Notice: Undefined offset: 1

saya ingin bertanya, Saya sedang membuat projek Kenapa muncul error seperti ini :

Notice: Undefined offset: 1

Saya punya script proses-pelatihan-datatesting.php

        <div class="form-group">
            <h3 align="center">Hasil Sorting Euclidian | K = <?php echo $kvall ?></h3>
        </div>
    </div>


    <div class="table-responsive col-md-8 col-md-offset-2" style="height:350px;overflow:auto;">
        <table class="table table-bordered">
            <thead>
                <tr>
                    <th>Array [DataTesting] [DataTraining]</th>
                    <th>Jarak Euclidian</th>
                    <th>Klasifikasi</th>
                </tr>
            </thead>
            <tbody>
                <tr>

    <?php

    $query_klas_del = "TRUNCATE TABLE temp";
    mysqli_query($koneksi,$query_klas_del);
        for($k=0;$k<$banyak_data2;$k++){
            for($l=0;$l<$kvall;$l++){
                echo "<tr>";
                echo "<td>"."[".$k."][".${'keys'.$k}[$l]."]"."</td>";
                echo "<td>".${'val'.$k}[$l]."</td>";
                echo "<td>".${'klasifikasi'.$l}[${'keys'.$k}[$l]]."</td>";
                echo "</tr>";
                //echo ${'keys'.$k}[$l];

        if(${'klasifikasi'.$l}[${'keys'.$k}[$l]] == $kelas1) {
            $counterClass1++;
        }
        else if(${'klasifikasi'.$l}[${'keys'.$k}[$l]] == $kelas2) {
            $counterClass2++;
        }
        else {
            $counterClass3++;
        }
    }

        if($counterClass1>$counterClass2 && $counterClass1>$counterClass3) {
            $query_klas1 = "insert into temp (klasifikasi) values ('$kelas1')";
            $result_klas1 = mysqli_query($koneksi,$query_klas1);
        }
        else if($counterClass2>$counterClass1 && $counterClass2>$counterClass3) {
            $query_klas2 = "insert into temp (klasifikasi) values ('$kelas2')";
            $result_klas2 = mysqli_query($koneksi,$query_klas2);
        }
        else if($counterClass3>$counterClass1 && $counterClass3>$counterClass2) {
            $query_klas3 = "insert into temp (klasifikasi) values ('$kelas3')";
            $result_klas3 = mysqli_query($koneksi,$query_klas3);
        }

            $counterClass1=0;
            $counterClass2=0;
            $counterClass3=0;

        }

    ?>
            </tr>
        </tbody>
    </table>
</div>

berikut bagian script yang error :

<?php

    $query_klas_del = "TRUNCATE TABLE temp";
    mysqli_query($koneksi,$query_klas_del);
        for($k=0;$k<$banyak_data2;$k++){
            for($l=0;$l<$kvall;$l++){
                echo "<tr>";
                echo "<td>"."[".$k."][".${'keys'.$k}[$l]."]"."</td>";
                echo "<td>".${'val'.$k}[$l]."</td>";
                echo "<td>".${'klasifikasi'.$l}[${'keys'.$k}[$l]]."</td>";
                echo "</tr>";
                //echo ${'keys'.$k}[$l];

        if(${'klasifikasi'.$l}[${'keys'.$k}[$l]] == $kelas1) {
            $counterClass1++;
        }
        else if(${'klasifikasi'.$l}[${'keys'.$k}[$l]] == $kelas2) {
            $counterClass2++;
        }
        else {
            $counterClass3++;
        }
    }

        if($counterClass1>$counterClass2 && $counterClass1>$counterClass3) {
            $query_klas1 = "insert into temp (klasifikasi) values ('$kelas1')";
            $result_klas1 = mysqli_query($koneksi,$query_klas1);
        }
        else if($counterClass2>$counterClass1 && $counterClass2>$counterClass3) {
            $query_klas2 = "insert into temp (klasifikasi) values ('$kelas2')";
            $result_klas2 = mysqli_query($koneksi,$query_klas2);
        }
        else if($counterClass3>$counterClass1 && $counterClass3>$counterClass2) {
            $query_klas3 = "insert into temp (klasifikasi) values ('$kelas3')";
            $result_klas3 = mysqli_query($koneksi,$query_klas3);
        }

            $counterClass1=0;
            $counterClass2=0;
            $counterClass3=0;

        }

    ?>

mohon bantuannya master?