python sklearn中的fit方法 [英] fit method in python sklearn

问题描述

我问自己关于sklearn中的fit方法的各种问题.

I am asking myself various questions about the fit method in sklearn.

问题1:当我这样做时:

Question 1: when I do:

from sklearn.decomposition import TruncatedSVD
model = TruncatedSVD()
svd_1 = model.fit(X1)
svd_2 = model.fit(X2)

在此过程中，变量模型的内容是否发生任何变化?

Is the content of the variable model changing whatsoever during the process?

问题2:当我这样做时:

Question 2: when I do:

from sklearn.decomposition import TruncatedSVD
model = TruncatedSVD()
svd_1 = model.fit(X1)
svd_2 = svd_1.fit(X2)

svd_1发生了什么?换句话说，svd_1已经安装好了，我又重新安装了它，那么它的组件正在发生什么事情?

What is happening to svd_1? In other words, svd_1 has already been fitted and I fit it again, so what is happenning to its component?

推荐答案

问题1:变量模型的内容在此过程中是否发生任何变化?

Question 1: Is the content of the variable model changing whatsoever during the process?

是的. fit方法修改对象.并且它返回对该对象的引用.因此，保重！在第一个示例中，所有三个变量model，svd_1和svd_2实际上都引用相同的对象.

Yes. The fit method modifies the object. And it returns a reference to the object. Thus, take care! In the first example all three variables model, svd_1, and svd_2 actually refer to the same object.

from sklearn.decomposition import TruncatedSVD
model = TruncatedSVD()
svd_1 = model.fit(X1)
svd_2 = model.fit(X2)
print(model is svd_1 is svd_2)  # prints True

问题2: svd_1发生了什么事?

Question 2: What is happening to svd_1?

model和svd_1指的是同一对象，因此第一个示例和第二个示例之间绝对没有区别.

model and svd_1 refer to the same object, so there is absolutely no difference between the first and the second example.

最后的评论: 在两个示例中发生的都是fit(X1)的结果被fit(X2)覆盖，如David Maust的答案<.如果要将两个不同的模型拟合到两个不同的数据集，则需要执行以下操作:

Final Remark: What happens in both examples is that the result of fit(X1) is overwritten by fit(X2), as pointed out in the answer by David Maust. If you want to have two different models fitted to two different sets of data you need to do something like this:

svd_1 = TruncatedSVD().fit(X1)
svd_2 = TruncatedSVD().fit(X2)

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