我有一個保存的模型，我設法加載，運行并獲得1行9個特征的預測。（輸入）現在我試圖預測100行這樣的行，但是當嘗試從Tensor.copyTo（）讀取結果數組時，我得到了不兼容的形狀java.lang.IllegalArgumentException: cannot copy Tensor with shape [1, 1] into object with shape [100, 1]顯然，我設法在循環中運行了這個預測 - 但這比一次運行100的等效python執行慢20倍。這里是 /saved_model_cli.py 報告的已保存模型信息MetaGraphDef with tag-set: 'serve' contains the following SignatureDefs:signature_def['serving_default']: The given SavedModel SignatureDef contains the following input(s): inputs['input'] tensor_info: dtype: DT_FLOAT shape: (-1, 9) name: dense_1_input:0 The given SavedModel SignatureDef contains the following output(s): outputs['output'] tensor_info: dtype: DT_FLOAT shape: (-1, 1) name: dense_4/BiasAdd:0 Method name is: tensorflow/serving/predict問題是 - 我是否需要為我想預測的每一行運行（），就像這里的問題一樣
1 回答
catspeake
TA貢獻1111條經驗 獲得超0個贊
好吧，所以我發現了一個問題，我無法為我想要的所有行（預測）運行一次?？赡苁且粋€張量流新手問題，我搞砸了輸入和輸出矩陣。當報告工具（python）說你有一個形狀（-1，9）的輸入張量映射到java long[]{1，9}時，這并不意味著你不能傳遞long[]{1000，9}的輸入張量 - 這意味著1000行用于預測。在此輸入之后，定義為 [1，1] 的輸出張量可以是 [1000，1]。
這個代碼實際上比python運行得快得多（1.2秒對7秒），這是代碼（也許會解釋得更好）
public Tensor prepareData(){
Random r = new Random();
float[]inputArr = new float[NUMBER_OF_KEWORDS*NUMBER_OF_FIELDS];
for (int i=0;i<NUMBER_OF_KEWORDS * NUMBER_OF_FIELDS;i++){
inputArr[i] = r.nextFloat();
}
FloatBuffer inputBuff = FloatBuffer.wrap(inputArr, 0, NUMBER_OF_KEWORDS*NUMBER_OF_FIELDS);
return Tensor.create(new long[]{NUMBER_OF_KEWORDS,NUMBER_OF_FIELDS}, inputBuff);
}
public void predict (Tensor inputTensor){
try ( Session s = savedModelBundle.session()) {
Tensor result;
long globalStart = System.nanoTime();
result = s.runner().feed("dense_1_input", inputTensor).fetch("dense_4/BiasAdd").run().get(0);
final long[] rshape = result.shape();
if (result.numDimensions() != 2 || rshape[0] <= NUMBER_OF_KEWORDS) {
throw new RuntimeException(
String.format(
"Expected model to produce a [N,1] shaped tensor where N is the number of labels, instead it produced one with shape %s",
Arrays.toString(rshape)));
}
float[][] resultArray = (float[][]) result.copyTo(new float[NUMBER_OF_KEWORDS][1]);
System.out.println(String.format("Total of %d, took : %.4f ms", NUMBER_OF_KEWORDS, ((double) System.nanoTime() - globalStart) / 1000000));
for (int i=0;i<10;i++){
System.out.println(resultArray[i][0]);
}
}
}
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