Question

我目前有很多具有相同字段但不同行业的excel文件。我正在尝试创建一个函数，使我可以从所有Excel工作表的“硫酸盐”字段中获取总平均值。

这是我当前拥有的代码：

Mean_Pollution<-function(directory,pollutant,id = 1:332){
  directory <- c("001","002","003","004")
  for (x in directory){
    print(paste("Reading",x,"file"))
    temp = read.csv(paste(directory.path,x,".csv",sep = ""))
    print(paste("Finished reading",x,"file"))
    i = print(mean(temp$sulfate,na.rm = TRUE))
  }
}

Answer 1

您似乎拥有了所需的一切，并且有很多类似的问题，但我将提供一个示例。这里的假设是所有内容都在1个文件夹中。我将使用自己的文件夹：

setwd(dir = "C:/Users/Evan Friedland/Documents")
# Put some fake data in a fake folder
dir.create("Test_Folder")
setwd(dir = "C:/Users/Evan Friedland/Documents/Test_Folder")
n <- 5 # let's write 5 csv files
for(i in 1:n){
  write.csv(data.frame(madeupX = LETTERS[sample(1:24, 100, T)], sulfate = rnorm(100)), # fake data
            paste0(sprintf("%03d", i), ".csv")) # fake names
}
csvnames <- paste0(sprintf("%03d", 1:n), ".csv")

现在要“堆叠”每个文件的平均值，您需要做的是初始化一个空向量，遍历它们并将每个结果保存到向量元素中。

means <- numeric(n) # intialize numeric type vector, n long
names(means) <- csvnames # name each element for fun
means <- sapply(1:n, function(x){ # used sapply instead of a for loop but either is fine
  cat("+ ",csvnames[x],"\n") # print which csv is running
  mean(read.csv(paste0(csvnames[x]))$sulfate) # return the mean of the sulfate col
  })
#+  001.csv 
#+  002.csv 
#+  003.csv 
#+  004.csv 
#+  005.csv 
means # print results
#[1]  0.007859499  0.077447995  0.048796633 -0.101449790 0.224429258

在* for *函数中，我可以保持堆积值吗？

1 个答案: