R：强制引入的 Kaggle Titanic 数据集随机森林 NA

Question

我目前正在使用 Titanic 数据集在 Kaggle 上练习 R 我正在使用随机森林算法

下面是代码

fit <- randomForest(as.factor(Survived) ~ Pclass + Sex + Age_Bucket + Embarked
                + Age_Bucket + Fare_Bucket + F_Name + Title + FamilySize + FamilyID, 
                data=train, importance=TRUE, ntree=5000)

我收到以下错误

Error in randomForest.default(m, y, ...) : 
  NA/NaN/Inf in foreign function call (arg 1)
In addition: Warning messages:
1: In data.matrix(x) : NAs introduced by coercion
2: In data.matrix(x) : NAs introduced by coercion
3: In data.matrix(x) : NAs introduced by coercion
4: In data.matrix(x) : NAs introduced by coercion

我的数据如下所示

$ Survived   : int  0 1 1 1 0 0 0 0 1 1 ...
$ Pclass     : int  3 1 3 1 3 3 1 3 3 2 ...
$ Sex        : Factor w/ 2 levels "female","male": 2 1 1 1 2 2 2 2 1 1...
$ Age_Bucket : chr  "20-25" "30-40" "25-30" "30-40" ...
$ Fare_Bucket: chr  "<10" "30+" "<10" "30+" ...
$ Title      : Factor w/ 11 levels "Col","Dr","Lady",..: 7 8 5 8 7 7 7 4 8 8 ...
$ F_Name     : chr  "Braund" "Cumings" "Heikkinen" "Futrelle" ...
$ FamilySize : num  2 2 1 2 1 1 1 5 3 2 ...
$ Embarked   : Factor w/ 3 levels "C","Q","S": 3 1 3 3 3 2 3 3 3 1 ...
$ FamilyID   : chr  "Small" "Small" "Alone" "Small" ...

如果我只输入以下内容，我没有强制问题，据我所知，这是唯一发生强制以创建 NA 值的地方

as.factor(Survived)

谁能看到问题

感谢您的宝贵时间

Answer 1

您需要将char 列转换为因子。因子在内部被视为整数，而字符字段则不是。请看下面的小示范：

数据：

df <- data.frame(y = sample(0:1, 26, rep=T), x1=runif(26), x2=letters, stringsAsFactors=F)

df$y <- as.factor(df$y)

> str(df)
'data.frame':   26 obs. of  3 variables:
 $ y : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 2 2 1 ...
 $ x1: num  0.457 0.296 0.517 0.478 0.764 ...
 $ x2: chr  "a" "b" "c" "d" ...

现在，如果我运行 randomForest 函数：

> randomForest(y ~ x1 + x2, data=df)
Error in randomForest.default(m, y, ...) : 
  NA/NaN/Inf in foreign function call (arg 1)
In addition: Warning message:
In data.matrix(x) : NAs introduced by coercion

我遇到了同样的错误。

而如果我将char 列转换为factor：

df$x2 <- as.factor(df$x2)

> randomForest(y ~ x1 + x2, data=df)

Call:
 randomForest(formula = y ~ x1 + x2, data = df) 
               Type of random forest: classification
                     Number of trees: 500
No. of variables tried at each split: 1

        OOB estimate of  error rate: 61.54%
Confusion matrix:
  0  1 class.error
0 0 16           1
1 0 10           0

效果很好！