Refine the solution for the top-10 in each namespace considering that services can be in different clusters but in the same namespace (need to merge)

If services can be in different clusters but belong to the same namespace, then for the top-10 in each namespace, data should be aggregated by namespace, combining results from all clusters.

Approach:

1. Collect metrics or data on services from all clusters.
2. Merge data by the namespace key.
3. For each namespace, combine all services from different clusters.
4. Sort services by the desired criterion (e.g., load) and select the top-10.

Example in Go (simplified):

Type ServiceData struct {
    Namespace string
    Service   string
    Metric    int
}

// servicesFromClusters — slice of data from different clusters
func Top10ByNamespace(servicesFromClusters [][]ServiceData) map[string][]ServiceData {
    combined := make(map[string][]ServiceData)

    // Merge data by namespace
    for _, clusterData := range servicesFromClusters {
        for _, svc := range clusterData {
            combined[svc.Namespace] = append(combined[svc.Namespace], svc)
        }
    }

    result := make(map[string][]ServiceData)
    for ns, svcs := range combined {
        // Sort by metric in descending order
        sort.Slice(svcs, func(i, j int) bool {
            return svcs[i].Metric > svcs[j].Metric
        })
        // Take top-10
        limit := 10
        if len(svcs) < 10 {
            limit = len(svcs)
        }
        result[ns] = svcs[:limit]
    }

    return result
}

This approach allows aggregating and ranking services by namespace, regardless of which clusters they are in.