我在做coursea的作业的时候发现如果我用mean求的答案跟用普通的数学公式求的均值不同，大神能帮我看看哪里出错了吗，以下是两个程序：红色部分就是有疑问的地方，根据答案来看，第二个程序的答案才是对的
pollutantmean<- function(directory, pollutant, id=1:332) {

total=0
observations=0
for (i in id){
if (i<10){
data<-read.csv(paste("0","0",as.character(i),".csv",sep=""),
header =TRUE,
na.strings=c("NA","NaN","")
)
}
else if (i>=10 & i<100) {
data<-read.csv(paste("0",as.character(i),".csv",sep=""),
header =TRUE,
na.strings=c("NA","NaN","")
)
}

else {
data<-read.csv(paste(as.character(i),".csv",sep=""),
header =TRUE,
na.strings=c("NA","NaN",""))
}
data=na.omit(data)
observations = observations + nrow(data)
if (pollutant == "sulfate") { return(mean(data$sulfate))}
else {return(mean(data$nitrate))}
}

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pollutantmean <- function(directory, pollutant, id = 1:332) {


total = 0
observations = 0

for (i in id)
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if (i <10) {
data <- read.csv(paste("0","0", as.character(i), ".csv", sep=""),## for example, if 'id' =7, we get 007.csv
header = T,
na.strings=c("NA","NaN", " "))
}

else if (i>=10 & i<100) {
data <- read.csv(paste("0", as.character(i), ".csv", sep=""),## for example, if 'id' = 17, we get 017.csv
header = T,
na.strings=c("NA","NaN", " ")
)
}



else {
data <- read.csv(paste(as.character(i), ".csv", sep=""), ## Normal
header = T,
na.strings=c("NA","NaN", " ")
)
}

data = na.omit(data)
observations = observations + nrow(data)
if (pollutant == "sulfate") {total = total + sum(data$sulfate)}
else {total = total + sum(data$nitrate)}
}
return(total/observations)

}