我有一个字典列表,每个字典可以嵌套一个字典列表,如:
mydictlist = [{'name':'foo'}, {'name':'bar','next-level':[{'name':'next-level-foo'}, {'name':'next-level-bar'}] } ]
我试图像这样在列表理解中扁平化诸如名称之类的属性:
flattened = [ *['{}_{}'.format(iter['name'],next['name']) for next in iter] if 'next-level' in iter else '{}'.format(iter['name']) for iter in mydictlist]
得到类似的东西:
['foo', 'bar_next-level-foo', 'bar_next-level-bar']
但是这会导致错误!我可以使用for,如果没有列表理解,也可以这样做(并且已经做到了),但是我想知道使用列表(或元组拆包)和列表理解来做到这一点的正确语法是什么?
mydictlist = [{'name':'foo'}, {'name':'bar','next-level':[{'name':'next-level-foo'}, {'name':'next-level-bar'}] } ]
def flatten(currlist, prefix = None, delimiter = "_"):
l = []
for x in currlist:
print (l)
if "next-level" in x:
if prefix:
prefix = delimiter.join([prefix, x["name"]])
else:
prefix = x["name"]
c = flatten(currlist=x["next-level"], prefix=prefix, delimiter=delimiter)
l.extend(c)
else:
if not prefix:
item = x["name"]
else:
item = delimiter.join([prefix, x["name"]])
l.append(item)
return l
flatten(currlist=mydictlist)
这里是一个相当通用的解决方案,它将以dict和list的任何组合递归。这将返回一个简单的名称值列表。
def get_leafs_as_list(obj, key):
l = []
if isinstance(obj, list):
for v in obj:
l.extend(get_leafs_as_list(v, key))
elif isinstance(obj, dict):
for k, v in obj.items():
if k == key:
l.append(v)
else:
l.extend(get_leafs_as_list(v, key))
return l
mydictlist = [{'name':'foo'}, {'name':'bar','next-level':[{'name':'next-level-foo'}, {'name':'next-level-bar'}] } ]
print(get_leafs_as_list(mydictlist, 'name'))
这将返回:
['foo', 'bar', 'next-level-foo', 'next-level-bar']
result = [x['name'] if "next-level" not in x else [ j['name'] for j in x['next-level']] for x in mydictlist ]
print(result)
# output: ['foo', ['next-level-foo', 'next-level-bar']]
result2 = []
def flettern_list(lst):
for s in lst:
if isinstance(s, str):
result2.append(s)
elif isinstance(s, list):
flettern_list(s)
flettern_list(result)
print(result2) #output: ['foo', 'next-level-foo', 'next-level-bar']
我能做的最接近的是在必要时提供“伪”第二级,然后忽略它:
mydictlist = [{'name':'foo'}, {'name':'bar','next-level':[{'name':'next-level-foo'}, {'name':'next-level-bar'}] } ]
flattened = [ '{}_{}'.format(item['name'],next['name']) if next['name'] else item['name'] for item in mydictlist for next in (item['next-level'] if 'next-level' in item else [{'name':None}]) ]
输出:
['foo', 'bar_next-level-foo', 'bar_next-level-bar']
似乎没有一种方法可以有条件地循环使用列表理解。因此,如果有两个for
-s,它们都将运行,但是可以有条件地生成第二个循环的“主题”(最初我使用lambda进行了此操作,但显然带括号的表达式就足够了。)
您的尝试有一个错误,for next in iter
应该是for next in iter['next-level']
(而且iter()
是内置函数,所以我将其重命名为item
)。而且,如果尝试了一些不太雄心勃勃的操作,您将遇到一条明确的错误消息:无法理解的可重复拆包。