I have thousands of text files and would like to know how to check if a particular file is empty. I am reading all the files using this line of code
我有数千个文本文件，我想知道如何检查某个文件是否为空。我正在用这一行代码读取所有的文件
Y<-grep("*.txt", list.files(), value = TRUE)
I would like a list of names of all the blank files. Have to do it in R.
我想要一张所有空白文件的名单。必须用R表示。
Thanks.
谢谢。
2 个解决方案
#1
19
You can use file.info for that:
你可以使用文件。info:
info = file.info(filenames)
empty = rownames(info[info$size == 0, ])
Incidentally, there’s a better way of listing text files than using grep: specify the pattern argument to list.files:
顺便说一句，列出文本文件的方法比使用grep:为list.files指定模式参数更好:
list.files(pattern = '\\.txt$')
Notice that the pattern needs to be a regular expression, not a glob – the same is true for grep!
请注意，模式需要是正则表达式，而不是glob——对于grep也是如此!
#2
0
find . -empty
or
或
find . -empty |awk -F\/ '{print $FN}'
If you want to limit just txt files:
如果你只想限制txt文件:
find . -empty -name "*.txt"
If you want only asci files (not just .txt)
如果您只想要asci文件(不只是.txt)
find . -empty -type f
put it all together:
把这一切放在一起:
find . -empty -type f -name "*.txt" |awk -F\/ '{print $NF}'
#1
19
You can use file.info for that:
你可以使用文件。info:
info = file.info(filenames)
empty = rownames(info[info$size == 0, ])
Incidentally, there’s a better way of listing text files than using grep: specify the pattern argument to list.files:
顺便说一句，列出文本文件的方法比使用grep:为list.files指定模式参数更好:
list.files(pattern = '\\.txt$')
Notice that the pattern needs to be a regular expression, not a glob – the same is true for grep!
请注意，模式需要是正则表达式，而不是glob——对于grep也是如此!
#2
0
find . -empty
or
或
find . -empty |awk -F\/ '{print $FN}'
If you want to limit just txt files:
如果你只想限制txt文件:
find . -empty -name "*.txt"
If you want only asci files (not just .txt)
如果您只想要asci文件(不只是.txt)
find . -empty -type f
put it all together:
把这一切放在一起:
find . -empty -type f -name "*.txt" |awk -F\/ '{print $NF}'