如何定義一個具有相同值的多個子字典的Python字典？

我想創建一個Python dict其成員本身dict秒，他們兩個是一樣的：如何定義一個具有相同值的多個子字典的Python字典？

servers = { 
    "test": { 
    "auth_token": "some-auth-token", 
    "client_id": "some-client-id", 
    "client_secret": "some-client-secret", 
    "scope": "some-scope" 
    }, 
    "live": { 
    "auth_token": "auth-token-for-live", 
    "client_id": "client-id-for-live", 
    "client_secret": "client-secret-for-live", 
    "scope": "scope-for-live" 
    }, 
    "demo": servers["test"] # this doesn't work, but this is what I need 
}

我demo服務器定義是一樣的我test服務器定義，但我不希望不得不重複一遍 - 有沒有辦法做到這一點？

來源

2017-03-09 code-tinkerer

在一條語句中無法 – timgeb

讓我們做一些改變，我的演示配置=>'服務器[「示範」] [「client_secret」] =「foo'' =>進入的堆棧溢出問題，」我的測試服務器突然失敗而沒有改變，爲什麼？「我真的不認爲複製粘貼是一件壞事。 – polku

你能先用測試和活鍵創建字典嗎？然後，只需更新與演示作爲關鍵字和測試爲值例如：'servers.update（{'demo'：servers ['test']}）' –

x = { 
    "auth_token": "some-auth-token", 
    "client_id": "some-client-id", 
    "client_secret": "some-client-secret", 
    "scope": "some-scope" 
} 

servers = { 
    "test": x, 

    "live": { 
    "auth_token": "auth-token-for-live", 
    "client_id": "client-id-for-live", 
    "client_secret": "client-secret-for-live", 
    "scope": "scope-for-live" 
    }, 

    "demo": x 

}

例如， https://repl.it/GO6Y/0

來源

2017-03-09 16:28:03 TessellatingHeckler

是的，當然我可以做到這一點;我想知道在「服務器」本身的定義中是否有辦法做到這一點。 –

「我該如何做X？」「像這樣」「當然就是這樣，我實際上是在問一個不同的問題，但保持它的祕密，因爲原因」 – TessellatingHeckler

servers = { 
    "test": { 
    "auth_token": "some-auth-token", 
    "client_id": "some-client-id", 
    "client_secret": "some-client-secret", 
    "scope": "some-scope" 
    }, 
    "live": { 
    "auth_token": "auth-token-for-live", 
    "client_id": "client-id-for-live", 
    "client_secret": "client-secret-for-live", 
    "scope": "scope-for-live" 
    }} 

servers['demo'] = servers['test']

來源

2017-03-09 16:29:07 harryscholes

請注意，更改爲「測試」會自動更新「演示」，反之亦然。 – CaptainKinematics

如何定義一個具有相同值的多個子字典的Python字典？

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