写这篇文章时
Python新手。
之所以出现，是因为我希望用户能够从目录(以及任何子目录)中选择一组文件，不幸的是，Tkinter在文件对话框中选择多个文件的默认功能在Windows 7上被破坏了(http://bugs.python.org/issue8010)。
所以我尝试用另一种方法(仍然使用Tkinter)来表示目录结构:构造一个目录结构的副本，由标记和缩进的复选框组成(以树状结构组织)。目录如下:
SomeRootDirectory
foo.txt
bar.txt
Stories
Horror
scary.txt
Trash
notscary.txt
Cyberpunk
Poems
doyoureadme.txt
看起来像这样(其中#代表一个复选按钮):
SomeRootDirectory
# foo.txt
# bar.txt
Stories
Horror
# scary.txt
Trash
# notscary.txt
Cyberpunk
Poems
# doyoureadme.txt
根据目录结构构建原始字典很容易，使用我在ActiveState上找到的某个配方(见下文)，但是当我试图遍历剩下的嵌套良好的字典时，我遇到了瓶颈。
这是一个打印所有文件名的函数。它遍历字典中的所有键，如果它们映射到不是字典的东西(在你的例子中是文件名)，我们打印出名称。否则，在映射到的字典上调用函数。
def print_all_files(directory):
for filename in directory.keys():
if not isinstance(directory[filename], dict):
print filename
else:
print_all_files(directory[filename])
所以这段代码可以修改为你想做的任何事情，但它只是一个例子，如何避免通过使用递归来固定深度。
要理解的关键是，每次调用print_all_files时，它不知道它在树中的深度。它只是查看那里的文件，并打印出名称。如果有目录，它就在上面运行。
这是一个初步代码。看一遍，告诉我你遇到的问题在哪里。
Parents={-1:"Root"}
def add_dir(level, parent, index, k):
print "Directory"
print "Level=%d, Parent=%s, Index=%d, value=%s" % (level, Parents[parent], index, k)
def add_file(parent, index, k):
print "File"
print "Parent=%s, Index=%d, value=%s" % (Parents[parent], index, k)
def f(level=0, parent=-1, index=0, di={}):
for k in di:
index +=1
if di[k]:
Parents[index]=k
add_dir(level, parent, index, k)
f(level+1, index, index, di[k])
else:
add_file(parent, index, k)
a={
'SomeRootDirectory': {
'foo.txt': None,
'bar.txt': None,
'Stories': {
'Horror': {
'scary.txt' : None,
'Trash' : {
'notscary.txt' : None,
},
},
'Cyberpunk' : None
},
'Poems' : {
'doyoureadme.txt' : None
}
}
}
f(di=a)
我知道这是一个老问题，但我只是在寻找一种简单、干净的方法来遍历嵌套字典，这是我有限的搜索中最接近的东西。如果你想要的不仅仅是文件名，那么Oadams的答案就不够有用了，而spicavigo的答案看起来很复杂。
我最终只是滚动我自己的行为类似于os。Walk处理目录，除了它返回所有键/值信息。
它返回一个迭代器，对于"树"中的每个目录对于嵌套字典，迭代器返回(path, subdicts, values)，其中:
- 路径是字典 的路径。
- subdicts是该字典 中每个子字典的(key,dict)对的元组。
- values是该字典 中每个(非字典)项的(键，值)对的元组。
def walk(d):
'''
Walk a tree (nested dicts).
For each 'path', or dict, in the tree, returns a 3-tuple containing:
(path, sub-dicts, values)
where:
* path is the path to the dict
* sub-dicts is a tuple of (key,dict) pairs for each sub-dict in this dict
* values is a tuple of (key,value) pairs for each (non-dict) item in this dict
'''
# nested dict keys
nested_keys = tuple(k for k in d.keys() if isinstance(d[k],dict))
# key/value pairs for non-dicts
items = tuple((k,d[k]) for k in d.keys() if k not in nested_keys)
# return path, key/sub-dict pairs, and key/value pairs
yield ('/', [(k,d[k]) for k in nested_keys], items)
# recurse each subdict
for k in nested_keys:
for res in walk(d[k]):
# for each result, stick key in path and pass on
res = ('/%s' % k + res[0], res[1], res[2])
yield res
import simplejson as json
from collections import defaultdict
# see https://gist.github.com/2012250
tree = lambda: defaultdict(tree)
def walk(d):
'''
Walk a tree (nested dicts).
For each 'path', or dict, in the tree, returns a 3-tuple containing:
(path, sub-dicts, values)
where:
* path is the path to the dict
* sub-dicts is a tuple of (key,dict) pairs for each sub-dict in this dict
* values is a tuple of (key,value) pairs for each (non-dict) item in this dict
'''
# nested dict keys
nested_keys = tuple(k for k in d.keys() if isinstance(d[k],dict))
# key/value pairs for non-dicts
items = tuple((k,d[k]) for k in d.keys() if k not in nested_keys)
# return path, key/sub-dict pairs, and key/value pairs
yield ('/', [(k,d[k]) for k in nested_keys], items)
# recurse each subdict
for k in nested_keys:
for res in walk(d[k]):
# for each result, stick key in path and pass on
res = ('/%s' % k + res[0], res[1], res[2])
yield res
# use fancy tree to store arbitrary nested paths/values
mem = tree()
root = mem['SomeRootDirectory']
root['foo.txt'] = None
root['bar.txt'] = None
root['Stories']['Horror']['scary.txt'] = None
root['Stories']['Horror']['Trash']['notscary.txt'] = None
root['Stories']['Cyberpunk']
root['Poems']['doyoureadme.txt'] = None
# convert to json string
s = json.dumps(mem, indent=2)
#print mem
print s
print
# json.loads converts to nested dicts, need to walk them
for (path, dicts, items) in walk(json.loads(s)):
# this will print every path
print '[%s]' % path
for key,val in items:
# this will print every key,value pair (skips empty paths)
print '%s = %s' % (path+key,val)
print
输出如下:
{
"SomeRootDirectory": {
"foo.txt": null,
"Stories": {
"Horror": {
"scary.txt": null,
"Trash": {
"notscary.txt": null
}
},
"Cyberpunk": {}
},
"Poems": {
"doyoureadme.txt": null
},
"bar.txt": null
}
}
[/]
[/SomeRootDirectory/]
/SomeRootDirectory/foo.txt = None
/SomeRootDirectory/bar.txt = None
[/SomeRootDirectory/Stories/]
[/SomeRootDirectory/Stories/Horror/]
/SomeRootDirectory/Stories/Horror/scary.txt = None
[/SomeRootDirectory/Stories/Horror/Trash/]
/SomeRootDirectory/Stories/Horror/Trash/notscary.txt = None
[/SomeRootDirectory/Stories/Cyberpunk/]
[/SomeRootDirectory/Poems/]
/SomeRootDirectory/Poems/doyoureadme.txt = None
可以使用递归遍历嵌套字典
def walk_dict(dictionary):
for key in dictionary:
if isinstance(dictionary[key], dict):
walk_dict(dictionary[key])
else:
#do something with dictionary[k]
pass
希望有帮助
a={
'SomeRootDirectory': {
'foo.txt': None,
'bar.txt': None,
'Stories': {
'Horror': {
'scary.txt' : None,
'Trash' : {
'notscary.txt' : None,
},
},
'Cyberpunk' : None
},
'Poems' : {
'doyoureadme.txt' : None
}
}
}
def dict_paths(dictionary, level=0, parents=[], paths=[]):
for key in dictionary:
parents = parents[0:level]
paths.append(parents + [key])
if dictionary[key]:
parents.append(key)
dict_paths(dictionary[key], level+1, parents, paths)
return paths
dp = dict_paths(a)
for p in dp:
print '/'.join(p)