基于多個字符的ifelse陳述句作為R中的條件

我有以下功能：

brt <- gbm.step(data = train,
                          gbm.x = 5:9, # column indices for covariates
                          gbm.y = 4, # column index for response (presence-absence)
                          family = "bernoulli",
                          tree.complexity = ifelse(prNum < 50, 1, 5),
                          learning.rate = 0.001,
                          bag.fraction = 0.75,
                          max.trees = ifelse(s == "Jacaranda.puberula" | s == "Cestrum.intermedium", 50, 10000),
                          n.trees = 50,
                          n.folds = 5, # 5-fold cross-validation
                          silent = TRUE) # avoid printing the cv results

我正在嘗試運行上述函式，但他們使用ifelse內部max.trees引數的方式回傳錯誤。這是正確的編碼方式嗎？我想要完成的只是max.trees對這兩個物種使用 50，而不是 10000：Jacaranda.puberula或Cestrum.intermedium。提醒一下，該物件s來自另一個名為 的向量sp，它是一個物種名稱串列。每個s都是我在這個函式中回圈的不同物種。

問題出在我撰寫陳述句的方式上，還是我使用 OR 作為邏輯的事實？

編輯：我認為這可能與功能有關。我在這里重新運行了分析，它回傳：

Error in UseMethod("predict") :
no applicable method for 'predict' applied to an object of class "NULL"
In addition: There were 50 or more warnings (use warnings() to see the first 50)

uj5u.com熱心網友回復：

廣義gbm提升max.trees回歸樹n.trees（max_depth

n.trees = ifelse(s == "Jacaranda.puberula" | s == "Cestrum.intermedium", 50, 10000)

編輯：

如果無法訪問您的資料，就很難準確猜測出了什么問題。上次我假設你使用的包是gbm. 但是，使用示例資料集（BostonHousing來自 package mlbench），下面的示例顯示了the和packages 的gbm實作如何按預期作業：gbmdismo

library(mlbench)
data(BostonHousing)
head(BostonHousing)
# crim zn indus chas   nox    rm  age    dis rad tax ptratio      b lstat medv
#1 0.00632 18  2.31    0 0.538 6.575 65.2 4.0900   1 296    15.3 396.90  4.98 24.0
#2 0.02731  0  7.07    0 0.469 6.421 78.9 4.9671   2 242    17.8 396.90  9.14 21.6
#3 0.02729  0  7.07    0 0.469 7.185 61.1 4.9671   2 242    17.8 392.83  4.03 34.7
#4 0.03237  0  2.18    0 0.458 6.998 45.8 6.0622   3 222    18.7 394.63  2.94 33.4
#5 0.06905  0  2.18    0 0.458 7.147 54.2 6.0622   3 222    18.7 396.90  5.33 36.2
#6 0.02985  0  2.18    0 0.458 6.430 58.7 6.0622   3 222    18.7 394.12  5.21 28.7

library(gbm)
set.seed(102) # for reproducibility
# predict target medv given other predictors
gbm1 <- gbm(medv ~ ., data = BostonHousing, #var.monotone = c(0, 0, 0, 0, 0, 0),
            distribution = "gaussian", shrinkage = 0.1,
            n.trees = 100,
            interaction.depth = 3, bag.fraction = 0.5, train.fraction = 0.5,
            n.minobsinnode = 10, cv.folds = 5, keep.data = TRUE,
            verbose = FALSE, n.cores = 1)
#predict(gbm1)

library(dismo)
# predict target medv given other predictors
gbm2 <- gbm.step(data=BostonHousing, gbm.x = 1:13, gbm.y = 14, family = "gaussian",
                            max.trees = 10000,
                            tree.complexity = 5, learning.rate = 0.01, bag.fraction = 0.5)



#predict(gbm2)

# compare predictions
library(tidyverse)
data.frame(index=1:nrow(BostonHousing),
           actual=BostonHousing$medv, 
           pred.gbm1=predict(gbm1), 
           pred.gbm2=predict(gbm2)) %>% 
  gather('pred', 'medv', -index) %>%
  ggplot(aes(index, medv, group=pred, color=pred))   geom_line()

標籤：r if 语句 机器学习 gbm

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