New clustering metrics

### Describe the workflow you want to enable

Scikit-learn defines three popular metrics for evaluating clustering performance when there are no ground-truth cluster labels: [sklearn.metrics.silhouette_score](https://scikit-learn.org/stable/modules/generated/sklearn.metrics.silhouette_score.html), [sklearn.metrics.calinski_harabasz_score](https://scikit-learn.org/stable/modules/generated/sklearn.metrics.calinski_harabasz_score.html) and [sklearn.metrics.davies_bouldin_score](https://scikit-learn.org/stable/modules/generated/sklearn.metrics.davies_bouldin_score.html). But there are lots of others, and it's previously been [discussed](https://github.com/scikit-learn/scikit-learn/discussions/21164) whether to integrate more into scikit-learn.

### Describe your proposed solution

I've implemented four relatively popular ones, using the same interface and code style as in https://github.com/scikit-learn/scikit-learn/blob/main/sklearn/metrics/cluster/_unsupervised.py. Would there be interest in integrating these into scikit-learn?

```python
import numpy as np
from itertools import combinations
from sklearn.utils import check_X_y
from sklearn.metrics.cluster._unsupervised import check_number_of_labels
from sklearn.preprocessing import LabelEncoder


def log_ss_ratio(X, labels):
    X, labels = check_X_y(X, labels)
    le = LabelEncoder()
    labels = le.fit_transform(labels)

    n_samples, _ = X.shape
    n_labels = len(le.classes_)

    check_number_of_labels(n_labels, n_samples)

    extra_disp, intra_disp = 0.0, 0.0
    mean = X.mean(axis=0)
    for k in range(n_labels):
        cluster_k = X[labels == k]
        mean_k = cluster_k.mean(axis=0)
        extra_disp += len(cluster_k) * ((mean_k - mean) ** 2).sum()
        intra_disp += ((cluster_k - mean_k) ** 2).sum()

    return np.log(extra_disp / intra_disp)


def ball_hall(X, labels):
    X, labels = check_X_y(X, labels)
    le = LabelEncoder()
    labels = le.fit_transform(labels)

    n_samples, _ = X.shape
    n_labels = len(le.classes_)

    check_number_of_labels(n_labels, n_samples)

    sum_mean_dispersions = 0

    for k in range(n_labels):
        cluster_k = X[labels == k]
        mean_k = cluster_k.mean(axis=0)
        intra_disp_cluster = ((cluster_k - mean_k) ** 2).sum()
        sum_mean_dispersions += intra_disp_cluster / len(cluster_k)

    return sum_mean_dispersions / n_labels


def banfeld_raftery(X, labels):
    X, labels = check_X_y(X, labels)
    le = LabelEncoder()
    labels = le.fit_transform(labels)

    n_samples, _ = X.shape
    n_labels = len(le.classes_)

    check_number_of_labels(n_labels, n_samples)

    br_index = 0
    for k in range(n_labels):
        cluster_k = X[labels == k]
        mean_k = cluster_k.mean(axis=0)
        intra_disp_cluster = ((cluster_k - mean_k) ** 2).sum()
        br_index += len(cluster_k) * \
                    np.log(intra_disp_cluster / len(cluster_k))

    return br_index


def ray_turi(X, labels):
    X, labels = check_X_y(X, labels)
    le = LabelEncoder()
    labels = le.fit_transform(labels)

    n_samples, _ = X.shape
    n_labels = len(le.classes_)

    check_number_of_labels(n_labels, n_samples)

    intra_disp = 0.0

    cluster_means = []
    for k in range(n_labels):
        cluster_k = X[labels == k]
        mean_k = cluster_k.mean(axis=0)
        intra_disp += ((cluster_k - mean_k) ** 2).sum()
        cluster_means.append(mean_k)

    min_cluster_mean_diff = min(
        ((mean_i - mean_j) ** 2).sum()
        for mean_i, mean_j in combinations(cluster_means, 2))

    return intra_disp / (min_cluster_mean_diff * n_samples)
```

### Describe alternatives you've considered, if relevant

_No response_

### Additional context

_No response_

COMMENT:
> let me try to be honest, your comment https://github.com/scikit-learn/scikit-learn/issues/27259#issuecomment-1961290296 sounded very demanding.

@lesteve I'm sorry it came off that way, certainly not what I intended. What I meant to communicate was not that I feel entitled to a review and therefore demand to receive one, but instead that **under the assumption** that there is still interest in the issue at hand, I'm anticipating a review in the near future. If I hadn't gotten any response or an explicit no, that would have been fine with me as well. At least in that case I can "cut my losses" and move on. I'd prefer that outcome over ambiguity which results in me unnecessarily remaining on standby.

> I really like Brett Cannon talk on open-source participation:
https://snarky.ca/setting-expectations-for-open-source-participation/. In particular, I find the "PR as a puppy" metaphor great.

I read through the talk in its article form. Looking at PRs as puppys seems like a good way to think about it. I like to think that I personally [have](https://github.com/vega/vega-lite/pull/9214) a [history](https://github.com/scikit-learn/scikit-learn/issues/28243) of [interacting with maintainers](https://github.com/demisto/content/pull/22616#issuecomment-1333920875) in a [polite](https://github.com/Neo23x0/Raccine/pull/134) way that's [consistent](https://github.com/scikit-learn-contrib/hdbscan/pull/552#issuecomment-1193848434) with that.

> Personally when I don't get any feed-back on an issue, I kind of try to say things like "of course I completely understand if you have other priorities" or "let me know if there is anything I can do to help this push forward". A recent example where I did not have any feed-back in roughly three weeks: https://github.com/colesbury/nogil-wheels/pull/6#issuecomment-1954427705

that seems like a good way of going about it

Labels: New Feature, Needs Decision - Include Feature