如何在Python中打印深度为~4的字典?我尝试使用pprint()
进行相当打印,但它不起作用:
import pprint
pp = pprint.PrettyPrinter(indent=4)
pp.pprint(mydict)
我只想为每个嵌套添加缩进("\t"
),以便得到这样的结果:
key1
value1
value2
key2
value1
value2
等。
我该怎么做?
答案 0 :(得分:415)
我的第一个想法是JSON序列化程序可能非常擅长嵌套字典,所以我会欺骗并使用它:
>>> import json
>>> print json.dumps({'a':2, 'b':{'x':3, 'y':{'t1': 4, 't2':5}}},
... sort_keys=True, indent=4)
{
"a": 2,
"b": {
"x": 3,
"y": {
"t1": 4,
"t2": 5
}
}
}
答案 1 :(得分:114)
我不确定你想要格式化的样子,但你可以从这样的函数开始:
def pretty(d, indent=0):
for key, value in d.items():
print('\t' * indent + str(key))
if isinstance(value, dict):
pretty(value, indent+1)
else:
print('\t' * (indent+1) + str(value))
答案 2 :(得分:40)
您可以通过YAML尝试PyYAML。它的输出可以微调。我建议从以下开始:
print yaml.dump(data, allow_unicode=True, default_flow_style=False)
结果是非常可读;如果需要,它也可以解析回Python。
修改强>
示例:
>>> import yaml
>>> data = {'a':2, 'b':{'x':3, 'y':{'t1': 4, 't2':5}}}
>>> print yaml.dump(data, default_flow_style=False)
a: 2
b:
x: 3
y:
t1: 4
t2: 5
答案 3 :(得分:31)
至于已经完成的工作,我没有看到任何漂亮的打印机,至少模仿python解释器的输出,格式非常简单,所以这里是我的:
class Formatter(object):
def __init__(self):
self.types = {}
self.htchar = '\t'
self.lfchar = '\n'
self.indent = 0
self.set_formater(object, self.__class__.format_object)
self.set_formater(dict, self.__class__.format_dict)
self.set_formater(list, self.__class__.format_list)
self.set_formater(tuple, self.__class__.format_tuple)
def set_formater(self, obj, callback):
self.types[obj] = callback
def __call__(self, value, **args):
for key in args:
setattr(self, key, args[key])
formater = self.types[type(value) if type(value) in self.types else object]
return formater(self, value, self.indent)
def format_object(self, value, indent):
return repr(value)
def format_dict(self, value, indent):
items = [
self.lfchar + self.htchar * (indent + 1) + repr(key) + ': ' +
(self.types[type(value[key]) if type(value[key]) in self.types else object])(self, value[key], indent + 1)
for key in value
]
return '{%s}' % (','.join(items) + self.lfchar + self.htchar * indent)
def format_list(self, value, indent):
items = [
self.lfchar + self.htchar * (indent + 1) + (self.types[type(item) if type(item) in self.types else object])(self, item, indent + 1)
for item in value
]
return '[%s]' % (','.join(items) + self.lfchar + self.htchar * indent)
def format_tuple(self, value, indent):
items = [
self.lfchar + self.htchar * (indent + 1) + (self.types[type(item) if type(item) in self.types else object])(self, item, indent + 1)
for item in value
]
return '(%s)' % (','.join(items) + self.lfchar + self.htchar * indent)
初始化它:
pretty = Formatter()
它可以支持为已定义的类型添加格式化程序,您只需要为此创建一个函数,并使用set_formater将其绑定到所需的类型:
from collections import OrderedDict
def format_ordereddict(self, value, indent):
items = [
self.lfchar + self.htchar * (indent + 1) +
"(" + repr(key) + ', ' + (self.types[
type(value[key]) if type(value[key]) in self.types else object
])(self, value[key], indent + 1) + ")"
for key in value
]
return 'OrderedDict([%s])' % (','.join(items) +
self.lfchar + self.htchar * indent)
pretty.set_formater(OrderedDict, format_ordereddict)
由于历史原因,我保留了以前漂亮的打印机,这是一个函数而不是类,但它们都可以使用相同的方式,类版本只允许更多:
def pretty(value, htchar='\t', lfchar='\n', indent=0):
nlch = lfchar + htchar * (indent + 1)
if type(value) is dict:
items = [
nlch + repr(key) + ': ' + pretty(value[key], htchar, lfchar, indent + 1)
for key in value
]
return '{%s}' % (','.join(items) + lfchar + htchar * indent)
elif type(value) is list:
items = [
nlch + pretty(item, htchar, lfchar, indent + 1)
for item in value
]
return '[%s]' % (','.join(items) + lfchar + htchar * indent)
elif type(value) is tuple:
items = [
nlch + pretty(item, htchar, lfchar, indent + 1)
for item in value
]
return '(%s)' % (','.join(items) + lfchar + htchar * indent)
else:
return repr(value)
使用它:
>>> a = {'list':['a','b',1,2],'dict':{'a':1,2:'b'},'tuple':('a','b',1,2),'function':pretty,'unicode':u'\xa7',("tuple","key"):"valid"}
>>> a
{'function': <function pretty at 0x7fdf555809b0>, 'tuple': ('a', 'b', 1, 2), 'list': ['a', 'b', 1, 2], 'dict': {'a': 1, 2: 'b'}, 'unicode': u'\xa7', ('tuple', 'key'): 'valid'}
>>> print(pretty(a))
{
'function': <function pretty at 0x7fdf555809b0>,
'tuple': (
'a',
'b',
1,
2
),
'list': [
'a',
'b',
1,
2
],
'dict': {
'a': 1,
2: 'b'
},
'unicode': u'\xa7',
('tuple', 'key'): 'valid'
}
与其他版本相比:
答案 4 :(得分:10)
最有效的方法之一就是使用已经建立的pprint模块。
定义打印深度所需的参数如您所料depth
import pprint
pp = pprint.PrettyPrinter(depth=4)
pp.pprint(mydict)
就是这样!
答案 5 :(得分:7)
带有yapf
的另一个选项:
from pprint import pformat
from yapf.yapflib.yapf_api import FormatCode
dict_example = {'1': '1', '2': '2', '3': [1, 2, 3, 4, 5], '4': {'1': '1', '2': '2', '3': [1, 2, 3, 4, 5]}}
dict_string = pformat(dict_example)
formatted_code, _ = FormatCode(dict_string)
print(formatted_code)
输出:
{
'1': '1',
'2': '2',
'3': [1, 2, 3, 4, 5],
'4': {
'1': '1',
'2': '2',
'3': [1, 2, 3, 4, 5]
}
}
答案 6 :(得分:4)
正如其他人发布的那样,你可以使用recursion / dfs来打印嵌套的字典数据,如果是字典则可以递归调用;否则打印数据。
def print_json(data):
if type(data) == dict:
for k, v in data.items():
print k
print_json(v)
else:
print data
答案 7 :(得分:3)
我也必须像这样传递default
参数:
print(json.dumps(my_dictionary, indent=4, default=str))
如果您希望对键进行排序,则可以:
print(json.dumps(my_dictionary, sort_keys=True, indent=4, default=str))
为了解决此类型错误:
TypeError: Object of type 'datetime' is not JSON serializable
这是由于日期时间是字典中的某些值造成的。
答案 8 :(得分:3)
您可以使用print-dict
from print_dict import pd
dict1 = {
'key': 'value'
}
pd(dict1)
输出:
{
'key': 'value'
}
this Python代码的输出:
{
'one': 'value-one',
'two': 'value-two',
'three': 'value-three',
'four': {
'1': '1',
'2': '2',
'3': [1, 2, 3, 4, 5],
'4': {
'method': <function custom_method at 0x7ff6ecd03e18>,
'tuple': (1, 2),
'unicode': '✓',
'ten': 'value-ten',
'eleven': 'value-eleven',
'3': [1, 2, 3, 4]
}
},
'object1': <__main__.Object1 object at 0x7ff6ecc588d0>,
'object2': <Object2 info>,
'class': <class '__main__.Object1'>
}
安装:
$ pip install print-dict
披露:我是print-dict
答案 9 :(得分:3)
我拿了sth's answer并略微修改它以满足我对嵌套词典和列表的需求:
def pretty(d, indent=0):
if isinstance(d, dict):
for key, value in d.iteritems():
print '\t' * indent + str(key)
if isinstance(value, dict) or isinstance(value, list):
pretty(value, indent+1)
else:
print '\t' * (indent+1) + str(value)
elif isinstance(d, list):
for item in d:
if isinstance(item, dict) or isinstance(item, list):
pretty(item, indent+1)
else:
print '\t' * (indent+1) + str(item)
else:
pass
然后给我输出如下:
>>>
xs:schema
@xmlns:xs
http://www.w3.org/2001/XMLSchema
xs:redefine
@schemaLocation
base.xsd
xs:complexType
@name
Extension
xs:complexContent
xs:restriction
@base
Extension
xs:sequence
xs:element
@name
Policy
@minOccurs
1
xs:complexType
xs:sequence
xs:element
...
答案 10 :(得分:2)
我尝试了以下方法并得到了我想要的结果
方法一:
第 1 步:通过在 print_dict
cmd
pip install print_dict
第 2 步:将 print_dict 导入为
from print_dict import pd
第 3 步:使用 pd
打印
pd(your_dictionary_name)
示例输出:
{
'Name': 'Arham Rumi',
'Age': 21,
'Movies': ['adas', 'adfas', 'fgfg', 'gfgf', 'vbxbv'],
'Songs': ['sdfsd', 'dfdgfddf', 'dsdfd', 'sddfsd', 'sdfdsdf']
}
方法 2:
我们也可以使用 for
循环使用 items
方法打印字典
for key, Value in your_dictionary_name.items():
print(f"{key} : {Value}")
答案 11 :(得分:1)
这里的现代解决方案是使用 rich。安装
pip install rich
并用作
from rich import print
d = {
"Alabama": "Montgomery",
"Alaska": "Juneau",
"Arizona": "Phoenix",
"Arkansas": "Little Rock",
"California": "Sacramento",
"Colorado": "Denver",
"Connecticut": "Hartford",
"Delaware": "Dover",
"Florida": "Tallahassee",
"Georgia": "Atlanta",
"Hawaii": "Honolulu",
"Idaho": "Boise",
}
print(d)
输出很好缩进:
答案 12 :(得分:1)
通过这种方式,您可以以漂亮的方式打印它,例如您的词典名称是yasin
import json
print (json.dumps(yasin, indent=2))
答案 13 :(得分:1)
pout可以很好地打印您扔给它的任何东西,例如(从另一个答案中借用data
)
data = {'a':2, 'b':{'x':3, 'y':{'t1': 4, 't2':5}}}
pout.vs(data)
将导致输出打印到屏幕上,例如:
{
'a': 2,
'b':
{
'y':
{
't2': 5,
't1': 4
},
'x': 3
}
}
或者您可以返回对象的格式化字符串输出:
v = pout.s(data)
它的主要用例是用于调试,因此它不会阻塞对象实例或任何事物,并且可以处理unicode输出,可以在python 2.7和3中工作。
披露:我是pout的作者和维护者。
答案 14 :(得分:1)
我编写了这个简单的代码来打印Python中json对象的一般结构。
def getstructure(data, tab = 0):
if type(data) is dict:
print ' '*tab + '{'
for key in data:
print ' '*tab + ' ' + key + ':'
getstructure(data[key], tab+4)
print ' '*tab + '}'
elif type(data) is list and len(data) > 0:
print ' '*tab + '['
getstructure(data[0], tab+4)
print ' '*tab + ' ...'
print ' '*tab + ']'
以下数据的结果
a = {'list':['a','b',1,2],'dict':{'a':1,2:'b'},'tuple':('a','b',1,2),'function':'p','unicode':u'\xa7',("tuple","key"):"valid"}
getstructure(a)
非常紧凑,看起来像这样:
{
function:
tuple:
list:
[
...
]
dict:
{
a:
2:
}
unicode:
('tuple', 'key'):
}
答案 15 :(得分:1)
stt.execute("CREATE TABLE IF NOT EXISTS bus"
+ "(id INT(11) NOT NULL AUTO_INCREMENT PRIMARY KEY,"
+ "mac VARCHAR(30) NOT NULL UNIQUE,"
+ "route int(11) NOT NULL,"
+ "latitude FLOAT(10,6) NOT NULL,"
+ "longitude FLOAT(10,6) NOT NULL,"
+ "created_at TIMESTAMP DEFAULT CURRENT_TIMESTAMP)");
stt.execute("CREATE EVENT IF NOT EXISTS AutoDelete "
+ "ON SCHEDULE EVERY 3 MINUTE "
+ "DO "
+ "DELETE FROM bus WHERE created_at < (NOW() - INTERVAL 3 MINUTE)");
stt.execute("SET GLOBAL event_scheduler = ON");
答案 16 :(得分:1)
def pretty(d, indent=0):
for key, value in d.iteritems():
if isinstance(value, dict):
print '\t' * indent + (("%30s: {\n") % str(key).upper())
pretty(value, indent+1)
print '\t' * indent + ' ' * 32 + ('} # end of %s #\n' % str(key).upper())
elif isinstance(value, list):
for val in value:
print '\t' * indent + (("%30s: [\n") % str(key).upper())
pretty(val, indent+1)
print '\t' * indent + ' ' * 32 + ('] # end of %s #\n' % str(key).upper())
else:
print '\t' * indent + (("%30s: %s") % (str(key).upper(),str(value)))
答案 17 :(得分:0)
我在回答sth的答案并做出一个小但非常有用的修改后回到这个问题。此函数打印JSON树中的所有键以及该树中<叶子节点的大小。
def print_JSON_tree(d, indent=0):
for key, value in d.iteritems():
print ' ' * indent + unicode(key),
if isinstance(value, dict):
print; print_JSON_tree(value, indent+1)
else:
print ":", str(type(d[key])).split("'")[1], "-", str(len(unicode(d[key])))
当你拥有大型JSON对象并且想要找出肉的位置时,它真的很棒。 示例强>:
>>> print_JSON_tree(JSON_object)
key1
value1 : int - 5
value2 : str - 16
key2
value1 : str - 34
value2 : list - 5623456
这会告诉您,您关心的大多数数据可能都在JSON_object['key1']['key2']['value2']
内,因为格式化为字符串的值的长度非常大。
答案 18 :(得分:0)
这里可以打印任何类型的嵌套字典,同时跟踪沿途的“父”字典。
dicList = list()
def prettierPrint(dic, dicList):
count = 0
for key, value in dic.iteritems():
count+=1
if str(value) == 'OrderedDict()':
value = None
if not isinstance(value, dict):
print str(key) + ": " + str(value)
print str(key) + ' was found in the following path:',
print dicList
print '\n'
elif isinstance(value, dict):
dicList.append(key)
prettierPrint(value, dicList)
if dicList:
if count == len(dic):
dicList.pop()
count = 0
prettierPrint(dicExample, dicList)
这是一个良好的起点,用于根据不同的格式进行打印,例如OP中指定的格式。您真正需要做的就是围绕 Print 块进行操作。请注意,它会查看值是否为'OrderedDict()'。根据您是否使用Container datatypes Collections中的内容,您应该制作这类失败保护程序,以便 elif 块因其名称而不会将其视为附加字典。截至目前,像
这样的字典example_dict = {'key1': 'value1',
'key2': 'value2',
'key3': {'key3a': 'value3a'},
'key4': {'key4a': {'key4aa': 'value4aa',
'key4ab': 'value4ab',
'key4ac': 'value4ac'},
'key4b': 'value4b'}
将打印
key3a: value3a
key3a was found in the following path: ['key3']
key2: value2
key2 was found in the following path: []
key1: value1
key1 was found in the following path: []
key4ab: value4ab
key4ab was found in the following path: ['key4', 'key4a']
key4ac: value4ac
key4ac was found in the following path: ['key4', 'key4a']
key4aa: value4aa
key4aa was found in the following path: ['key4', 'key4a']
key4b: value4b
key4b was found in the following path: ['key4']
lastDict = list()
dicList = list()
def prettierPrint(dic, dicList):
global lastDict
count = 0
for key, value in dic.iteritems():
count+=1
if str(value) == 'OrderedDict()':
value = None
if not isinstance(value, dict):
if lastDict == dicList:
sameParents = True
else:
sameParents = False
if dicList and sameParents is not True:
spacing = ' ' * len(str(dicList))
print dicList
print spacing,
print str(value)
if dicList and sameParents is True:
print spacing,
print str(value)
lastDict = list(dicList)
elif isinstance(value, dict):
dicList.append(key)
prettierPrint(value, dicList)
if dicList:
if count == len(dic):
dicList.pop()
count = 0
使用相同的示例代码,它将打印以下内容:
['key3']
value3a
['key4', 'key4a']
value4ab
value4ac
value4aa
['key4']
value4b
这不是完全 OP中请求的内容。不同之处在于仍然打印父^ n,而不是缺少并用空格替换。要获得OP的格式,您需要执行以下操作:迭代地将 dicList 与 lastDict 进行比较。您可以通过创建一个新字典并将dicList的内容复制到它来执行此操作,检查复制的字典中的 i 是否与lastDict中的 i 相同,并且 - 如果是是 - 使用字符串乘数函数将空格写入 i 位置。
答案 19 :(得分:0)
使用此功能:
def pretty_dict(d, n=1):
for k in d:
print(" "*n + k)
try:
pretty_dict(d[k], n=n+4)
except TypeError:
continue
这样称呼它:
pretty_dict(mydict)
答案 20 :(得分:0)
这是我在处理需要在.txt文件中编写字典的类时想到的:
@staticmethod
def _pretty_write_dict(dictionary):
def _nested(obj, level=1):
indentation_values = "\t" * level
indentation_braces = "\t" * (level - 1)
if isinstance(obj, dict):
return "{\n%(body)s%(indent_braces)s}" % {
"body": "".join("%(indent_values)s\'%(key)s\': %(value)s,\n" % {
"key": str(key),
"value": _nested(value, level + 1),
"indent_values": indentation_values
} for key, value in obj.items()),
"indent_braces": indentation_braces
}
if isinstance(obj, list):
return "[\n%(body)s\n%(indent_braces)s]" % {
"body": "".join("%(indent_values)s%(value)s,\n" % {
"value": _nested(value, level + 1),
"indent_values": indentation_values
} for value in obj),
"indent_braces": indentation_braces
}
else:
return "\'%(value)s\'" % {"value": str(obj)}
dict_text = _nested(dictionary)
return dict_text
现在,如果我们有这样的字典:
some_dict = {'default': {'ENGINE': [1, 2, 3, {'some_key': {'some_other_key': 'some_value'}}], 'NAME': 'some_db_name', 'PORT': '', 'HOST': 'localhost', 'USER': 'some_user_name', 'PASSWORD': 'some_password', 'OPTIONS': {'init_command': 'SET foreign_key_checks = 0;'}}}
我们这样做:
print(_pretty_write_dict(some_dict))
我们得到:
{
'default': {
'ENGINE': [
'1',
'2',
'3',
{
'some_key': {
'some_other_key': 'some_value',
},
},
],
'NAME': 'some_db_name',
'OPTIONS': {
'init_command': 'SET foreign_key_checks = 0;',
},
'HOST': 'localhost',
'USER': 'some_user_name',
'PASSWORD': 'some_password',
'PORT': '',
},
}
答案 21 :(得分:0)
来自this link:
def prnDict(aDict, br='\n', html=0,
keyAlign='l', sortKey=0,
keyPrefix='', keySuffix='',
valuePrefix='', valueSuffix='',
leftMargin=0, indent=1 ):
'''
return a string representive of aDict in the following format:
{
key1: value1,
key2: value2,
...
}
Spaces will be added to the keys to make them have same width.
sortKey: set to 1 if want keys sorted;
keyAlign: either 'l' or 'r', for left, right align, respectively.
keyPrefix, keySuffix, valuePrefix, valueSuffix: The prefix and
suffix to wrap the keys or values. Good for formatting them
for html document(for example, keyPrefix='<b>', keySuffix='</b>').
Note: The keys will be padded with spaces to have them
equally-wide. The pre- and suffix will be added OUTSIDE
the entire width.
html: if set to 1, all spaces will be replaced with ' ', and
the entire output will be wrapped with '<code>' and '</code>'.
br: determine the carriage return. If html, it is suggested to set
br to '<br>'. If you want the html source code eazy to read,
set br to '<br>\n'
version: 04b52
author : Runsun Pan
require: odict() # an ordered dict, if you want the keys sorted.
Dave Benjamin
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/161403
'''
if aDict:
#------------------------------ sort key
if sortKey:
dic = aDict.copy()
keys = dic.keys()
keys.sort()
aDict = odict()
for k in keys:
aDict[k] = dic[k]
#------------------- wrap keys with ' ' (quotes) if str
tmp = ['{']
ks = [type(x)==str and "'%s'"%x or x for x in aDict.keys()]
#------------------- wrap values with ' ' (quotes) if str
vs = [type(x)==str and "'%s'"%x or x for x in aDict.values()]
maxKeyLen = max([len(str(x)) for x in ks])
for i in range(len(ks)):
#-------------------------- Adjust key width
k = {1 : str(ks[i]).ljust(maxKeyLen),
keyAlign=='r': str(ks[i]).rjust(maxKeyLen) }[1]
v = vs[i]
tmp.append(' '* indent+ '%s%s%s:%s%s%s,' %(
keyPrefix, k, keySuffix,
valuePrefix,v,valueSuffix))
tmp[-1] = tmp[-1][:-1] # remove the ',' in the last item
tmp.append('}')
if leftMargin:
tmp = [ ' '*leftMargin + x for x in tmp ]
if html:
return '<code>%s</code>' %br.join(tmp).replace(' ',' ')
else:
return br.join(tmp)
else:
return '{}'
'''
Example:
>>> a={'C': 2, 'B': 1, 'E': 4, (3, 5): 0}
>>> print prnDict(a)
{
'C' :2,
'B' :1,
'E' :4,
(3, 5):0
}
>>> print prnDict(a, sortKey=1)
{
'B' :1,
'C' :2,
'E' :4,
(3, 5):0
}
>>> print prnDict(a, keyPrefix="<b>", keySuffix="</b>")
{
<b>'C' </b>:2,
<b>'B' </b>:1,
<b>'E' </b>:4,
<b>(3, 5)</b>:0
}
>>> print prnDict(a, html=1)
<code>{
'C' :2,
'B' :1,
'E' :4,
(3, 5):0
}</code>
>>> b={'car': [6, 6, 12], 'about': [15, 9, 6], 'bookKeeper': [9, 9, 15]}
>>> print prnDict(b, sortKey=1)
{
'about' :[15, 9, 6],
'bookKeeper':[9, 9, 15],
'car' :[6, 6, 12]
}
>>> print prnDict(b, keyAlign="r")
{
'car':[6, 6, 12],
'about':[15, 9, 6],
'bookKeeper':[9, 9, 15]
}
'''
答案 22 :(得分:0)
我使用你们教我的东西加上装饰器的力量来重载经典的打印功能。只需根据您的需要更改缩进。我在 github 中将它添加为 gist,以防您想为它加注星标(保存)。
def print_decorator(func):
"""
Overload Print function to pretty print Dictionaries
"""
def wrapped_func(*args,**kwargs):
if isinstance(*args, dict):
return func(json.dumps(*args, sort_keys=True, indent=2, default=str))
else:
return func(*args,**kwargs)
return wrapped_func
print = print_decorator(print)
现在只需像往常一样使用打印。
答案 23 :(得分:-1)
这是我根据什么评论写的函数。它与带缩进的json.dump的工作方式相同,但我使用制表符而不是空格来缩进。在Python 3.2+中,您可以直接将缩进指定为'\ t',但不能在2.7中指定。
def pretty_dict(d):
def pretty(d, indent):
for i, (key, value) in enumerate(d.iteritems()):
if isinstance(value, dict):
print '{0}"{1}": {{'.format( '\t' * indent, str(key))
pretty(value, indent+1)
if i == len(d)-1:
print '{0}}}'.format( '\t' * indent)
else:
print '{0}}},'.format( '\t' * indent)
else:
if i == len(d)-1:
print '{0}"{1}": "{2}"'.format( '\t' * indent, str(key), value)
else:
print '{0}"{1}": "{2}",'.format( '\t' * indent, str(key), value)
print '{'
pretty(d,indent=1)
print '}'
例如:
>>> dict_var = {'a':2, 'b':{'x':3, 'y':{'t1': 4, 't2':5}}}
>>> pretty_dict(dict_var)
{
"a": "2",
"b": {
"y": {
"t2": "5",
"t1": "4"
},
"x": "3"
}
}
答案 24 :(得分:-1)
我自己是一个相对蟒蛇新手,但过去几周我一直在使用嵌套词典,这就是我想出来的。
你应该尝试使用堆栈。将根词典中的键放入列表列表中:
stack = [ root.keys() ] # Result: [ [root keys] ]
从最后到第一个按相反的顺序,查找字典中的每个键,看它的值是否(也)是一个字典。如果没有,请打印密钥然后将其删除。但是,如果键的值是字典,则打印该键,然后将该值的键附加到堆栈的末尾,并以相同的方式开始处理该列表,对每个新的重复递归钥匙清单。
如果每个列表中第二个键的值都是字典,那么在几轮后你会得到类似的东西:
[['key 1','key 2'],['key 2.1','key 2.2'],['key 2.2.1','key 2.2.2'],[`etc.`]]
这种方法的优点是缩进只是堆栈长度的\t
倍:
indent = "\t" * len(stack)
缺点是,为了检查你需要散列到相关子词典的每个键,虽然这可以通过列表理解和简单的for
循环轻松处理:
path = [li[-1] for li in stack]
# The last key of every list of keys in the stack
sub = root
for p in path:
sub = sub[p]
if type(sub) == dict:
stack.append(sub.keys()) # And so on
请注意,此方法将要求您清理尾随空列表,和以删除任何列表中的最后一个键,后跟一个空列表(当然可能会创建另一个空列表,因此上)。
还有其他方法可以实现这种方法,但希望这能为您提供如何实现此方法的基本知识。
编辑:如果您不想完成所有这些操作,pprint
模块会以漂亮的格式打印嵌套字典。