Neptune ML 中的 Gremlin 節點分類查詢 - Amazon Neptune
更多節點分類使用歸納推論

本文為英文版的機器翻譯版本，如內容有任何歧義或不一致之處，概以英文版為準。

Neptune ML 中的 Gremlin 節點分類查詢

對於 Neptune ML 中的 Gremlin 節點分類：

  • 模型根據頂點的一個屬性進行訓練。這個屬性的唯一值集被稱為節點類別集，或者簡稱為類別。

  • 頂點屬性的節點類別屬性值可從節點分類模型中推斷出來。這在此屬性尚未附加至頂點時很有用。

  • 為了從節點分類模型中擷取一個或多個類別，您需要使用 with() 步驟搭配述詞 Neptune#ml.classification 來設定 properties() 步驟。如果輸出格式是頂點屬性，則輸出格式與您所期望的格式相似。

注意

節點分類僅會使用字串屬性值。這表示不支援數值屬性值 (例如 0 或 1)，儘管字串等同於 "0" 和 "1"。同樣地，布林屬性值 true 和 false 不起作用，但 "true" 和 "false" 可以。

以下是範例節點分類查詢：

g.with( "Neptune#ml.endpoint","node-classification-movie-lens-endpoint" )
 .with( "Neptune#ml.iamRoleArn","arn:aws:iam::0123456789:role/sagemaker-role" )
 .with( "Neptune#ml.limit", 2 )
 .with( "Neptune#ml.threshold", 0.5D )
 .V( "movie_1", "movie_2", "movie_3" )
 .properties("genre").with("Neptune#ml.classification")

此查詢的輸出如下所示：

==>vp[genre->Action]
==>vp[genre->Crime]
==>vp[genre->Comedy]

在上述查詢中，V() 和 properties() 步驟的使用方式如下：

V() 步驟包含您要從節點分類模型中擷取其類別的頂點集：

 .V( "movie_1", "movie_2", "movie_3" )

properties() 步驟包含模型訓練時根據的索引鍵，而且具有 .with("Neptune#ml.classification") 指示這是節點分類 ML 推論查詢。

properties().with("Neptune#ml.classification") 步驟中目前不支援多個內容索引鍵。例如，下列查詢會導致例外狀況：

g.with("Neptune#ml.endpoint", "node-classification-movie-lens-endpoint")
 .with("Neptune#ml.iamRoleArn","arn:aws:iam::0123456789:role/sagemaker-role")
 .V( "movie_1", "movie_2", "movie_3" )
 .properties("genre", "other_label").with("Neptune#ml.classification")

如需特定錯誤訊息，請參閱 Neptune ML 例外狀況清單。

properties().with("Neptune#ml.classification") 步驟可與下列任一步驟搭配使用：

  • value()

  • value().is()

  • hasValue()

  • has(value,"")

  • key()

  • key().is()

  • hasKey()

  • has(key,"")

  • path()

其他節點分類查詢

如果推論端點和對應的IAM角色都已儲存在您的資料庫叢集參數群組中，節點分類查詢可以像這樣簡單：

g.V("movie_1", "movie_2", "movie_3").properties("genre").with("Neptune#ml.classification")

您可以使用 union() 步驟，在查詢中混合頂點屬性和類別：

g.with("Neptune#ml.endpoint","node-classification-movie-lens-endpoint")
 .with("Neptune#ml.iamRoleArn","arn:aws:iam::0123456789:role/sagemaker-role")
 .V( "movie_1", "movie_2", "movie_3" )
 .union(
   properties("genre").with("Neptune#ml.classification"),
   properties("genre")
 )

您也可以建立無限制的查詢，如下所示：

g.with("Neptune#ml.endpoint","node-classification-movie-lens-endpoint")
 .with("Neptune#ml.iamRoleArn","arn:aws:iam::0123456789:role/sagemaker-role")
 .V()
 .properties("genre").with("Neptune#ml.classification")

您可以使用 select() 步驟搭配 as() 步驟來擷取節點類別以及頂點：

g.with("Neptune#ml.endpoint","node-classification-movie-lens-endpoint")
 .with("Neptune#ml.iamRoleArn","arn:aws:iam::0123456789:role/sagemaker-role")
 .V( "movie_1", "movie_2", "movie_3" ).as("vertex")
 .properties("genre").with("Neptune#ml.classification").as("properties")
 .select("vertex","properties")

您也可以根據節點類別篩選，如下列範例所示：

g.with("Neptune#ml.endpoint", "node-classification-movie-lens-endpoint")
 .with("Neptune#ml.iamRoleArn","arn:aws:iam::0123456789:role/sagemaker-role")
 .V( "movie_1", "movie_2", "movie_3" )
 .properties("genre").with("Neptune#ml.classification")
 .has(value, "Horror")

g.with("Neptune#ml.endpoint","node-classification-movie-lens-endpoint")
 .with("Neptune#ml.iamRoleArn","arn:aws:iam::0123456789:role/sagemaker-role")
 .V( "movie_1", "movie_2", "movie_3" )
 .properties("genre").with("Neptune#ml.classification")
 .has(value, P.eq("Action"))

g.with("Neptune#ml.endpoint","node-classification-movie-lens-endpoint")
 .with("Neptune#ml.iamRoleArn","arn:aws:iam::0123456789:role/sagemaker-role")
 .V( "movie_1", "movie_2", "movie_3" )
 .properties("genre").with("Neptune#ml.classification")
 .has(value, P.within("Action", "Horror"))

您可以使用 Neptune#ml.score 述詞，取得節點分類可信度分數：

 g.with("Neptune#ml.endpoint","node-classification-movie-lens-endpoint")
 .with("Neptune#ml.iamRoleArn","arn:aws:iam::0123456789:role/sagemaker-role")
 .V( "movie_1", "movie_2", "movie_3" )
 .properties("genre", "Neptune#ml.score").with("Neptune#ml.classification")

回應看起來像這樣：

==>vp[genre->Action]
==>vp[Neptune#ml.score->0.01234567]
==>vp[genre->Crime]
==>vp[Neptune#ml.score->0.543210]
==>vp[genre->Comedy]
==>vp[Neptune#ml.score->0.10101]

在節點分類查詢中使用歸納推論

假設您要在 Jupyter 筆記本中將新節點新增至現有圖形，如下所示：

%%gremlin
g.addV('label1').property(id,'101').as('newV')
 .V('1').as('oldV1')
 .V('2').as('oldV2')
 .addE('eLabel1').from('newV').to('oldV1')
 .addE('eLabel2').from('oldV2').to('newV')

您接著可以使用歸納推論查詢，取得反映了新節點的類型和可信度分數：

%%gremlin
g.with("Neptune#ml.endpoint", "nc-ep")
 .with("Neptune#ml.iamRoleArn", "arn:aws:iam::123456789012:role/NeptuneMLRole")
 .V('101').properties("genre", "Neptune#ml.score")
 .with("Neptune#ml.classification")
 .with("Neptune#ml.inductiveInference")

不過，如果您執行了多次查詢，則可能會得到有些不同的結果：

# First time
==>vp[genre->Action]
==>vp[Neptune#ml.score->0.12345678]

# Second time
==>vp[genre->Action]
==>vp[Neptune#ml.score->0.21365921]

你可以使相同的查詢具有確定性：

%%gremlin
g.with("Neptune#ml.endpoint", "nc-ep")
 .with("Neptune#ml.iamRoleArn", "arn:aws:iam::123456789012:role/NeptuneMLRole")
 .V('101').properties("genre", "Neptune#ml.score")
 .with("Neptune#ml.classification")
 .with("Neptune#ml.inductiveInference")
 .with("Neptune#ml.deterministic")

在這種情況下，每次結果將大致相同：

# First time
==>vp[genre->Action]
==>vp[Neptune#ml.score->0.12345678]
# Second time
==>vp[genre->Action]
==>vp[Neptune#ml.score->0.12345678]