我试图获得需要用于分类的组件数量.我已经阅读了类似的问题使用scikit-learn PCA和关于此的scikit文档查找具有最高方差的维度:
但是,这仍然没有解决我的问题.我的所有PCA组件都非常大,因为我可以选择所有这些组件,但如果我这样做,PCA将毫无用处.
我还在scikit中阅读了PCA库 http://scikit-learn.org/stable/modules/generated/sklearn.decomposition.PCA.html 它表明了L:
如果n_components =='mle',如果0 <n_components <1,则使用Minka的MLE猜测维度,选择组件数量,使得需要解释的方差量大于n_components指定的百分比
但是,我无法找到有关使用此技术分析PCA的n_components的更多信息
这是我的PCA分析代码:
from sklearn.decomposition import PCA
pca = PCA()
pca.fit(x_array_train)
print(pca.explained_variance_)
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结果:
[ 6.58902714e+50 6.23266555e+49 2.93568652e+49 2.25418736e+49
1.10063872e+49 3.25107359e+40 4.72113817e+39 1.40411862e+39
4.03270198e+38 1.60662882e+38 3.20028861e+28 2.35570241e+27
1.54944915e+27 8.05181151e+24 1.42231553e+24 5.05155955e+23
2.90909468e+23 2.60339206e+23 1.95672973e+23 1.22987336e+23
9.67133111e+22 7.07208772e+22 4.49067983e+22 3.57882593e+22
3.03546737e+22 2.38077950e+22 2.18424235e+22 1.79048845e+22
1.50871735e+22 1.35571453e+22 1.26605081e+22 1.04851395e+22
8.88191944e+21 6.91581346e+21 5.43786989e+21 5.05544020e+21
4.33110823e+21 3.18309135e+21 3.06169368e+21 2.66513522e+21
2.57173046e+21 2.36482212e+21 2.32203521e+21 2.06033130e+21
1.89039408e+21 1.51882514e+21 1.29284842e+21 1.26103770e+21
1.22012185e+21 1.07857244e+21 8.55143095e+20 4.82321416e+20
2.98301261e+20 2.31336276e+20 1.31712446e+20 1.05253795e+20
9.84992112e+19 8.27574150e+19 4.66007620e+19 4.09687463e+19
2.89855823e+19 2.79035170e+19 1.57015298e+19 1.39218538e+19
1.00594159e+19 7.31960049e+18 5.29043685e+18 5.29043685e+18
5.29043685e+18 5.29043685e+18 5.29043685e+18 5.29043685e+18
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5.29043685e+18 5.24952686e+18 2.09685699e+18 4.16588190e+17]
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我尝试过PCA(n_components ='mle')但是我遇到了这些错误..
Traceback (most recent call last):
File "xx", line 166, in <module>
pca.fit(x_array_train)
File "xx", line 225, in fit
self._fit(X)
File "/Users/lib/python2.7/site-packages/sklearn/decomposition/pca.py", line 294, in _fit
n_samples, n_features)
File "/Users/lib/python2.7/site-packages/sklearn/decomposition/pca.py", line 98, in _infer_dimension_
ll[rank] = _assess_dimension_(spectrum, rank, n_samples, n_features)
File "/Users/lib/python2.7/site-packages/sklearn/decomposition/pca.py", line 83, in _assess_dimension_
(1. / spectrum_[j] - 1. / spectrum_[i])) + log(n_samples)
ValueError: math domain error
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非常感谢任何帮助......
我没有使用Python
,但我没有你需要的东西C++
:opencv
.希望您成功将其转换为任何语言.
// choose how many eigenvectors you want:
int nEigensOfInterest = 0;
float sum = 0.0;
for (int i = 0; i < mEiVal.rows; ++i)
{
sum += mEiVal.at<float>(i, 0);
if (((sum * 100) / (sumOfEigens)) > 80)
{
nEigensOfInterest = i;
break;
}
}
logfile << "No of Eigens of interest: " << nEigensOfInterest << std::endl << std::endl;
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基本的想法是决定你需要的"任何%"组件.我选择那些80
.mEiVal
是按降序排序的特征值的列矩阵.sumOfEigens
是所有特征值的总和.
我没有经验scikit-learn
,请让我知道,我会删除答案.
我自己刚刚学习这一点,但在我看来,对 using 的引用0 < n_components < 1
表明您可以设置n_components
为 0.85,并且将使用解释 85% 方差所需的确切组件数量。您还可以通过打印来验证是否选择了正确数量的组件sum(pca.explained_variance_)
。您应该获得数据可能达到的超过 0.85(或您选择的任何值)的最小方差百分比总和。
当然,还有更复杂的方法来选择多个组件,但根据经验,70% - 90% 是一个合理的开始。