scikit lda與R.MASS.lda相同的報告（痕跡比例）

Question

在R中運行LDA，如下所示：

iris = read.table("https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data", sep=",")
library("MASS")
mylda = lda(iris$V5~iris$V1+iris$V2+iris$V3+iris$V4)
mylda

給出以下輸出報告：

Call:
lda(iris$V5 ~ iris$V1 + iris$V2 + iris$V3 + iris$V4)

Prior probabilities of groups:
    Iris-setosa Iris-versicolor  Iris-virginica 
      0.3333333       0.3333333       0.3333333 

Group means:
                iris$V1 iris$V2 iris$V3 iris$V4
Iris-setosa       5.006   3.418   1.464   0.244
Iris-versicolor   5.936   2.770   4.260   1.326
Iris-virginica    6.588   2.974   5.552   2.026

Coefficients of linear discriminants:
               LD1         LD2
iris$V1  0.8192685  0.03285975
iris$V2  1.5478732  2.15471106
iris$V3 -2.1849406 -0.93024679
iris$V4 -2.8538500  2.80600460

Proportion of trace:
   LD1    LD2 
0.9915 0.0085 
> mylda
Call:
lda(iris$V5 ~ iris$V1 + iris$V2 + iris$V3 + iris$V4)

Prior probabilities of groups:
    Iris-setosa Iris-versicolor  Iris-virginica 
      0.3333333       0.3333333       0.3333333 

Group means:
                iris$V1 iris$V2 iris$V3 iris$V4
Iris-setosa       5.006   3.418   1.464   0.244
Iris-versicolor   5.936   2.770   4.260   1.326
Iris-virginica    6.588   2.974   5.552   2.026

Coefficients of linear discriminants:
               LD1         LD2
iris$V1  0.8192685  0.03285975
iris$V2  1.5478732  2.15471106
iris$V3 -2.1849406 -0.93024679
iris$V4 -2.8538500  2.80600460

Proportion of trace:
   LD1    LD2 
0.9915 0.0085 

通過scikit在python中做LDA：

from sklearn.discriminant_analysis import LinearDiscriminantAnalysis
mylda = LinearDiscriminantAnalysis().fit(iris.loc[:, 0:3], iris.loc[:, 4])
print(mylda.priors_)
print(mylda.means_)
print(mylda.scalings_)

到目前為止，將產生以下輸出：

array([ 0.33333333,  0.33333333,  0.33333333])
array([[ 5.006,  3.418,  1.464,  0.244],
       [ 5.936,  2.77 ,  4.26 ,  1.326],
       [ 6.588,  2.974,  5.552,  2.026]])
array([[-0.81926852,  0.03285975],
       [-1.5478732 ,  2.15471106],
       [ 2.18494056, -0.93024679],
       [ 2.85385002,  2.8060046 ]])

如何獲取/打印R的LDA在python中打印的“軌跡比例：”？

Answer 1

線性描述的系數可以使用以下公式找到：

trans=mylda.transform(iris.loc[:, 0:3])