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

这是我在编写一个需要在.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,如果我们有一个这样的字典:

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': '',
    },
}

其他回答

我把sth的答案稍微修改一下,以适应我的嵌套字典和列表的需要:

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
                                            ...

晚了,但是这个答案不需要任何额外的库。类似于STH的答案,但在格式上更健壮一点,并返回一个完整的字符串,然后可以打印:

def pretty_print_dict(
        input_dictionary,
        indent=1,
        depth=0
):
    # Bool flag to add comma's after first item in dict.
    needs_comma = False
    # String for any dict will start with a '{'
    return_string = '\t' * depth + '{\n'
    # Iterate over keys and values, building the full string out.
    for key, value in input_dictionary.items():
        # Start with key. If key follows a previous item, add comma.
        if needs_comma:
            return_string = return_string + ',\n' + '\t' * (depth + 1) + str(key) + ': '
        else:
            return_string = return_string + '\t' * (depth + 1) + str(key) + ': '
        # If the value is a dict, recursively call function.
        if isinstance(value, dict):
            return_string = return_string + '\n' + pretty_print_dict(value, depth=depth+2)
        else:
            return_string = return_string + '\t' * indent + str(value)
        # After first line, flip bool to True to make sure commas make it.
        needs_comma = True
    # Complete the dict with a '}'
    return_string = return_string + '\n' + '\t' * depth + '}'
    # Return dict string.
    return return_string

让我们看看它如何处理像test_dict={1,2,3:{4:{5:6}, 7:8}, 9:10}这样的字典。

字符串的样子:“{\ n \ t1: \ t2, t3: \ n \ \ n \ t \ {\ n \ t \ \ t4: \ n \ t \ t \ \ {\ n \ t \ t \ \ \ t5: \ t6 \ n \ t \ t \ \ t}, \ n \ t \ \ t7: \ t8 \ n \ t \ t}, \ n \ t9: \ t10 \ n}”。

打印该字符串会得到:

{
    1:  2,
    3: 
        {
            4: 
                {
                    5:  6
                },
            7:  8
        },
    9:  10
}

这里的现代解决方案是使用富。安装与

pip install rich

使用as

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)

输出很好地缩进:

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))

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。