Scikit Learn GridSearchCV 无需交叉验证(无监督学习) [英] Scikit Learn GridSearchCV without cross validation (unsupervised learning)

问题描述

是否可以在没有交叉验证的情况下使用 GridSearchCV?我正在尝试通过网格搜索优化 KMeans 聚类中的聚类数量，因此我不需要也不想要交叉验证.

Is it possible to use GridSearchCV without cross validation? I am trying to optimize the number of clusters in KMeans clustering via grid search, and thus I don't need or want cross validation.

文档 也让我感到困惑，因为在fit() 方法，它有一个无监督学习的选项(说对无监督学习使用 None).但是如果你想做无监督学习，你需要在没有交叉验证的情况下进行，而且似乎没有办法摆脱交叉验证.

The documentation is also confusing me because under the fit() method, it has an option for unsupervised learning (says to use None for unsupervised learning). But if you want to do unsupervised learning, you need to do it without cross validation and there appears to be no option to get rid of cross validation.

推荐答案

经过多方搜索，我找到了 这个话题.如果您使用:

After much searching, I was able to find this thread. It appears that you can get rid of cross validation in GridSearchCV if you use:

cv=[(slice(None), slice(None))]

我已经在没有交叉验证的情况下针对我自己的网格搜索编码版本对此进行了测试，并且我从两种方法中得到了相同的结果.我将这个答案发布到我自己的问题中，以防其他人遇到同样的问题.

I have tested this against my own coded version of grid search without cross validation and I get the same results from both methods. I am posting this answer to my own question in case others have the same issue.

在评论中回答 jjrr 的问题，这是一个示例用例:

to answer jjrr's question in the comments, here is an example use case:

from sklearn.metrics import silhouette_score as sc

def cv_silhouette_scorer(estimator, X):
    estimator.fit(X)
    cluster_labels = estimator.labels_
    num_labels = len(set(cluster_labels))
    num_samples = len(X.index)
    if num_labels == 1 or num_labels == num_samples:
        return -1
    else:
        return sc(X, cluster_labels)

cv = [(slice(None), slice(None))]
gs = GridSearchCV(estimator=sklearn.cluster.MeanShift(), param_grid=param_dict, 
                  scoring=cv_silhouette_scorer, cv=cv, n_jobs=-1)
gs.fit(df[cols_of_interest])

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