Я столкнулся с проблемами при попытке вывести вероятности предсказания SVM с помощью mlr3.
library(mlr3)
task = mlr_tasks$get("iris")
svm_learner = mlr_learners$get("classif.svm")
train_set = sample(task$nrow, 0.8 * task$nrow)
test_set = setdiff(seq_len(task$nrow), train_set)
svm_learner$train(task, row_ids = task$row_ids[train_set])
svm_learner$predict_type<-"prob"
prediction<-svm_learner$predict(task,row_ids = task$row_ids[test_set])
prediction
Warning message:
In predict.svm(self$model, newdata = newdata, probability = (self$predict_type == :
SVM has not been trained using `probability = TRUE`, probabilities not available for predictions.
Session info
> sessionInfo(package = NULL)
R version 3.6.2 (2019-12-12)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 10 x64 (build 17763)
Matrix products: default
Random number generation:
RNG: Mersenne-Twister
Normal: Inversion
Sample: Rounding
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] precrec_0.10.1 forcats_0.4.0 stringr_1.4.0 purrr_0.3.3 readr_1.3.1
[6] tidyr_1.0.0 tibble_2.1.3 tidyverse_1.2.1 dplyr_0.8.3 mlr3learners_0.1.5
[11] GGally_1.4.0 ggplot2_3.2.1 mlr3_0.1.6 mlr3viz_0.1.0 e1071_1.7-3
[16] biomaRt_2.38.0
loaded via a namespace (and not attached):
[1] Biobase_2.42.0 httr_1.4.1 bit64_0.9-7 jsonlite_1.6
[5] modelr_0.1.4 assertthat_0.2.1 lgr_0.3.3 stats4_3.6.2
[9] blob_1.2.0 cellranger_1.1.0 mlr3misc_0.1.6 progress_1.2.2
[13] pillar_1.4.3 RSQLite_2.1.2 backports_1.1.5 lattice_0.20-38
[17] glue_1.3.1 uuid_0.1-2 digest_0.6.23 RColorBrewer_1.1-2
[21] checkmate_1.9.4 rvest_0.3.3 colorspace_1.4-1 plyr_1.8.5
[25] XML_3.98-1.20 pkgconfig_2.0.3 mlr3measures_0.1.1 broom_0.5.2
[29] haven_2.1.0 scales_1.0.0 generics_0.0.2 IRanges_2.16.0
[33] withr_2.1.2 BiocGenerics_0.28.0 lazyeval_0.2.2 cli_2.0.0
[37] magrittr_1.5 crayon_1.3.4 readxl_1.3.1 paradox_0.1.0
[41] memoise_1.1.0 fansi_0.4.0 nlme_3.1-142 xml2_1.2.0
[45] class_7.3-15 tools_3.6.2 data.table_1.12.8 prettyunits_1.0.2
[49] hms_0.5.2 lifecycle_0.1.0 S4Vectors_0.20.1 munsell_0.5.0
[53] AnnotationDbi_1.44.0 compiler_3.6.2 rlang_0.4.1 grid_3.6.2
[57] RCurl_1.95-4.12 rstudioapi_0.10 bitops_1.0-6 labeling_0.3
[61] gtable_0.3.0 DBI_1.0.0 reshape_0.8.8 reshape2_1.4.3
[65] R6_2.4.1 lubridate_1.7.4 bit_1.1-14 zeallot_0.1.0
[69] stringi_1.4.3 parallel_3.6.2 Rcpp_1.0.2 vctrs_0.2.1
[73] tidyselect_0.2.5
Я понимаю, что SVM не будет выводить вероятность, но SVM может подогнать данные предсказания к функции гиперплоскости разделения и получить мера расстояния со знаком от гиперплоскости. Я хочу получить подписанное расстояние, а затем использовать их для расчета AU C. Но с predict_type<-"response" я могу получить только предсказанный класс, а не расстояние со знаком. С predict_type<-"probability" я получил ошибку выше.