python习语:列表理解和项目限制

假设 xml 是一个 ElementTree 对象 findall() 方法返回一个列表,因此只需对该列表进行切片:

limit = 10
limited_artists = xml.findall('artist')[:limit]
results = [xml_to_dict(artist) for artist in limited_artists]

你在用吗? lxml ?可以使用xpath限制查询级别中的项目,例如

>>> from lxml import etree
>>> from io import StringIO
>>> xml = etree.parse(StringIO('<foo><bar>1</bar><bar>2</bar><bar>4</bar><bar>8</bar></foo>'))
>>> [bar.text for bar in xml.xpath('bar[position()<=3]')]
['1', '2', '4']

你也可以 use itertools.islice to limit any iterable ,例如

>>> from itertools import islice
>>> [bar.text for bar in islice(xml.iterfind('bar'), 3)]
['1', '2', '4']
>>> [bar.text for bar in islice(xml.iterfind('bar'), 5)]
['1', '2', '4', '8']

limit = 10
limited_artists = [artist in xml.findall('artist')][:limit]
results = [xml_to_dict(artist) for limited_artists]

这就避免了切片的问题:它不会改变操作的顺序,也不会构建新的列表,如果您过滤列表理解,这对于大列表很重要。

def first(it, count):
    it = iter(it)
    for i in xrange(0, count):
        yield next(it)
    raise StopIteration

print [i for i in first(range(1000), 5)]

它还可以与生成器表达式一起正常工作,因为内存使用,切片将丢失:

exp = (i for i in first(xrange(1000000000), 10000000))
for i in exp:
    print i

对于其他所有因为试图限制从无限生成器返回的项而发现此问题的人:

from itertools import takewhile
ltd = takewhile(lambda x: x[0] < MY_LIMIT, enumerate( MY_INFINITE_GENERATOR ))
# ^ This is still an iterator. 
# If you want to materialize the items, e.g. in a list, do:
ltd_m = list( ltd )
# If you don't want the enumeration indices, you can strip them as usual:
ltd_no_enum = [ v for i,v in ltd_m ]