libsvm预测方法混乱

发布于 2024-10-09 12:37:57 字数 811 浏览 4 评论 0原文

我对 libsvm 中的 svm_predict() 方法有疑问。

自述文件有以下快速入门示例代码：

>>> y, x = [1,-1], [{1:1, 3:1}, {1:-1,3:-1}]
>>> prob  = svm_problem(y, x)
>>> param = svm_parameter('-c 4 -b 1')
>>> m = svm_train(prob, param)

>>> p_label, p_acc, p_val = svm_predict(y, x, m)

现在我明白 y 是与 x 中的词典关联的类别列表。我也理解 svm_train 部分。

没有意义的部分是，在 svm_predict 中，我需要提供 y 中的“真实值”以及 x 中的测试数据。我认为这个想法是我不提前知道测试数据的分类。

如果我的训练数据是：

y = [1, 2, 3]
x = [{1:1}, {1:10}, {1:20}]

但我的测试数据是：

z = [{1:4}, {1:12}, {1:19}]

那么为什么我需要将 z 的真实值传递到 svm_predict() 中，例如：

a, b, c = svm_predict(y, z, m)

我不会知道 z 的真实值——这就是预测的目的。当我执行预测时，我应该为 y 输入任意分类值，还是我完全遗漏了一些东西？

谢谢大家

原文

I have a question about the svm_predict() method in libsvm.

The README has this quickstart example code:

>>> y, x = [1,-1], [{1:1, 3:1}, {1:-1,3:-1}]
>>> prob  = svm_problem(y, x)
>>> param = svm_parameter('-c 4 -b 1')
>>> m = svm_train(prob, param)

>>> p_label, p_acc, p_val = svm_predict(y, x, m)

Now I understand that y is a list of categories that are associated with the dictionaries in x. I also understand the svm_train part.

The part that does not make sense is that in svm_predict, I am required to provide the 'true values' from y, along with the test data in x. I thought the idea was that I do not know the classifications of the test data ahead of time.

if my training data is:

y = [1, 2, 3]
x = [{1:1}, {1:10}, {1:20}]

but my test data is:

z = [{1:4}, {1:12}, {1:19}]

Then why am I required to pass in true values of z into svm_predict() like:

a, b, c = svm_predict(y, z, m)

I'm not going to know the true values for z--that's what the prediction is for. Should I just put arbitrary classification values for y when I perform a prediction, or am I completely missing something?

Thanks all

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