如何计算 100s 不同 csv 文件中相同列（具有相同名称）的平均值，其中部分文件名相同？

Question

我有一堆结构如下的 csv 文件：

df <- data.frame (first_column  = c(3, 2, 6, 7),
                  second_column = c(7, 5, 1, 8))

所有的 csv 文件都有一个名字

"type1_1.csv"
"type1_2.csv"
...
"type2_1.csv"
"type2_2.csv"
...

这些 csv 中的每一个都有 first_column 和 second_column。我想要的是创建一个如下所示的新数据框：

# name        meanofsecond_column
# type1_1     5.25
# ...

我已经开始做的是分别写出每一个：

type1_1 <- read_csv("type1_1.csv")
type1_1mean <- mean(type1_1$second_column)
...
df <- data.frame (name  = c(type1_1, type1_2...),
                  meanofsecondcolumn = c(type1_1mean, type1_2mean...))

然而，由于有超过 100 个 csv 文件，这种方法不是很有效或干净。我怎样才能让它更浓缩？

Answer 1

# path where your csv files are (here current working directory)
CSV_FOLDER <- "."

# list all csv files in given directory
# second parameter is a regex meaning ends with .csv
# third parameter make function return file names with path
csv_files <- list.files(CSV_FOLDER, "\.csv$", full.names=TRUE)

# apply given function on each file and collect results in a list
res <- lapply(csv_files, function(csv_file) {
  # read current file
  tmp <- read.csv(csv_file)

  # build a data.frame from filename (without path) and mean of second column
  return(data.frame(
    name = basename(csv_file),
    meanofsecondcolumn = mean(tmp$second_column)
  ))
})

# rbind all single line data.frames in a single data.frame
res <- do.call("rbind", res)