Question

我有一个现有的词典列表如下：

FA = [{u'child': [{u'cdesc': u'Audit'},
                  {u'cdesc': u'Equity Research'},
                  {u'cdesc': u'Finance / Accounts / Tax'},
                  {u'cdesc': u'Investment Banking / M&A'}],
       u'pdesc': u'Finance / Accounts / Investment Banking',
       u'pid': 10007}]

我想把它转换成这样的东西：

FA = {u'Audit':2,
     u'Equity Research':2,
     u'Finance / Accounts / Tax':2,
     u'Investment Banking / M&A':2,
     u'Finance / Accounts / Investment Banking':2}

我可以使用嵌套循环轻松完成此操作，其代码如下所示。有没有办法使用词典理解来做到这一点？

a = dict()
for fa in FA:
    a.update({slugify(fa['pdesc']):2})


    for c in fa['child']: 
        a.update({slugify(c['cdesc']):2})

Answer 1

字典理解在这里看起来很丑陋......无论如何......

# METHOD 1
FA_dict1 = {d:2 for v in FA[0][u'child'] for d in v.values()}
FA_dict1.update({FA[0][u'pdesc']: 2})


# METHOD 2
from itertools import chain
FA_dict = {d:2 for v in FA[0][u'child'] for d in chain(v.values(), [FA[0][u'pdesc']])}


# METHOD 3
FA_DICT = {d:2 for v in FA[0][u'child'] for d in list(v.values())+[FA[0][u'pdesc']]}

Answer 2

理解解决方案确实是一只丑小鸭！

在上述数据上运行：

{ k:2 for k in reduce(lambda x,y: x+y, [ [ chld[u'cdesc'] for chld in FA[0]['child'] ], [ FA[0][u'pdesc'] ] ] ) }

{u'Audit': 2,
 u'Equity Research': 2,
 u'Finance / Accounts / Investment Banking': 2,
 u'Finance / Accounts / Tax': 2,
 u'Investment Banking / M&A': 2}

Answer 3

这会让你开始吗？

{ chld[u'cdesc']:2 for chld in FA[0]['child'] }

产生（不是100％但接近）：

{u'Audit': 2,
 u'Equity Research': 2,
 u'Finance / Accounts / Tax': 2,
 u'Investment Banking / M&A': 2}

字典理解中的嵌套循环

3 个答案: