R 选择一个具有特定名称和最近更新的文件

我想选择一个像下面这行一样的文件，但它并没有按照我的意愿执行

which(substr(rownames(fInfo),1,8) == "mySource" ) & which.max(fInfo$mtime)

在一个英语句子中，我想选择名称以“mySource”开头的文件，在选择的文件中，我想选择最近更新的文件

我下面的脚本已经足够了，但是太长了。有人能缩短我的脚本吗

# create dummy files under Folder "scriptFld"
ifelse(!dir.exists(file.path("scriptFld")), dir.create(file.path("scriptFld")), FALSE)
strTime = format(Sys.time(), "%H%M")
file.create(NA, paste0("scriptFld/mySource1_", strTime,".R")); Sys.sleep(1)
file.create(NA, paste0("scriptFld/mySource2_", strTime,".R")); Sys.sleep(1)
file.create(NA, paste0("scriptFld/notMySource3_", strTime,".R"))


# read source R files
setwd("scriptFld")
fInfo = file.info(list.files()) # find all files under the folder "scriptFld"
iCandidate = which(substr(rownames(fInfo),1,8) == "mySource") # focus on file names starting with "source"
iCandidateMax = iCandidate[ which.max(fInfo$mtime[iCandidate]) ] # choose the most recent file
fSourceName = rownames(fInfo)[iCandidateMax]
source(file = fSourceName) # This is what I want, except the script is too long.
setwd("..")
(fSourceName)

你可以做：

library(dplyr)
files <- list.files("scriptFld", pattern = "^mySource", full.names = TRUE)
files %>%
  file.info() %>%
  pull(mtime) %>%
  which.max() %>%
  files[.] %>%
  source()

库（dplyr）
文件%
file.info（）%%>%
拉动（mtime）%>%
哪个.max（）%>%
文件[]%>%
资料来源（）

x这正是我想要的。谢谢