将字典转换为OrderedDict

我在使用collections.OrderedDict类时遇到了一些麻烦。我在Raspbian上使用Python 2.7，这是Raspberry Pi的Debian发行版。我试图打印两个字典，以便进行文本冒险比较（并排）。该顺序对于准确比较至关重要。无论我尝试用他们惯用的无序方式打印字典。将字典转换为OrderedDict

这里就是我得到的，当我做我RPI：

import collections 

ship = {"NAME": "Albatross", 
     "HP":50, 
     "BLASTERS":13, 
     "THRUSTERS":18, 
     "PRICE":250} 

ship = collections.OrderedDict(ship) 

print ship 
# OrderedDict([('PRICE', 250), ('HP', 50), ('NAME', 'Albatross'), ('BLASTERS', 13), ('THRUSTERS', 18)])

显然有什么不对的，因为它是打印函数调用，并把键和值组到嵌套列表...

这是我得到通过运行在我的电脑上类似的东西：

import collections 

Joe = {"Age": 28, "Race": "Latino", "Job": "Nurse"} 
Bob = {"Age": 25, "Race": "White", "Job": "Mechanic", "Random": "stuff"} 

#Just for clarity: 
Joe = collections.OrderedDict(Joe) 
Bob = collections.OrderedDict(Bob) 

print Joe 
# OrderedDict([('Age', 28), ('Race', 'Latino'), ('Job', 'Nurse')]) 
print Bob 
# OrderedDict([('Age', 25), ('Race', 'White'), ('Job', 'Mechanic'), ('Random', 'stuff')])

这一次，这是为了，但它不应该被打印在其他的事情，虽然对不对？（把它放入列表并显示函数调用。）

我在哪里犯我的错误？它不应该与Python的pi版本有关，因为它只是Linux版本。

来源

2013-03-29 pythonpiboy

注：'OrderedDict'按插入顺序排序，而不是字母数字键顺序。 –

142

您正在创建字典第一个，然后将该字典传递给OrderedDict。当你这样做时，排序不再是正确的。 dict本质上不是有序的。

通行证的元组序列来代替：

ship = [("NAME", "Albatross"), 
     ("HP", 50), 
     ("BLASTERS", 13), 
     ("THRUSTERS", 18), 
     ("PRICE", 250)] 
ship = collections.OrderedDict(ship)

你所看到的，当你打印OrderedDict是它的表示，这是完全正确的。 OrderedDict([('PRICE', 250), ('HP', 50), ('NAME', 'Albatross'), ('BLASTERS', 13), ('THRUSTERS', 18)])只是给你看，在重现表示，内容是OrderedDict。

来源

2013-03-29 21:40:59

好吧OrderedDict会为我创建字典吗？此外，要打印没有可重现的表示，我可以打印变量OrderedDict用于？例如： 'print ship' ?? 谢谢:) – pythonpiboy

@pythonpiboy：打印'ship'将打印它引用的值，因此它将打印OrderedDict'表示。 –

问题：按照你所描述的方式，OrderedDict正在制作一个有序的“列表”。这不是破坏目的吗？它只是创建一个元素列表，而不是字典与键和值... – pythonpiboy

多数时候我们去OrderedDict当我们需要一个自定义的顺序不是一个普通的人喜欢ASC等

这里的旨意解决时间：

import collections 
ship = {"NAME": "Albatross", 
     "HP":50, 
     "BLASTERS":13, 
     "THRUSTERS":18, 
     "PRICE":250} 

ship = collections.OrderedDict(ship) 

print ship 


new_dict = collections.OrderedDict() 
new_dict["NAME"]=ship["NAME"] 
new_dict["HP"]=ship["HP"] 
new_dict["BLASTERS"]=ship["BLASTERS"] 
new_dict["THRUSTERS"]=ship["THRUSTERS"] 
new_dict["PRICE"]=ship["PRICE"] 


print new_dict

这将是输出：

OrderedDict([('PRICE', 250), ('HP', 50), ('NAME', 'Albatross'), ('BLASTERS', 13), ('THRUSTERS', 18)]) 
OrderedDict([('NAME', 'Albatross'), ('HP', 50), ('BLASTERS', 13), ('THRUSTERS', 18), ('PRICE', 250)])

注意：当删除条目时新的已排序字典保持其排序顺序。（official doc）

来源

2017-04-28 07:51:54

我知道是一个旧的帖子，但我最近不得不处理类似这样的事情，只是它更复杂。

这两个答案都很好，取决于你的最终目标是什么。我想提一下这是@Martijn和@Abdul提供的两个答案之间的重要区别。

如果你看他们的答案，他们构造字典的方式不同。 @Martijn使用元组列表，而@Abdul则采用更为标准的方式来定义Python中的字典。

在那里你可以看到这是你有多个级别字典时，像这样：

dict = {"key": value, 
      {"key_2": value_2, 
       {"key_3": value_3 } 
      } 
     }

如果你去了@Abdul回答您可以轻松地只是通过跟踪迭代子字典键，但@Martijn的答案会给你一个列表里面的列表，其中不容易迭代（你需要循环等）。

同样，这取决于应用程序的复杂性。我最近遇到了一种情况，我需要一个OrderedDict并在任何级别遍历其子字符键。 Python中的字典的全部重点就是通过搜索特定的密钥来实现价值。

以下示例是由我通过解析几个XML文件并附加到一个空字典构造的。在你看到字典后，我必须检查每个子字典的某些条件。

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       "_id" : "unionhse_key_formats__t_1a0b387adc00cb2e8ab10c36bf56a811d4", 
       "refid" : "union", 
       "type_name" : "keyFormat", 
       "name" : "keyFormat" 
      } 
     }, 
     "member_2" : { 
      "_id" : "structhse_load_plain_key_cmd__t_1a1c4ec69f75f3e8dcd1666ede97d65022", 
      "refid" : "structhse_key_info_flags__t", 
      "type_name" : "hseKeyInfoFlags_t", 
      "name" : "keyInfo", 
      "member_0" : { 
       "_id" : "structhse_key_info_flags__t_1a9cc5afac809e83a7bfdc55531fed9c53", 
       "refid" : "group___h_s_e___types_1ga76a0191e789df1f47b4ac84ea612f7bb", 
       "type_name" : "hseKeyMemTarget_t", 
       "name" : "keyMemTarget" 
      }, 
      "member_1" : { 
       "_id" : "structhse_key_info_flags__t_1ac3c1d57b55a13979d8d521a266ff7f58", 
       "refid" : "group___h_s_e___types_1ga1134b580f8da4de94ca6b1de4d37975e", 
       "type_name" : "uint32", 
       "name" : "keyUsage" 
      }, 
      "member_2" : { 
       "_id" : "structhse_key_info_flags__t_1a46a26062e5eaa0bd1fdf17596409cb56", 
       "refid" : "group___h_s_e___types_1gaa2e9e15da1391eafeeb2c5c57dfee278", 
       "type_name" : "hseKeyRA_t", 
       "name" : "ra" 
      }, 
      "member_3" : { 
       "_id" : "structhse_key_info_flags__t_1a5479c6b76ad1716839561f56bb7c62b8", 
       "refid" : "group___h_s_e___types_1ga3534a0c18a0013fa5f8cbce405ed4713", 
       "type_name" : "hseKeyWA_t", 
       "name" : "wa" 
      } 
     }, 
     "member_3" : { 
      "_id" : "structhse_load_plain_key_cmd__t_1aa901f13e8ba971e748258ca603a8e7fb", 
      "refid" : "group___h_s_e___types_1ga7e573d098b9f292a1794a56b02a05b15", 
      "type_name" : "hseKeyId_t", 
      "name" : "keyId_Addr" 
     }, 
     "last_modified" : ISODate("2017-08-29T07:18:36.796Z") 
    } 
}

来源

2017-08-29 08:03:32

如果您不能编辑的地方被定义你的字典这部分代码仍可以在任何你想要的任何点订购，像这样：

from collections import OrderedDict 

order_of_keys = ["key1", "key2", "key3", "key4", "key5"] 
list_of_tuples = [(key, your_dict[key]) for key in order_of_keys] 
your_dict = OrderedDict(list_of_tuples)

来源

2017-09-27 12:33:16

将字典转换为OrderedDict

回答

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