为什么 SVM 的所有结果在 scikit learn 中都是一样的?
Why are all of my results from SVM the same in scikit learn?
我正在尝试使用 scikit learn 计算多 class 数据集的概率。但是,出于某种原因,我对每个示例都获得了相同的概率。知道发生了什么吗?这与我的模型、我对库的使用或其他什么有关吗?感谢任何帮助!
svm_model = svm.SVC(probability=True, kernel='rbf',C=1, decision_function_shape='ovr', gamma=0.001,verbose=100)
svm_model.fit(train_X,train_y)
preds= svm_model.predict_proba(test_X)
train_X 看起来像这样
array([[2350, 5550, 2750.0, ..., 23478, 1, 3],
[2500, 5500, 3095.5, ..., 23674, 0, 3],
[3300, 6900, 3600.0, ..., 6529, 0, 3],
...,
[2150, 6175, 2500.0, ..., 11209, 0, 3],
[2095, 5395, 2595.4, ..., 10070, 0, 3],
[1650, 2850, 2000.0, ..., 25463, 1, 3]], dtype=object)
train_y 看起来像这样
0 1
1 2
10 2
100 2
1000 2
10000 2
10001 2
10002 2
10003 2
10004 2
10005 2
10006 2
10007 2
10008 1
10009 1
1001 2
10010 2
test_X 看起来像这样
array([[2190, 3937, 2200.5, ..., 24891, 1, 5],
[2695, 7000, 2850.0, ..., 5491, 1, 4],
[2950, 12000, 4039.5, ..., 22367, 0, 4],
...,
[2850, 5200, 3000.0, ..., 15576, 1, 1],
[3200, 16000, 4100.0, ..., 1320, 0, 3],
[2100, 3750, 2400.0, ..., 6022, 0, 1]], dtype=object)
我的结果看起来像
array([[ 0.07819139, 0.22727628, 0.69453233],
[ 0.07819139, 0.22727628, 0.69453233],
[ 0.07819139, 0.22727628, 0.69453233],
...,
[ 0.07819139, 0.22727628, 0.69453233],
[ 0.07819139, 0.22727628, 0.69453233],
[ 0.07819139, 0.22727628, 0.69453233]])
从预处理开始!
将数据标准化为零均值和单位方差非常重要。
scikit-learn 文档说 this:
Support Vector Machine algorithms are not scale invariant, so it is highly recommended to scale your data. For example, scale each attribute on the input vector X to [0,1] or [-1,+1], or standardize it to have mean 0 and variance 1. Note that the same scaling must be applied to the test vector to obtain meaningful results. See section Preprocessing data for more details on scaling and normalization
- sklearns Section on Preprocessing
- sklearns StandardScaler.
此后的下一步是参数调整(C、伽玛和系数)。这通常由 GridSearch 完成。 但我通常希望人们在尝试 Kernel-SVM 之前先尝试一个简单的 LinearSVM(更少的超参数,更少的计算时间,更好地概括非最佳参数选择).
我正在尝试使用 scikit learn 计算多 class 数据集的概率。但是,出于某种原因,我对每个示例都获得了相同的概率。知道发生了什么吗?这与我的模型、我对库的使用或其他什么有关吗?感谢任何帮助!
svm_model = svm.SVC(probability=True, kernel='rbf',C=1, decision_function_shape='ovr', gamma=0.001,verbose=100)
svm_model.fit(train_X,train_y)
preds= svm_model.predict_proba(test_X)
train_X 看起来像这样
array([[2350, 5550, 2750.0, ..., 23478, 1, 3],
[2500, 5500, 3095.5, ..., 23674, 0, 3],
[3300, 6900, 3600.0, ..., 6529, 0, 3],
...,
[2150, 6175, 2500.0, ..., 11209, 0, 3],
[2095, 5395, 2595.4, ..., 10070, 0, 3],
[1650, 2850, 2000.0, ..., 25463, 1, 3]], dtype=object)
train_y 看起来像这样
0 1
1 2
10 2
100 2
1000 2
10000 2
10001 2
10002 2
10003 2
10004 2
10005 2
10006 2
10007 2
10008 1
10009 1
1001 2
10010 2
test_X 看起来像这样
array([[2190, 3937, 2200.5, ..., 24891, 1, 5],
[2695, 7000, 2850.0, ..., 5491, 1, 4],
[2950, 12000, 4039.5, ..., 22367, 0, 4],
...,
[2850, 5200, 3000.0, ..., 15576, 1, 1],
[3200, 16000, 4100.0, ..., 1320, 0, 3],
[2100, 3750, 2400.0, ..., 6022, 0, 1]], dtype=object)
我的结果看起来像
array([[ 0.07819139, 0.22727628, 0.69453233],
[ 0.07819139, 0.22727628, 0.69453233],
[ 0.07819139, 0.22727628, 0.69453233],
...,
[ 0.07819139, 0.22727628, 0.69453233],
[ 0.07819139, 0.22727628, 0.69453233],
[ 0.07819139, 0.22727628, 0.69453233]])
从预处理开始!
将数据标准化为零均值和单位方差非常重要。 scikit-learn 文档说 this:
Support Vector Machine algorithms are not scale invariant, so it is highly recommended to scale your data. For example, scale each attribute on the input vector X to [0,1] or [-1,+1], or standardize it to have mean 0 and variance 1. Note that the same scaling must be applied to the test vector to obtain meaningful results. See section Preprocessing data for more details on scaling and normalization
- sklearns Section on Preprocessing
- sklearns StandardScaler.
此后的下一步是参数调整(C、伽玛和系数)。这通常由 GridSearch 完成。 但我通常希望人们在尝试 Kernel-SVM 之前先尝试一个简单的 LinearSVM(更少的超参数,更少的计算时间,更好地概括非最佳参数选择).