2012-07-24 53 views
0

我有一对夫妇的txt文件的目录遍历目录中的文件,并创建变量与文件名

dayton1.txt
datton2.txt
...
dakton50.txt

我想用文件名作为名称创建变量并读取相应的文件。我如何在R中做到这一点?

谢谢!

+0

像这样做,而不是:http://stackoverflow.com/a/9565095/602276 – Andrie 2012-07-24 20:02:57

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谢谢。我认为它会做我想做的。 – Seen 2012-07-24 20:12:31

回答

6

你想要list.files(),看到它的帮助页面?list.files。例如: -

R> getwd() 
[1] "/blah/blah/packages/analogue/analogue/pkg" 
R> list.files() 
[1] "data"   "DESCRIPTION" "DESCRIPTION~" "inst"   "man"   
[6] "NAMESPACE" "NAMESPACE~" "R"   "src"   "tests"  
[11] "vignettes" 
R> list.files("./inst") 
[1] "ChangeLog" "ChangeLog~" "CITATION" "CITATION~" "COPYING" 
[6] "doc"  "THANKS"  "TODO"