当我只使用mtry参数作为tuingrid,它工作,但当我添加ntree参数时,错误变为Error in train.default(x, y, weights = w, ...): The tuning parameter grid should have columns mtry.代码如下:
require(RCurl)
require(prettyR)
library(caret)
url <- "https://raw.githubusercontent.com/gastonstat/CreditScoring/master/CleanCreditScoring.csv"
cs_data <- getURL(url)
cs_data <- read.csv(textConnection(cs_data))
classes <- cs_data[, "Status"]
predictors <- cs_data[, -match(c("Status", "Seniority", "Time", "Age", "Expenses",
"Income", "Assets", "Debt", "Amount", "Price", "Finrat", "Savings"), colnames(cs_data))]
train_set <- createDataPartition(classes, p = 0.8, list = FALSE)
set.seed(123)
cs_data_train = cs_data[train_set, ]
cs_data_test = cs_data[-train_set, ]
# Define the tuned parameter
grid …我的问题是我的pandas数据框中有这么多列,我正在尝试使用sklearn-pandas库中的dataframe mapper应用sklearn预处理,例如
mapper= DataFrameMapper([
('gender',sklearn.preprocessing.LabelBinarizer()),
('gradelevel',sklearn.preprocessing.LabelEncoder()),
('subject',sklearn.preprocessing.LabelEncoder()),
('districtid',sklearn.preprocessing.LabelEncoder()),
('sbmRate',sklearn.preprocessing.StandardScaler()),
('pRate',sklearn.preprocessing.StandardScaler()),
('assn1',sklearn.preprocessing.StandardScaler()),
('assn2',sklearn.preprocessing.StandardScaler()),
('assn3',sklearn.preprocessing.StandardScaler()),
('assn4',sklearn.preprocessing.StandardScaler()),
('assn5',sklearn.preprocessing.StandardScaler()),
('attd1',sklearn.preprocessing.StandardScaler()),
('attd2',sklearn.preprocessing.StandardScaler()),
('attd3',sklearn.preprocessing.StandardScaler()),
('attd4',sklearn.preprocessing.StandardScaler()),
('attd5',sklearn.preprocessing.StandardScaler()),
('sbm1',sklearn.preprocessing.StandardScaler()),
('sbm2',sklearn.preprocessing.StandardScaler()),
('sbm3',sklearn.preprocessing.StandardScaler()),
('sbm4',sklearn.preprocessing.StandardScaler()),
('sbm5',sklearn.preprocessing.StandardScaler())
])
我只是想知道是否有另一种更简洁的方法让我一次预处理许多变量而不用明确地写出来.
我发现有点烦人的另一件事是当我将所有pandas数据帧转换为sklearn可以使用的数组时,它们将丢失列名特征,这使得选择非常困难.有人知道如何在将pandas数据帧更改为np数组时保留列名作为键吗？
非常感谢!
我的问题是我如何能够为do.call函数添加更多参数.例如,我想绘制多面grid图grid.arrange,如何添加更多参数,例如ncol=3和main="main title"命令do.call(grid.arrange,plots)？
我不知道为什么代码不起作用
dd = function(x){
t = which.max(x[,'Sepal.Length'])
data = x[-t,]
m= max(data[,'Sepal.Length'])
return(m)
}
iris %>% group_by (Species) %>% do(dd(.))