I have a data frame, which contains the "date variable". (the test data and code is available here时,为什么兄弟元素和父元素会移动
但是,我使用“function = caretFunc”。它显示错误消息。
Error in { : task 1 failed - "missing value where TRUE/FALSE needed"
In addition: Warning messages:
1: In FUN(newX[, i], ...) : NAs introduced by coercion
2: In FUN(newX[, i], ...) : NAs introduced by coercion
3: In FUN(newX[, i], ...) : NAs introduced by coercion
4: In FUN(newX[, i], ...) : NAs introduced by coercion
5: In FUN(newX[, i], ...) : NAs introduced by coercion
6: In FUN(newX[, i], ...) : NAs introduced by coercion
7: In FUN(newX[, i], ...) : NAs introduced by coercion
8: In FUN(newX[, i], ...) : NAs introduced by coercion
9: In FUN(newX[, i], ...) : NAs introduced by coercion
10: In FUN(newX[, i], ...) : NAs introduced by coercion
我该怎么办?
重现错误的代码:
library(mlbench)
library(caret)
library(maps)
library(rgdal)
library(raster)
library(sp)
library(spdep)
library(GWmodel)
library(e1071)
library(plyr)
library(kernlab)
library(zoo)
mydata <- read.csv("Realestatedata_all_delete_date.csv", header=TRUE)
mydata$estate_TransDate <- as.Date(paste(mydata$estate_TransDate,1,sep="-"),format="%Y-%m-%d")
mydata$estate_HouseDate <- as.Date(mydata$estate_HouseDate,format="%Y-%m-%d")
rfectrl <- rfeControl(functions=caretFuncs, method="cv",number=10,verbose=TRUE,returnResamp = "final")
results <- rfe(mydata[,1:48],mydata[,49],sizes = c(1:48),rfeControl=rfectrl,method = "svmRadial")
print(results)
predictors(results)
plot(results, type=c("g", "o"))
答案 0 :(得分:0)
以下输入变量(您提供给分类器)NAs
中mydata
(缺少值):
colnames(mydata)[unique(which(is.na(mydata[,1:48]), arr.ind = TRUE)[,2])]
给出:
[1] "Aport_Distance" "Univ_Distance" "ParkR_Distance"
[4] "TRA_StationDistance" "THSR_StationDistance" "Schools_Distance"
[7] "Lib_Distance" "Sport_Distance" "ParkS_Distance"
[10] "Hyper_Distance" "Shop_Distance" "Post_Distance"
[13] "Hosp_Distance" "Gas_Distance" "Incin_Distance"
[16] "Mort_Distance"
此外,您的日期变量(交易日期和房屋日期)似乎在NAs
内转换为rfe(..)
。
SVM回归量似乎无法按原样处理NAs
。
我会将日期转换为自给定参考资料以来的年份:
:mydata$estate_TransAge <- as.numeric(as.Date("2015-11-01") - mydata$estate_TransDate) / 365.25
mydata$estate_HouseAge <- as.numeric(as.Date("2015-11-01") - mydata$estate_HouseDate) / 365.25
# define the set of input variables
inputVars = setdiff(colnames(mydata),
# exclude these
c("estate_TransDate", "estate_HouseDate", "estate_TotalPrice")
)
并且还会删除您用作回归量输入的任何列中任何NA
的条目:
traindata <- mydata[complete.cases(mydata[,inputVars]),]
然后用:
运行rferfectrl <- rfeControl(functions=caretFuncs, method="cv",number=10,verbose=TRUE,returnResamp = "final")
results <- rfe(
traindata[,inputVars],
traindata[,"estate_TotalPrice"],
rfeControl=rfectrl,
method = "svmRadial"
)
就我而言,这需要很长时间才能完成,所以我只使用以下百分之一的数据对其进行了测试:
traindata <- sample_frac(traindata, 0.01)
如果您获得数据来预测某些输入变量的价格NA
,那么问题仍然是该怎么办。