Example:
The code is as follows:
d = { "root": { "folder2": { "item2": None, "item1": None }, "folder1": { "subfolder1": { "item2": None, "item1": None }, "subfolder2": { "item3": None } } } }The beautiful output is:

1. The keys of the same level are left aligned, that is, the indent is the same.
d = { "root": { "folder2": { "item2": None, "item1": None }, "folder1": { "subfolder1": { "item2": None, "item1": None }, "subfolder2": { "item3": None } } } }3. If value is a dictionary, that is, a nested dictionary, then the nested dictionary is at the next level, and the indent of the key at each level is different.
This is a "splicing string" problem. The elements include ""{}:,\n and space indent.
Traverse each (k, v) key-value pair, splice them together for yield, and recurse when encountering a nested dictionary, that is, recursion + yield.
Upload the code.
#coding=utf-8
def pretty_dict(obj, indent=' '):
def _pretty(obj, indent):
for i, tup in enumerate(obj.items()):
k, v = tup
#如果是字符串则拼上""
if isinstance(k, basestring): k = '"%s"'% k
if isinstance(v, basestring): v = '"%s"'% v
#如果是字典则递归
if isinstance(v, dict):
v = ''.join(_pretty(v, indent + ' '* len(str(k) + ': {')))#计算下一层的indent
#case,根据(k,v)对在哪个位置确定拼接什么
if i == 0:#开头,拼左花括号
if len(obj) == 1:
yield '{%s: %s}'% (k, v)
else:
yield '{%s: %s,\n'% (k, v)
elif i == len(obj) - 1:#结尾,拼右花括号
yield '%s%s: %s}'% (indent, k, v)
else:#中间
yield '%s%s: %s,\n'% (indent, k, v)
print ''.join(_pretty(obj, indent))
d = { "root": { "folder2": { "item2": None, "item1": None }, "folder1": { "subfolder1": { "item2": None, "item1": None }, "subfolder2": { "item3": None } } } }
pretty_dict(d)