Python &引用;OrderedDict()”的缩写;使用OrderedDict()时自动打印
我试图使用OrderedDict打印一个有序的字典,但是当我打印它时,“OrderedDict”也会打印出来。仅供参考,这只是一段代码,而不是整个代码。我能做些什么来解决这个问题?我正在使用Python 3.2 看起来像这样:Python &引用;OrderedDict()”的缩写;使用OrderedDict()时自动打印,python,python-3.x,dictionary,ordereddictionary,Python,Python 3.x,Dictionary,Ordereddictionary,我试图使用OrderedDict打印一个有序的字典,但是当我打印它时,“OrderedDict”也会打印出来。仅供参考,这只是一段代码,而不是整个代码。我能做些什么来解决这个问题?我正在使用Python 3.2 看起来像这样: def returnAllStats(ints): choices = ["Yes","No"] dictInfo = {"Calories":ints[2], "Servings per Container":ints[0], "Amount per S
def returnAllStats(ints):
choices = ["Yes","No"]
dictInfo = {"Calories":ints[2], "Servings per Container":ints[0], "Amount per Serving":ints[1], "Total Fat":(ints[3]/100)*ints[2], "Saturated Fat":(ints[4]/100)*(ints[3]/100)*ints[2], "Cholesterol":ints[5], "Fiber":ints[6], "Sugar":ints[7], "Protein":ints[8], "Sodium":ints[9], "USA":choices[ints[10]], "Caffeine":ints[11]}
dictInfo = collections.OrderedDict(dictInfo)
return dictInfo
我把它放在文本文件中,这是在写:
('snack', 'bananana')OrderedDict([('USA', 'No'), ('Sodium', 119), ('Calories', 479), ('Servings per Container', 7), ('Sugar', 49), ('Saturated Fat', 37.553599999999996), ('Total Fat', 234.71), ('Cholesterol', 87), ('Amount per Serving', 40), ('Fiber', 1), ('Caffeine', 7), ('Protein', 53)])
谢谢 您有几种选择 您可以使用列表理解并打印:
>>> od
OrderedDict([('one', 1), ('two', 2), ('three', 3)])
>>> [(k,v) for k,v in od.items()]
[('one', 1), ('two', 2), ('three', 3)]
或者,知道顺序可能会改变,如果需要输出,您可以转换为dict:
>>> dict(od)
{'one': 1, 'two': 2, 'three': 3}
(对于Python3.6,常规的dict
。使用Python3.6,顺序不会改变。将来可能会出现这种情况,但还不能保证。)
最后,您可以将OrderDict
子类化,并用所需的格式替换\uuuu str\uuuu
方法:
class Mydict(OrderedDict):
def __str__(self):
return ''.join([str((k, v)) for k,v in self.items()])
>>> md=Mydict([('one', 1), ('two', 2), ('three', 3)])
>>> md # repr
Mydict([('one', 1), ('two', 2), ('three', 3)])
>>> print(md)
('one', '1')('two', '2')('three', '3')
(如果希望repr的输出不同,请更改\uuuu repr\uuu
方法…)
最后说明: 为此:
def returnAllStats(ints):
choices = ["Yes","No"]
dictInfo = {"Calories":ints[2], "Servings per Container":ints[0], "Amount per Serving":ints[1], "Total Fat":(ints[3]/100)*ints[2], "Saturated Fat":(ints[4]/100)*(ints[3]/100)*ints[2], "Cholesterol":ints[5], "Fiber":ints[6], "Sugar":ints[7], "Protein":ints[8], "Sodium":ints[9], "USA":choices[ints[10]], "Caffeine":ints[11]}
dictInfo = collections.OrderedDict(dictInfo)
return dictInfo
实际上,您得到的是一个无序的dict结果,因为您是从无序的dict文本创建OrderedDict的
您可能希望执行以下操作:
def returnAllStats(ints):
choices = ["Yes","No"]
return collections.OrderedDict([("Calories",ints[2]), ("Servings per Container",ints[0]), ("Amount per Serving",ints[1]), ("Total Fat",(ints[3]/100)*ints[2]), ("Saturated Fat",(ints[4]/100)*(ints[3]/100)*ints[2]), ("Cholesterol",ints[5]), ("Fiber",ints[6]), ("Sugar",ints[7]), ("Protein",ints[8]), ("Sodium",ints[9]), ("USA",choices[ints[10]]), ("Caffeine",ints[11])]}
return dictInfo
你有几个选择 您可以使用列表理解并打印:
>>> od
OrderedDict([('one', 1), ('two', 2), ('three', 3)])
>>> [(k,v) for k,v in od.items()]
[('one', 1), ('two', 2), ('three', 3)]
或者,知道顺序可能会改变,如果需要输出,您可以转换为dict:
>>> dict(od)
{'one': 1, 'two': 2, 'three': 3}
(对于Python3.6,常规的dict
。使用Python3.6,顺序不会改变。将来可能会出现这种情况,但还不能保证。)
最后,您可以将OrderDict
子类化,并用所需的格式替换\uuuu str\uuuu
方法:
class Mydict(OrderedDict):
def __str__(self):
return ''.join([str((k, v)) for k,v in self.items()])
>>> md=Mydict([('one', 1), ('two', 2), ('three', 3)])
>>> md # repr
Mydict([('one', 1), ('two', 2), ('three', 3)])
>>> print(md)
('one', '1')('two', '2')('three', '3')
(如果希望repr的输出不同,请更改\uuuu repr\uuu
方法…)
最后说明: 为此:
def returnAllStats(ints):
choices = ["Yes","No"]
dictInfo = {"Calories":ints[2], "Servings per Container":ints[0], "Amount per Serving":ints[1], "Total Fat":(ints[3]/100)*ints[2], "Saturated Fat":(ints[4]/100)*(ints[3]/100)*ints[2], "Cholesterol":ints[5], "Fiber":ints[6], "Sugar":ints[7], "Protein":ints[8], "Sodium":ints[9], "USA":choices[ints[10]], "Caffeine":ints[11]}
dictInfo = collections.OrderedDict(dictInfo)
return dictInfo
实际上,您得到的是一个无序的dict结果,因为您是从无序的dict文本创建OrderedDict的
您可能希望执行以下操作:
def returnAllStats(ints):
choices = ["Yes","No"]
return collections.OrderedDict([("Calories",ints[2]), ("Servings per Container",ints[0]), ("Amount per Serving",ints[1]), ("Total Fat",(ints[3]/100)*ints[2]), ("Saturated Fat",(ints[4]/100)*(ints[3]/100)*ints[2]), ("Cholesterol",ints[5]), ("Fiber",ints[6]), ("Sugar",ints[7]), ("Protein",ints[8]), ("Sodium",ints[9]), ("USA",choices[ints[10]]), ("Caffeine",ints[11])]}
return dictInfo
如果您不关心订单,只需打印
dict(您的Ordereddict)
,如果您关心订单,您可以:
for key, value in yourOrderedDict.items():
print(key, value)
希望它能有所帮助如果您不关心订单,只需打印
dict(您的订单信息)
,如果您关心订单,您可以:
for key, value in yourOrderedDict.items():
print(key, value)
希望对您有所帮助您知道,在您的示例中,
dictInfo
的输出顺序是无序的,因为您首先创建了一个常规dict,然后才创建了一个OrderedDict
,对吗?您知道在您的示例中,dictInfo
的输出顺序是无序的,因为您首先创建了一个常规dict,然后再从中创建一个OrderedDict
,对吗?