TL、DR：

1）不，除非您顯式指定，或者它是估計器的默認.score方法。因為您沒有，它默認為DecisionTreeRegressor.score，它返回決定系數，即R^2。可能是負數。在

2）是的，這是個問題。這也解釋了為什么你會得到一個負的決定系數。在

細節：

您使用的函數如下：scores = cross_val_score(simple_tree, df.loc[:,'system':'gwno'], df['gdp_growth'], cv=cv)

所以你沒有顯式地傳遞一個“scoring”參數。讓我們看看docs：scoring : string, callable or None, optional, default: None

A string (see model evaluation documentation) or a scorer callable object / function with signature scorer(estimator, X, y).

所以它沒有明確說明，但這可能意味著它使用了估計器的默認.score方法。在

為了證實這個假設，讓我們深入研究source code。我們看到最終使用的記分器如下：

^{pr2}$

has_scoring = scoring is not None

if not hasattr(estimator, 'fit'):

raise TypeError("estimator should be an estimator implementing "

"'fit' method, %r was passed" % estimator)

if isinstance(scoring, six.string_types):

return get_scorer(scoring)

elif has_scoring:

# Heuristic to ensure user has not passed a metric

module = getattr(scoring, '__module__', None)

if hasattr(module, 'startswith') and \

module.startswith('sklearn.metrics.') and \

not module.startswith('sklearn.metrics.scorer') and \

not module.startswith('sklearn.metrics.tests.'):

raise ValueError('scoring value %r looks like it is a metric '

'function rather than a scorer. A scorer should '

'require an estimator as its first parameter. '

'Please use `make_scorer` to convert a metric '

'to a scorer.' % scoring)

return get_scorer(scoring)

elif hasattr(estimator, 'score'):

return _passthrough_scorer

elif allow_none:

return None

else:

raise TypeError(

"If no scoring is specified, the estimator passed should "

"have a 'score' method. The estimator %r does not." % estimator)

所以請注意，scoring=None已經完成，所以：has_scoring = scoring is not None

暗示has_scoring == False。另外，估計器有一個.score屬性，所以我們要通過這個分支：elif hasattr(estimator, 'score'):

return _passthrough_scorer

這很簡單：def _passthrough_scorer(estimator, *args, **kwargs):

"""Function that wraps estimator.score"""

return estimator.score(*args, **kwargs)

最后，我們現在知道scorer就是你的估計器默認的score。讓我們檢查一下docs for the estimator，它清楚地表明：Returns the coefficient of determination R^2 of the prediction.

The coefficient R^2 is defined as (1 - u/v), where u is the regression

sum of squares ((y_true - y_pred) ** 2).sum() and v is the residual

sum of squares ((y_true - y_true.mean()) ** 2).sum(). Best possible

score is 1.0 and it can be negative (because the model can be

arbitrarily worse). A constant model that always predicts the expected

value of y, disregarding the input features, would get a R^2 score of

0.0.

所以看起來你的分數實際上就是決定系數。所以，基本上，R^2為負值，意味著你的模型表現得很差。比我們僅僅預測每個輸入的期望值（即平均值）更糟糕。這是有道理的，因為正如你所說：I have a small sample of ~40 observations and ~70 variables. Might

this be the problem?

這是個問題。當你只有40個觀測值時，對一個70維的問題空間進行有意義的預測幾乎是沒有希望的。在

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