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
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:
其他回答
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'):
}
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))
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。
从这个链接:
def prnDict(aDict, br='\n', html=0,
keyAlign='l', sortKey=0,
keyPrefix='', keySuffix='',
valuePrefix='', valueSuffix='',
leftMargin=0, indent=1 ):
'''
return a string representive of aDict in the following format:
{
key1: value1,
key2: value2,
...
}
Spaces will be added to the keys to make them have same width.
sortKey: set to 1 if want keys sorted;
keyAlign: either 'l' or 'r', for left, right align, respectively.
keyPrefix, keySuffix, valuePrefix, valueSuffix: The prefix and
suffix to wrap the keys or values. Good for formatting them
for html document(for example, keyPrefix='<b>', keySuffix='</b>').
Note: The keys will be padded with spaces to have them
equally-wide. The pre- and suffix will be added OUTSIDE
the entire width.
html: if set to 1, all spaces will be replaced with ' ', and
the entire output will be wrapped with '<code>' and '</code>'.
br: determine the carriage return. If html, it is suggested to set
br to '<br>'. If you want the html source code eazy to read,
set br to '<br>\n'
version: 04b52
author : Runsun Pan
require: odict() # an ordered dict, if you want the keys sorted.
Dave Benjamin
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/161403
'''
if aDict:
#------------------------------ sort key
if sortKey:
dic = aDict.copy()
keys = dic.keys()
keys.sort()
aDict = odict()
for k in keys:
aDict[k] = dic[k]
#------------------- wrap keys with ' ' (quotes) if str
tmp = ['{']
ks = [type(x)==str and "'%s'"%x or x for x in aDict.keys()]
#------------------- wrap values with ' ' (quotes) if str
vs = [type(x)==str and "'%s'"%x or x for x in aDict.values()]
maxKeyLen = max([len(str(x)) for x in ks])
for i in range(len(ks)):
#-------------------------- Adjust key width
k = {1 : str(ks[i]).ljust(maxKeyLen),
keyAlign=='r': str(ks[i]).rjust(maxKeyLen) }[1]
v = vs[i]
tmp.append(' '* indent+ '%s%s%s:%s%s%s,' %(
keyPrefix, k, keySuffix,
valuePrefix,v,valueSuffix))
tmp[-1] = tmp[-1][:-1] # remove the ',' in the last item
tmp.append('}')
if leftMargin:
tmp = [ ' '*leftMargin + x for x in tmp ]
if html:
return '<code>%s</code>' %br.join(tmp).replace(' ',' ')
else:
return br.join(tmp)
else:
return '{}'
'''
Example:
>>> a={'C': 2, 'B': 1, 'E': 4, (3, 5): 0}
>>> print prnDict(a)
{
'C' :2,
'B' :1,
'E' :4,
(3, 5):0
}
>>> print prnDict(a, sortKey=1)
{
'B' :1,
'C' :2,
'E' :4,
(3, 5):0
}
>>> print prnDict(a, keyPrefix="<b>", keySuffix="</b>")
{
<b>'C' </b>:2,
<b>'B' </b>:1,
<b>'E' </b>:4,
<b>(3, 5)</b>:0
}
>>> print prnDict(a, html=1)
<code>{
'C' :2,
'B' :1,
'E' :4,
(3, 5):0
}</code>
>>> b={'car': [6, 6, 12], 'about': [15, 9, 6], 'bookKeeper': [9, 9, 15]}
>>> print prnDict(b, sortKey=1)
{
'about' :[15, 9, 6],
'bookKeeper':[9, 9, 15],
'car' :[6, 6, 12]
}
>>> print prnDict(b, keyAlign="r")
{
'car':[6, 6, 12],
'about':[15, 9, 6],
'bookKeeper':[9, 9, 15]
}
'''