dictをOrderedDictに変換する

Question

collections.OrderedDictクラスを使用する際に問題が発生します。私は、RaspbianのPython 2.7を使用しています。これは、Raspberry PiのDebianディストリビューションです。私は、テキストアドベンチャーのための比較のために2つの辞書を印刷しようとしています。順序は正確に比較するために不可欠です。 私が何を試しても、辞書は通常の順序どおりに印刷されません。ここで

は、私は私のRPIでそれを行うときに私が得るものです：それは関数呼び出しを印刷し、ネストされたリストにキーと値のグループを入れているので、

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)]) 

は明らかに適切ではない何かがあります...

これは私が私のPCに似た何かを実行することによって得たものである：

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')]) 

今回、それが順序であるが、右けれども、それは他のものを印刷すべきではありませんか？ （リストに入れて関数呼び出しを表示する）

ここでエラーが生じますか？それは単なるLinux版なので、PythonのPyバージョンとは関係ありません。

Answer 1

最初にという辞書を作成し、その辞書をOrderedDictに渡しています。あなたがそれをするときには、注文はもはや正しいとは言えません。 dictは本質的に発注されていません。代わりにタプルの順序で

パス：

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

何がOrderedDictを印刷するときに表示することが表現だし、それは完全に正しいです。 OrderedDict([('PRICE', 250), ('HP', 50), ('NAME', 'Albatross'), ('BLASTERS', 13), ('THRUSTERS', 18)])は、の再生可能なの表現で、内容はOrderedDictの内容を示しています。

Answer 2

など、我々はASCのようなカスタムオーダーではない一般的なものを必要と我々はOrderedDictのために行く時間のほとんど

ここでは、溶液目的とされています

：

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 

これが出力されます

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

注：エントリが削除されると、新しいソート済み辞書はソート順を維持します。しかし、新しいキーが追加されると、キーが最後に追加され、並べ替えが維持されません。

Answer 3

私は古い投稿ですが、私は最近このようなものに対処する必要がありました。

あなたの最終目標が何であるかによって、どちらの回答も良いです。 @Martijnと@Abdulによって提供される2つの回答の間に重要な違いがあることを言いたいと思います。

答えを見ると、辞書の構成方法が異なります。 @Martijnはタプルのリストを使用しますが、@AbdulはPythonで辞書を定義する標準的な方法になります。あなたは、このような複数のレベルで辞書を、持っているとき、これは見ることができます

：

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

あなたが@Abdulで行く場合は、あなたが簡単だけを追跡することにより、サブ辞書を反復処理することができます答えます@Martijnの答えは、あなたがリストを持つdictを与えますが、それは繰り返しが容易ではありません（ループなどが必要です）。

また、アプリの複雑さによって異なります。私は最近、OrderedDictが必要な状況と、どのレベルでもサブディットキーを反復する状況に遭遇しました。 Pythonの辞書の全ポイントは、特定のキーを検索するだけで値に行くことです。

以下の例は、私がいくつかのXMLファイルを解析し、空の辞書に追加することで作成されました。あなたが見るようにdictを取得した後、私は特定の条件のすべてのサブdictをチェックしなければならなかった。

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      "_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") 
    } 
} 

Answer 4

あなたのdictが定義されたコードのこの部分を編集することができない場合、あなたはまだこのように、あなたが望む任意の方法で任意の時点でそれを注文することができます。

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)