如何在Python中漂亮地打印深度为~4的字典?我尝试用pprint()进行漂亮的打印,但它不起作用:

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

我只是想要一个缩进(“\t”)为每个嵌套,这样我就会得到这样的东西:

key1
    value1
    value2
    key2
       value1
       value2

等。

我该怎么做呢?


当前回答

我必须传递默认参数,就像这样:

print(json.dumps(my_dictionary, indent=4, default=str))

如果你想要键排序,你可以这样做:

print(json.dumps(my_dictionary, sort_keys=True, indent=4, default=str))

为了修复此类型错误:

TypeError: Object of type 'datetime' is not JSON serializable

这是由于datetimes是字典中的一些值。

其他回答

我不确定你到底想要什么样的格式,但你可以从这样一个函数开始:

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))

你可以通过PyYAML尝试YAML。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))

结果非常易读;如果需要,还可以解析回Python。

编辑:

例子:

>>> 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

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!

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.

最简单的方法是安装IPython并使用如下所示的方法

from IPython.lib.pretty import pretty


class MyClass:
    __repr__(self):
       return pretty(data)  # replace data with what makes sense

在你的情况下

print(pretty(mydict))