我有这样的字典:
dirDict = {"DIR1" : {
"DIR11" : {
"DIR111" : "Maki111",
"DIR112" : "Maki112"
},
"DIR12" : "Maki12",
"DIR13" : {
"DIR131" : "Maki131"
}
}
}
想象一下这就像一个文件夹结构。我希望与os.walk
类似的文件夹结构。像这样:
["DIR1/DIR11/DIR111/Maki111",
"DIR1/DIR11/DIR112/Maki112",
"DIR1/DIR12/Maki12",
"DIR1/DIR13/DIR131/Maki131"]
所以它基本上是字典值的所有路径。我用递归函数尝试了很多方法,但我迷路了。
这是我最新的试用版:
def walk(input_dict, path_string = "", result = ""):
for key, value in input_dict.items():
if isinstance(value, dict):
path_string += "/" + key
print "==== DICT ====", "\nkey: ", key, "\nvalue: ", value, "\n\t\tpath_string: ", path_string
result = walk(value, path_string)
print "\t\t\t\tresulting: ", result
elif isinstance(value, str):
print "==== NOT DICT ===="
path_string += "/" + value
print "\t\tpath_string: ", path_string, "\nvalue: ", value
return path_string
else:
path_string = "/" + key
result += "\n" + result
return result
答案 0 :(得分:3)
使用Python 3:
dirDict = {"DIR1" : {
"DIR11" : {
"DIR111" : "Maki111",
"DIR112" : "Maki112"
},
"DIR12" : "Maki12",
"DIR13" : {
"DIR131" : "Maki131"
}
}
}
def recurse(d, prefix=None, sep='/'):
if prefix is None:
prefix = []
for key, value in d.items():
if isinstance(value, dict):
yield from recurse(value, prefix + [key])
else:
yield sep.join(prefix + [key, value])
print(list(recurse(dirDict)))
输出:
['DIR1/DIR13/DIR131/Maki131', 'DIR1/DIR11/DIR111/Maki111', 'DIR1/DIR11/DIR112/Maki112', 'DIR1/DIR12/Maki12']
答案 1 :(得分:1)
def walk(d, path):
paths = []
if len(d) == 0:
return path
for k, v in d.iteritems():
child_path = path + k + '/'
if isinstance(v, basestring):
paths.append(child_path + v)
else:
paths.extend(walk(v, child_path))
return paths
答案 2 :(得分:0)
我在https://gist.github.com/nvie/f304caf3b4f1ca4c3884#gistcomment-1597937发布的walk
函数可以用作问题的帮助:
def walk(obj, parent_first=True):
# Top down?
if parent_first:
yield (), obj
# For nested objects, the key is the path component.
if isinstance(obj, dict):
children = obj.items()
# For nested lists, the position is the path component.
elif isinstance(obj, (list, tuple)):
children = enumerate(obj)
# Scalar values have no children.
else:
children = []
# Recurse into children
for key, value in children:
for child_path, child in walk(value, parent_first):
yield (key,) + child_path, child
# Bottom up?
if not parent_first:
yield (), obj
您的问题可以通过以下方式解决:
for path, value in walk(obj):
if isinstance(value, str): # leaf node
path_with_value = path + (value,)
print("/".join(path_with_value))
答案 3 :(得分:0)
具有列表理解的紧凑型解决方案:
def f(v):
if isinstance(v, dict):
return dict_to_list(v)
elif isinstance(v, list):
return v
else:
return [v]
def dict_to_list(d):
return ['{}/{}'.format(k, i) for k, v in d.items() for i in f(v)]
lst = dict_to_list(dirDict)
lst.sort()
print('\n'.join(lst))