某物，我觉得很漂亮

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)))

例如，Pout可以漂亮地打印你扔给它的任何东西(借用另一个答案的数据):

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:

{
    'a': 2,
    'b':
    {
        'y':
        {
            't2': 5,
            't1': 4
        },
        'x': 3
    }
}

或者你可以返回对象的格式化字符串输出:

v = pout.s(data)

它的主要用途是调试，因此它不会阻塞对象实例或任何东西，它处理unicode输出，如你所期望的，在python 2.7和3中工作。

披露:我是撅嘴的作者和维护者。

我写了这段简单的代码，用Python打印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'):
}

这里有一些东西可以打印任何类型的嵌套字典，同时跟踪“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)

这是根据不同格式进行打印的一个很好的起点，就像op中指定的那样。你真正需要做的只是围绕打印块进行操作。注意，它将查看该值是否为'OrderedDict()'。这取决于你是否使用容器数据类型集合中的东西，你应该做这些故障保护，这样elif块不会因为它的名字而把它视为一个额外的字典。就像现在，一个例子字典

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']

~修改代码以适应问题的格式~

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

This isn't exactly what is requested in OP. The difference is that a parent^n is still printed, instead of being absent and replaced with white-space. To get to OP's format, you'll need to do something like the following: iteratively compare dicList with the lastDict. You can do this by making a new dictionary and copying dicList's content to it, checking if i in the copied dictionary is the same as i in lastDict, and -- if it is -- writing whitespace to that i position using the string multiplier function.

使用这个函数:

def pretty_dict(d, n=1):
    for k in d:
        print(" "*n + k)
        try:
            pretty_dict(d[k], n=n+4)
        except TypeError:
            continue

这样叫它:

pretty_dict(mydict)

最python化的方法之一是使用已经构建的pprint模块。

定义打印深度所需的参数与您预期的深度相同

import pprint
pp = pprint.PrettyPrinter(depth=4)
pp.pprint(mydict)

就是这样!