如何在while循环中使用列表理解

Question

I have such loop:

ex = [{u'white': []},
 {u'yellow': [u'9241.jpg', []]},
  {u'red': [u'241.jpg', []]},
   {u'blue': [u'59241.jpg', []]}]


for i in ex:
    while not len(i.values()[0]):
        break
    else:
        print i
        break

I need always to return first dict with lenght of values what is higher then 0 but i want to make it with list comprehension

Answer 1

A list comprehension would produce a whole list, while you want just one item .

Use a generator expression instead, and have the next() function iterate to the first value:

next((i for i in ex if i.values()[0]), None)

I've given next() a default to return as well; if there is no matching dictionary, None is returned instead.

Demo:

>>> ex = [{u'white': []},
...  {u'yellow': [u'9241.jpg', []]},
...   {u'red': [u'241.jpg', []]},
...    {u'blue': [u'59241.jpg', []]}]
>>> next((i for i in ex if i.values()[0]), None)
{u'yellow': [u'9241.jpg', []]}

You should, however, rethink your data structure. Dictionaries with just one key-value pair suggest to me you wanted a different type instead; tuples perhaps:

ex = [
    (u'white', []),
    (u'yellow', [u'9241.jpg', []]),
    (u'red', [u'241.jpg', []]),
    (u'blue', [u'59241.jpg', []]),
]