

skipcount = -1
song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
for sing in song:
    if sing == 'look' and skipcount <= 0:
        print sing
        skipcount = 3
    elif skipcount > 0:
        skipcount = skipcount - 1
        continue
    elif skipcount == 0:
        print 'a' + sing
        skipcount = skipcount - 1
    else:
        print sing
        skipcount = skipcount - 1

解决方案 5：

当然你也可以使用三次（这里我实际上使用了四次）

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
it = iter(song)
for sing in it:
    if sing == 'look':
        print sing
        try:
           sing = it.next(); sing = it.next(); sing = it.next(); sing=it.next()
        except StopIteration:
             break
        print 'a'+sing
    else:
        print sing

然后

always
look
aside
of
life

解决方案 6：

其实，使用 .next() 三次并不是无意义的。当您想要跳过 n 个值时，请调用 next() n+1 次（不要忘记将最后一次调用的值赋给某个​​值），然后“调用” continue。

要获取您发布的代码的精确副本：

song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']
songiter = iter(song)
for sing in songiter:
  if sing == 'look':
    print sing
    songiter.next()
    songiter.next()
    songiter.next()
    sing = songiter.next()
    print 'a' + sing
    continue
  print sing

解决方案 7：

我相信以下代码对我来说是最简单的。

# data list
song = ['always', 'look', 'on', 'the', 'bright', 'side', 'of', 'life']

# this is one possible way
for sing in song:
    if sing != 'look'\n            and sing != 'always' \n            and sing != 'side' \n            and sing != 'of'\n            and sing != 'life':
        continue
    if sing == 'side':
        sing = f'a{sing}'  # or sing = 'aside'
    print(sing)

# this is another possible way
songs_to_keep = ['always', 'look', 'of', 'side', 'of', 'life']
songs_to_change = ['side']
for sing in song:
    if sing not in songs_to_keep:
        continue
    if sing in songs_to_change:
        sing = f'a{sing}'
    print(sing)

这会产生您所寻找的结果。

always
look
aside
of
life