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?
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
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)))
其他回答
我自己是一个相对的python新手,但过去几周我一直在使用嵌套字典,这就是我想到的。
你应该尝试使用堆栈。将根字典中的键变成一个列表的列表:
stack = [ root.keys() ] # Result: [ [root keys] ]
按照从最后到第一个的相反顺序,查找字典中的每个键,看看它的值是否(也是)一个字典。如果不是,打印密钥,然后删除它。但是,如果键的值是一个字典,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,对每个新的键列表进行递归重复。
如果每个列表中第二个键的值是一个字典,那么在几轮之后,你会得到这样的结果:
[['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\t乘以堆栈的长度:
indent = "\t" * len(stack)
缺点是为了检查每个键,你需要散列到相关的子字典,尽管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
注意,这种方法将要求清除尾随的空列表,并删除后跟空列表的任何列表中的最后一个键(当然,这可能会创建另一个空列表,等等)。
还有其他方法来实现这个方法,但希望这能给你一个基本的想法。
编辑:如果您不想进行所有这些操作,pprint模块将以良好的格式打印嵌套字典。
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.
yapf的另一个选择:
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]
}
}
我的第一个想法是JSON序列化器可能很擅长嵌套字典,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
}
}
}
下面是我根据sth的注释写的函数。它的工作原理与json相同。转储与缩进,但我使用制表符而不是缩进的空间。在Python 3.2+中,您可以直接将缩进指定为'\t',但在2.7中不能。
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 '}'
Ex:
>>> 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"
}
}