我正在尝试在R中编写一个带3个输入的函数：
我的计算机上有一个目录，里面装满了CSV文件，超过300个。这个功能的功能如下所示：
pollutantmean <- function(directory, pollutant, id = 1:332) {
## 'directory' is a character vector of length 1 indicating
## the location of the CSV files
## 'pollutant' is a character vector of length 1 indicating
## the name of the pollutant for which we will calculate the
## mean; either "sulfate" or "nitrate".
## 'id' is an integer vector indicating the monitor ID numbers
## to be used
## Return the mean of the pollutant across all monitors list
## in the 'id' vector (ignoring NA values)
}
此函数的示例输出如下所示：
source("pollutantmean.R")
pollutantmean("specdata", "sulfate", 1:10)
## [1] 4.064
pollutantmean("specdata", "nitrate", 70:72)
## [1] 1.706
pollutantmean("specdata", "nitrate", 23)
## [1] 1.281
我可以一次性阅读整篇文章：
path = "C:/Users/Sean/Documents/R Projects/Data/specdata"
fileList = list.files(path=path,pattern="\\.csv$",full.names=T)
all.files.data = lapply(fileList,read.csv,header=TRUE)
DATA = do.call("rbind",all.files.data)
我的问题是：
来自所有文件的整个数据如下所示：
summary(DATA)
Date sulfate nitrate ID
2004-01-01: 250 Min. : 0.0 Min. : 0.0 Min. : 1.0
2004-01-02: 250 1st Qu.: 1.3 1st Qu.: 0.4 1st Qu.: 79.0
2004-01-03: 250 Median : 2.4 Median : 0.8 Median :168.0
2004-01-04: 250 Mean : 3.2 Mean : 1.7 Mean :164.5
2004-01-05: 250 3rd Qu.: 4.0 3rd Qu.: 2.0 3rd Qu.:247.0
2004-01-06: 250 Max. :35.9 Max. :53.9 Max. :332.0
(Other) :770587 NA's :653304 NA's :657738
任何想法如何制定这一点都将受到高度赞赏...
干杯
答案 0 :(得分：4)
所以，你可以像这样模拟你的情况;
# Simulate some data:
# Create 332 data frames
set.seed(1)
df.list<-replicate(332,data.frame(sulfate=rnorm(100),nitrate=rnorm(100)),simplify=FALSE)
# Generate names like 001.csv and 010.csv
file.names<-paste0('specdata/',sprintf('%03d',1:332),'.csv')
# Write them to disk
invisible(mapply(write.csv,df.list,file.names))
这是一个可以读取这些文件的函数：
pollutantmean <- function(directory, pollutant, id = 1:332) {
file.names <- list.files(directory)
file.numbers <- as.numeric(sub('\\.csv$','', file.names))
selected.files <- na.omit(file.names[match(id, file.numbers)])
selected.dfs <- lapply(file.path(directory,selected.files), read.csv)
mean(c(sapply(selected.dfs, function(x) x[ ,pollutant])), na.rm=TRUE)
}
pollutantmean('specdata','nitrate',c(1:100,141))
# [1] -0.005450574
答案 1 :(得分：2)
User enters id either atomic or in a range e.g.假设用户输入1，但文件名为001.csv，或者如果用户输入范围，该怎么办 1:10然后文件名是001.csv ... 010.csv
您可以使用正则表达式和gsub函数从文件名中删除前导零，然后创建一个字典（在r中，一个命名向量），将修改后的/ gsub'd文件名转换为实际的文件名。
例如：如果您的文件名是字符向量，fnames
fnames = c("001.csv","002.csv")
names(fnames) <- gsub(pattern="^[0]*", replacement="", x=fnames)
通过这个，矢量fnames被转换为字典，让你用001.csv的行调用名为fnames["1.csv"]的文件。您还可以使用gsub()删除文件名的.csv部分。
用户使用的是“硫酸盐”或“硝酸盐” 他/她有兴趣获得...的意思。有很多 这些列中缺少值（我需要从列中省略） 在计算平均值之前。
许多R函数都有一个忽略指示缺失值的特殊字符的选项。尝试在R命令提示符处输入help(mean)以查找有关此功能的信息。
答案 2 :(得分：2)
这是一个甚至你的祖母都能理解的解决方案：
pollutantmean <- function(directory, pollutant, id = 1:332) {
# Break this function up into a series of smaller functions
# that do exactly what you expect them to. Your friends
# will love you for it.
csvFiles = getFilesById(id, directory)
dataFrames = readMultipleCsvFiles(csvFiles)
dataFrame = bindMultipleDataFrames(dataFrames)
getColumnMean(dataFrame, column = pollutant)
}
getFilesById <- function(id, directory = getwd()) {
allFiles = list.files(directory)
file.path(directory, allFiles[id])
}
readMultipleCsvFiles <- function(csvFiles) {
lapply(csvFiles, read.csv)
}
bindMultipleDataFrames <- function(dataFrames) {
Reduce(function(x, y) rbind(x, y), dataFrames)
}
getColumnMean <- function(dataFrame, column, ignoreNA = TRUE) {
mean(dataFrame[ , column], na.rm = ignoreNA)
}
答案 3 :(得分：1)
这就是我修复它的方式：
pollutantmean <- function(directory, pollutant, id = 1:332) {
#set the path
path = directory
#get the file List in that directory
fileList = list.files(path)
#extract the file names and store as numeric for comparison
file.names = as.numeric(sub("\\.csv$","",fileList))
#select files to be imported based on the user input or default
selected.files = fileList[match(id,file.names)]
#import data
Data = lapply(file.path(path,selected.files),read.csv)
#convert into data frame
Data = do.call(rbind.data.frame,Data)
#calculate mean
mean(Data[,pollutant],na.rm=TRUE)
}
最后一个问题是我的功能应该调用＆＃34; specdata＆＃34; （所有csv所在的目录名称）作为目录，r中是否有目录类型对象？
假设我将该函数称为：
pollutantmean(specdata, "niterate", 1:10)
它应该在我的工作目录中找到specdata目录的路径......我该怎么做？
答案 4 :(得分：1)
这是一个用于计算文件列表中特定列的平均值的一般功能。不确定应该如何设置id，但现在它充当索引向量（即id = 1:3计算文件列表中前三个文件的平均值。）
multifile.means <- function(directory = getwd(), pollutant, id = NULL)
{
d <- match.arg(directory, list.files())
cn <- match.arg(pollutant, c('sulfate', 'nitrate'))
## get a vector of complete file paths in the given 'directory'
p <- dir(d, full.names = TRUE)
## subset 'p' based on 'id' values
if(!is.null(id)){
id <- id[!id > length(p)]
p <- p[id]
}
## read, store, and name the relevant columns
cl <- sapply(p, function(x){ read.csv(x)[,cn] }, USE.NAMES = FALSE)
colnames(cl) <- basename(p)
## return a named list of some results
list(values = cl,
mean = mean(cl, na.rm = TRUE),
colMeans = colMeans(cl, na.rm = TRUE))
}
将其用于试驾：
> multifile.means('testDir', 'sulfate')
# $values
# 001.csv 057.csv 146.csv 213.csv
# [1,] 5 10 NA 9
# [2,] 1 1 10 3
# [3,] 10 4 10 2
# [4,] 3 10 9 NA
# [5,] 4 1 5 5
# $mean
# [1] 5.666667
# $colMeans
# 001.csv 057.csv 146.csv 213.csv
# 4.60 5.20 8.50 4.75
答案 5 :(得分：1)
选择的答案看起来不错，但这是另一种选择。这个答案适用于JHU课程所涵盖的基础知识。
pollutantmean <- function(directory, pollutant, id = 1:332) {
csvfiles <- dir(directory, "*\\.csv$", full.names = TRUE)
data <- lapply(csvfiles[id], read.csv)
numDataPoints <- 0L
total <- 0L
for (filedata in data) {
d <- filedata[[pollutant]] # relevant column data
d <- d[complete.cases(d)] # remove NA values
numDataPoints <- numDataPoints + length(d)
total <- total + sum(d)
}
total / numDataPoints
}
答案 6 :(得分：1)
我花了几个小时来解决这个问题，但这是我的（较短的）版本
pollutmean<- function(dir, pollutant, id=1:332) {
dir<- list.files(dir, full.names = T) #list files
dat<- data.frame() #make empty df
for (i in id) {
dat <- rbind(dat, read.csv(dir[i])) #rbind all files
}
mean(dat[,pollutant], na.rm = TRUE) #calculate mean of given column
}
pollutmean("assign/specdata", "sulfate", id=1:60)
答案 7 :(得分：-1)
我也在阅读课程，并提出了以下解决方案：
pollutantmean <- function(directory="d:/dev/r/documents/specdata", pollutant,
id) {
myfilename = paste(directory,"/",formatC(id, width=3, flag="0"),".csv",
sep="")
master = lapply(myfilename, read.table, header=TRUE, sep=",")
masterfile = do.call("rbind", master)
head(masterfile[[2]], 100)
if (pollutant == "sulfate") {
#result=lapply(masterfile[[2]], mean, na.rm=TRUE)
result=mean(masterfile[[2]], na.rm=TRUE)
}
if (pollutant == "nitrate") {
result=mean(masterfile[[3]], na.rm=TRUE)
}
result
}