如何解决我可以使用sklearn IterativeImputer填写缺少的分类数据吗？
我有一个分类和连续性特征的数据集，其中许多缺少要素。我想知道我是否可以使用相应的计算机来填写连续数据和分类数据。
如果无法完成，什么是最好的进行方式？最好将数据分为连续特征和离散特征，然后对第一个集合使用IterativeImputer，对第二个集合使用KNN，然后将它们合并吗？
任何帮助将不胜感激。
数据包含65个特征：
x_train
age sex painloc painexer relrest cp trestbps htn chol smoke ... om1 om2 rcaprox rcadist lvx1 lvx2 lvx3 lvx4 lvf cathef
288 -1.109572 1.0 0.0 0.0 0.0 1.0 -0.655059 0.0 0.818661 NaN ... NaN NaN NaN NaN 1.0 1.0 1.0 1.0 2.0 0.568676
283 -0.180525 1.0 1.0 0.0 0.0 2.0 1.447445 0.0 -0.040919 NaN ... NaN NaN NaN NaN 1.0 1.0 1.0 1.0 1.0 NaN
230 -0.077297 1.0 1.0 1.0 0.0 3.0 0.659006 1.0 2.872604 NaN ... 2.0 NaN 2.0 NaN 1.0 1.0 1.0 1.0 1.0 NaN
380 -0.799890 0.0 1.0 1.0 1.0 4.0 -0.129433 0.0 0.339106 NaN ... NaN NaN NaN NaN 1.0 1.0 1.0 1.0 1.0 NaN
147 0.129157 1.0 1.0 1.0 1.0 4.0 NaN 0.0 0.031467 0.0 ... 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 -0.822164
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
847 -0.180525 0.0 NaN NaN NaN 3.0 0.185942 1.0 -0.040919 NaN ... 1.0 NaN 1.0 1.0 1.0 1.0 1.0 1.0 1.0 NaN
301 -0.283752 1.0 1.0 1.0 1.0 4.0 -0.129433 0.0 -0.194738 NaN ... NaN NaN NaN NaN 1.0 1.0 1.0 1.0 1.0 NaN
693 0.645295 1.0 NaN NaN NaN 4.0 -0.392246 1.0 0.520070 NaN ... 1.0 NaN 2.0 1.0 1.0 1.0 1.0 1.0 1.0 NaN
115 1.058204 1.0 1.0 1.0 1.0 4.0 NaN 0.0 0.954384 0.0 ... 1.0 1.0 2.0 1.0 1.0 1.0 1.0 1.0 1.0 -0.811925
155 1.574341 1.0 1.0 1.0 1.0 4.0 NaN 1.0 NaN 0.0 ... 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 NaN
我已经对连续变量进行了标准化。 有很多分类特征，例如'painloc'和'painexer'，具有缺失的值，也有连续的特征，例如'age'（我决定将其视为连续的）和'chol'，也具有缺失的元素。
x_mice=x_train
mice_impute = IterativeImputer(sample_posterior=True)
x_mice=pd.DataFrame(mice_impute.fit_transform(x_mice))
x_mice.columns=labels
x_mice
age sex painloc painexer relrest cp trestbps htn chol smoke ... om1 om2 rcaprox rcadist lvx1 lvx2 lvx3 lvx4 lvf cathef
0 1.049449 1.0 1.000000 1.000000 1.000000 4.0 0.444874 0.000000 0.540723 0.000000 ... 1.000000 1.000000 2.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 -0.891887
1 0.505617 1.0 1.000000 1.000000 0.000000 2.0 -0.266785 0.000000 -1.752150 0.000000 ... 1.000000 1.000000 2.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 -0.888760
2 0.831916 1.0 1.000000 0.000000 0.000000 4.0 -1.080109 0.764037 -1.752150 1.450166 ... 1.000000 1.000000 1.000000 1.000000 1.025761 0.879404 -0.400332 3.193691 3.267492 1.118696
3 -0.582047 1.0 1.000000 0.000000 0.000000 2.0 -1.588436 0.000000 -0.249794 0.000000 ... 1.383778 1.048614 -0.147575 1.942328 1.000000 1.000000 1.000000 1.000000 1.000000 0.802084
4 -1.452178 1.0 1.000000 0.000000 0.000000 3.0 0.444874 1.000000 5.232542 1.000000 ... 1.235595 1.249215 2.269437 1.155985 1.000000 1.000000 1.000000 1.000000 1.000000 -1.935223
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
624 0.179318 1.0 1.000000 1.000000 1.000000 4.0 -0.571781 0.000000 0.628910 -0.060307 ... 0.928614 0.830982 1.080936 1.185430 1.000000 1.000000 1.000000 1.000000 1.000000 -1.032691
625 1.702047 1.0 1.000000 0.000000 1.000000 3.0 0.444874 0.000000 -1.752150 0.000000 ... 1.000000 1.000000 2.000000 1.000000 1.000000 1.000000 1.000000 1.000000 2.000000 -0.895014
626 -0.364514 1.0 0.694690 1.738101 0.396025 4.0 0.953201 1.000000 0.390804 1.287500 ... 1.000000 0.739708 2.000000 1.000000 1.000000 1.000000 1.000000 1.000000 2.000000 -0.523902
627 0.723149 1.0 0.762459 0.038032 0.315826 4.0 0.444874 1.000000 0.831741 0.750375 ... 1.000000 0.912221 2.000000 1.000000 1.000000 1.000000 1.000000 1.000000 2.000000 0.730936
628 0.940682 1.0 1.000000 1.000000 1.000000 4.0 -0.000217 0.000000 -0.252964 0.000000 ... 1.000000 1.000000 2.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 -0.888134
它对于连续性功能很好，但对分类功能却不适用，因为它可以填写十进制数字，这显然是不正确的。
版权声明：本文内容由互联网用户自发贡献，该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 dio@foxmail.com 举报，一经查实，本站将立刻删除。