python - 如何在字典中循环提取值？

Question

我正在刮一页。我做了两个loops，但第一个循环只transcription_price取值而不是最后两个。为什么以及如何解决这个问题？

def start_requests(self):
    links = {'transcription_page': 'https://www.rev.com/freelancers/transcription',
             'captions_page': 'https://www.rev.com/freelancers/captions',
             'subtitles_page': 'https://www.rev.com/freelancers/subtitles'
            }
    call = [self.parse_transcription,self.parse_caption,self.parse_subtitles]

    for link in links.values():
        for n in range(0,3):
            return [scrapy.Request(link, callback=call[n])]

标签： pythonscrapy

Answer 1

Because return statement, well, returns the value and terminates¹ the function, passing the control flow to the caller. This way, your inner loop is terminated before it goes over all the values.

Perhaps what you wanted was yield:

>>> def f():
...  for x in (1, 2, 3):
...   yield x
...
>>> list(f())
[1, 2, 3]

Besides, using unnamed constants is a way to plant a bug which is often not so obvious, not to say non-Pythonic:

items = ["a", "b", "c"]

# will make an incomplete round when `items` is longer than 3 elements
def poor():
  for i in xrange(0, 3):
    yield items[i]

# will do just alright
def appropriate():
  for item in items:
    yield item

¹ Unless you are in try/except/finally block, in which case finally is always executed before return takes place:

def return_one():
  try:
    1/0
  except ZeroDivisionError:
    return 0
  finally:
    return 1