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.
其他回答
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.
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': '',
},
}
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!
正如其他人所发布的，你可以使用递归/dfs打印嵌套的字典数据，并递归调用如果它是一个字典;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
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
}
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