Purrr 映射在分割的训练数据帧上以获得每个模型的 auroc [英] Purrr map over splitted training-dataframe to get auroc for each model
本文介绍了Purrr 映射在分割的训练数据帧上以获得每个模型的 auroc的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
所以我有这个，有点废话的例子.我将 iris-dataframe 分成它们的物种，并尝试预测观察是否属于物种 setosa.
库(过程)iris[["setosa"]] = ifelse(iris$Species == "setosa", TRUE, FALSE)样本<-样本(c(真，假)，nrow(虹膜)，替换=T，概率=c(0.7，0.3))火车 = 虹膜 [样本，]测试 = 虹膜[！样品，]然后我想为每个模型获取 auroc.我想使用 purrr:map 方法，但我对如何做到这一点感到有些困惑.我有这个非工作伪类代码:
iris %>% split(., .$Species) %>%地图(~glm(setosa~Sepal.Length + Sepal.Width + Petal.Length + Petal.Width，数据=.x，家庭=二项式"))%>%{小费(预测 = map_dbl(~predict(., newdata = test, type="response")),参考 = test$setosa)}%>%地图(~roc(.x$setosa ~ .x$prediction))我想我犯了多个错误，也许最好手动完成.但我认为必须有一种巧妙的方法来使用 map 及其朋友来完成这项工作.
编辑(更好的数据)
这里有一个更好的例子:我想使用变量 p3 和 p15 预测 trigger 列.我想为每一类地质拟合一个模型并获得测试数据集的 auroc:
# 训练数据结构(列表(p3 = c(5, 0, 0, 2, 0, 0, 0, 2.8999999165535, 0,0, 0, 0, 17.5999999046326, 0, 0, 0.899999976158142, 10.0000000149012,1.79999995231628, 0, 8.80000019073486, 0, 0, 11.8999999761581,0, 9.60000026226044, 5.19999980926514, 11.7000002861023, 20.0999999046326,34.6000008583069, 0, 8.70000028610229, 33.8000000119209, 2.40000009536743,17, 27, 36.7999992370605, 0, 15.8999997973442, 0.300000011920929,7.69999980926514, 0, 0.899999976158142, 1.5, 0.300000011920929,3.50000002980232, 51.3999991416931, 6.09999990463257, 0.400000005960464,22, 65.3000020980835), p15 = c(34.2999999374151, 10.4999997392297,8.30000010877848, 69.4999992623925, 1.30000001192093, 0, 62.3999992161989,71.8999995738268, 32.6999994888902, 4, 0, 0.400000005960464,35.699999935925, 24.1000001206994, 0, 53.8999998271465, 41.8999992236495,37.8999994322658, 24.0999998524785, 65.5999999046326, 0, 0, 20.5000002905726,75.4000002145767, 68.1999989748001, 45.9000007808208, 180.999998480082,70.5000009462237, 112.099999666214, 11.3000001907349, 88.0999987274408,103.499998867512, 100.399998664856, 59.9999995827675, 200.699998855591,21.2999993562698, 47.1999997496605, 42.1999989748001, 58.6000000834465,161.299998879433, 43.3999999314547, 110.899999141693, 73.9000004529953,46.7999998703599, 25.7999995350838, 21.1000004559755, 86.100000500679,15.8999998569489, 5.3999999538064, 143.399998903275), 触发器 = c(FALSE,假，假，假，假，假，假，假，假，假，假，假，假，假，假，假，假，假，假，假，假，假，假，假，假，真，真，真，真，真、真、真、真、真、真、真、真、真、真、真、真、真、真、真、真、真、真、真、真、真)、geology = c(沉积岩"，结晶基底"，沉积岩"，沉积岩"，斑岩"，斑岩"，斑岩"、沉积岩"、沉积岩"、结晶基底"、带蛇绿岩的钙片岩"、结晶基底"、结晶基底"、沉积岩"、斑岩"、斑岩"、结晶基底"、结晶基底"、沉积岩"、结晶基底"、沉积岩"、斑岩"、结晶基底"、水晶地下室"，水晶地下室"，水晶地下室"，沉积岩"、斑岩"、沉积岩"、结晶基底"、水晶地下室"，水晶地下室"，斑岩"，带蛇绿岩的钙片岩"、结晶基底"、结晶基底"、沉积岩"、斑岩"、结晶基底"、斑岩"、结晶基底"、结晶基底"、结晶基底"、结晶基底"、带有蛇绿岩的钙片岩"，Plutonite"、水晶基底"、水晶基底"、斑岩"、沉积岩")), row.names = c(NA, -50L), class = c(tbl_df", tbl", data.frame"))和测试数据
# 测试数据# 和测试数据结构(列表(p3 = c(6.40000009536743，0，0，16.3000003397465，0, 0, 0, 0, 1, 1.5, 29.1000003814697, 7.49999982118607, 3.5,18.3999996185303, 2.69999990612268, 11.3000001907349, 0, 2, 9.5,0, 10.1999998092651, 0, 3.60000005364418, 0, 5.29999995231628,112.599998474121, 118.099997758865, 54.8999996185303, 72.8000011444092,79.9000015258789, 88.7000015377998, 0, 54.6000022888184, 144.599998474121,111.200000762939, 7.10000009834766, 32.0999999046326, 0.5, 5.3999999165535,0.300000011920929, 0, 36.7999982833862, 101.599998474121, 121.699998855591,31.0999994277954, 66.8000020980835, 139.200000762939, 9.50000011920929,135.300003051758, 110.900001525879), p15 = c(12.3999996185303,63.8000009655952, 20.7000007629395, 121.299998179078, 10.4000001549721,27.1999999880791, 49.5000003874302, 13.3000001907349, 31.3999998569489,15.4000002890825, 64.3999997377396, 25.1000001430511, 43.6999994516373,50.799999833107, 35.1999998092651, 35.1999998837709, 67.1000003442168,19.400000333786, 49.300000667572, 21.3999996706843, 75.600000411272,38.700000859797, 30.2999994754791, 14.9000003933907, 53.2000011727214,137.900000333786, 0.100000001490116, 119.300001859665, 139.700000107288,147.799997329712, 45.3000004068017, 56.5000000670552, 47.7999995946884,2.90000009536743, 139.499999403954, 26.6999999284744, 6.5, 149.700001835823,210.299998342991, 114.499999642372, 3.60000002384186, 60.099999524653,97.5999984890223, 153.100000120699, 245.299996376038, 123.49999922514,3.70000004768372, 90.5999985486269, 49.1000001132488, 138.599999785423), 触发器 = c(FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE,假，假，假，假，假，假，假，假，假，假，假，假，假，假，假，假，假，假，真、真、真、真、真、真、真、真、真、真、真、真、真、真、真、真、真、真、真、真、真、真、TRUE, TRUE, TRUE)), row.names = c(NA, -50L), groups = structure(list(trigger = c(FALSE, TRUE), .rows = structure(list(1:25, 26:50), ptype = integer(0), class = c(vctrs_list_of",vctrs_vctr", list"))), row.names = c(NA, -2L), class = c(tbl_df",tbl", data.frame"), .drop = TRUE), class = c(grouped_df",tbl_df"、tbl"、data.frame"))然后我会做类似上面的事情:
<预><代码>训练 %>% split(., .$geology) %>%地图(〜glm(触发器〜p3 + p15，数据= .x))％>％{小费(预测 = map_dfr(~predict(.x, newdata=test, type=response")),参考 = 测试 $ 触发器，auroc = as.numeric(roc(ref ~ 预测，数据 = 测试)[[1]]))}但这根本行不通...
解决方案
我建议使用以下代码:
train %>% group_by(geology) %>%group_modify(函数(x，y){模型 <- glm(trigger ~ p3 + p15, data=x)小费(预测=预测(模型，新数据=测试，类型=响应")，参考 = 测试 $ 触发器，auroc = as.numeric(pROC::roc(ref ~ 预测，数据 = 测试)[[1]]))})我认为不将预测数据与 KPI 结合以提高预测质量会更有意义.像这样:
train %>% group_by(geology) %>%group_modify(函数(x，y){模型 <- glm(trigger ~ p3 + p15, data=x)预测=预测(模型，新数据=测试，类型=响应")小费(auroc = as.numeric(pROC::roc(test$trigger ~ 预测，数据 = 测试)[[1]]))})So I have this, kind of nonsense, example. I split up the iris-dataframe into their species and try to predict if a observation is of species setosa.
library(proc)
iris[["setosa"]] = ifelse(iris$Species == "setosa", TRUE, FALSE)
sample <- sample(c(TRUE, FALSE), nrow(iris), replace = T, prob = c(0.7, 0.3))
train = iris[sample, ]
test = iris[!sample, ]
I would then like to get the auroc for each of the models. I wanted to use a purrr:map-approach, but I am a little confused about how I could do this. I have this, nonworking pseudo-like, code:
iris %>% split(., .$Species) %>%
map(~glm(setosa ~ Sepal.Length + Sepal.Width + Petal.Length + Petal.Width, data =.x, family="binomial")) %>% {
tibble(
prediction = map_dbl(~predict(., newdata = test, type="response")),
referece = test$setosa
)
} %>%
map(~roc(.x$setosa ~ .x$prediction))
I think I am making multiple errors and maybe it would be better to do it all manually. But I think there must be a neat way to get this done with map and its friends.
EDIT (better data)
Here is a little bit better example: I want to predict the trigger-column using the variables p3 and p15. I want to fit a model for each class of geology and get the auroc for the test dataset:
# the training data
structure(list(p3 = c(5, 0, 0, 2, 0, 0, 0, 2.8999999165535, 0,
0, 0, 0, 17.5999999046326, 0, 0, 0.899999976158142, 10.0000000149012,
1.79999995231628, 0, 8.80000019073486, 0, 0, 11.8999999761581,
0, 9.60000026226044, 5.19999980926514, 11.7000002861023, 20.0999999046326,
34.6000008583069, 0, 8.70000028610229, 33.8000000119209, 2.40000009536743,
17, 27, 36.7999992370605, 0, 15.8999997973442, 0.300000011920929,
7.69999980926514, 0, 0.899999976158142, 1.5, 0.300000011920929,
3.50000002980232, 51.3999991416931, 6.09999990463257, 0.400000005960464,
22, 65.3000020980835), p15 = c(34.2999999374151, 10.4999997392297,
8.30000010877848, 69.4999992623925, 1.30000001192093, 0, 62.3999992161989,
71.8999995738268, 32.6999994888902, 4, 0, 0.400000005960464,
35.699999935925, 24.1000001206994, 0, 53.8999998271465, 41.8999992236495,
37.8999994322658, 24.0999998524785, 65.5999999046326, 0, 0, 20.5000002905726,
75.4000002145767, 68.1999989748001, 45.9000007808208, 180.999998480082,
70.5000009462237, 112.099999666214, 11.3000001907349, 88.0999987274408,
103.499998867512, 100.399998664856, 59.9999995827675, 200.699998855591,
21.2999993562698, 47.1999997496605, 42.1999989748001, 58.6000000834465,
161.299998879433, 43.3999999314547, 110.899999141693, 73.9000004529953,
46.7999998703599, 25.7999995350838, 21.1000004559755, 86.100000500679,
15.8999998569489, 5.3999999538064, 143.399998903275), trigger = c(FALSE,
FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE,
FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE,
FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, TRUE, TRUE, TRUE, TRUE,
TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE,
TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE),
geology = c("Sedimentary rocks", "Crystalline basement",
"Sedimentary rocks", "Sedimentary rocks", "Porphyry", "Porphyry",
"Porphyry", "Sedimentary rocks", "Sedimentary rocks", "Crystalline basement",
"Calcschists with ophiolites", "Crystalline basement", "Crystalline basement",
"Sedimentary rocks", "Porphyry", "Porphyry", "Crystalline basement",
"Crystalline basement", "Sedimentary rocks", "Crystalline basement",
"Sedimentary rocks", "Porphyry", "Crystalline basement",
"Crystalline basement", "Crystalline basement", "Crystalline basement",
"Sedimentary rocks", "Porphyry", "Sedimentary rocks", "Crystalline basement",
"Crystalline basement", "Crystalline basement", "Porphyry",
"Calcschists with ophiolites", "Crystalline basement", "Crystalline basement",
"Sedimentary rocks", "Porphyry", "Crystalline basement",
"Porphyry", "Crystalline basement", "Crystalline basement",
"Crystalline basement", "Crystalline basement", "Calcschists with ophiolites",
"Plutonite", "Crystalline basement", "Crystalline basement",
"Porphyry", "Sedimentary rocks")), row.names = c(NA, -50L
), class = c("tbl_df", "tbl", "data.frame"))
and the test data
# test data
# and the test data
structure(list(p3 = c(6.40000009536743, 0, 0, 16.3000003397465,
0, 0, 0, 0, 1, 1.5, 29.1000003814697, 7.49999982118607, 3.5,
18.3999996185303, 2.69999990612268, 11.3000001907349, 0, 2, 9.5,
0, 10.1999998092651, 0, 3.60000005364418, 0, 5.29999995231628,
112.599998474121, 118.099997758865, 54.8999996185303, 72.8000011444092,
79.9000015258789, 88.7000015377998, 0, 54.6000022888184, 144.599998474121,
111.200000762939, 7.10000009834766, 32.0999999046326, 0.5, 5.3999999165535,
0.300000011920929, 0, 36.7999982833862, 101.599998474121, 121.699998855591,
31.0999994277954, 66.8000020980835, 139.200000762939, 9.50000011920929,
135.300003051758, 110.900001525879), p15 = c(12.3999996185303,
63.8000009655952, 20.7000007629395, 121.299998179078, 10.4000001549721,
27.1999999880791, 49.5000003874302, 13.3000001907349, 31.3999998569489,
15.4000002890825, 64.3999997377396, 25.1000001430511, 43.6999994516373,
50.799999833107, 35.1999998092651, 35.1999998837709, 67.1000003442168,
19.400000333786, 49.300000667572, 21.3999996706843, 75.600000411272,
38.700000859797, 30.2999994754791, 14.9000003933907, 53.2000011727214,
137.900000333786, 0.100000001490116, 119.300001859665, 139.700000107288,
147.799997329712, 45.3000004068017, 56.5000000670552, 47.7999995946884,
2.90000009536743, 139.499999403954, 26.6999999284744, 6.5, 149.700001835823,
210.299998342991, 114.499999642372, 3.60000002384186, 60.099999524653,
97.5999984890223, 153.100000120699, 245.299996376038, 123.49999922514,
3.70000004768372, 90.5999985486269, 49.1000001132488, 138.599999785423
), trigger = c(FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE,
FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE,
FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE,
TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE,
TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE,
TRUE, TRUE, TRUE)), row.names = c(NA, -50L), groups = structure(list(
trigger = c(FALSE, TRUE), .rows = structure(list(1:25, 26:50), ptype = integer(0), class = c("vctrs_list_of",
"vctrs_vctr", "list"))), row.names = c(NA, -2L), class = c("tbl_df",
"tbl", "data.frame"), .drop = TRUE), class = c("grouped_df",
"tbl_df", "tbl", "data.frame"))
I then would do something like the above:
train %>% split(., .$geology) %>%
map(~glm(trigger ~ p3 + p15, data = .x)) %>% {
tibble(
prediction = map_dfr(~predict(.x, newdata=test, type="response")),
ref = test$trigger,
auroc = as.numeric(roc(ref ~ prediction, data = test)[[1]])
)
}
But this does not work at all...
解决方案
I would suggest the following code:
train %>% group_by(geology) %>%
group_modify( function(x, y) {
model <- glm(trigger ~ p3 + p15, data=x)
tibble(
prediction = predict(model, newdata=test, type="response"),
ref = test$trigger,
auroc = as.numeric(pROC::roc(ref ~ prediction, data = test)[[1]])
)
})
I think it would make more sense to not combine prediction data with KPIs for prediction quality. Something like this:
train %>% group_by(geology) %>%
group_modify( function(x, y) {
model <- glm(trigger ~ p3 + p15, data=x)
prediction = predict(model, newdata=test, type="response")
tibble(
auroc = as.numeric(pROC::roc(test$trigger ~ prediction, data = test)[[1]])
)
})
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