How can I match this pattern of file name in a directory, and output the matched?

Question

there are many of files in this directory:

[ichen@ui01 data]$ ls
data.list
data.root
ntuple.data15_13TeV.00276262.DAOD_FTAG2.root
ntuple.data15_13TeV.00276329.DAOD_FTAG2.root
ntuple.data15_13TeV.00276336.DAOD_FTAG2.root
ntuple.data15_13TeV.00276416.DAOD_FTAG2.root
ntuple.data15_13TeV.00276511.DAOD_FTAG2.root

and i want to make a list which just contains those files which have the pattern of:

    [many chars].[many chars].[many numbers].[many chars].root

to match the file names such like:

ntuple.data15_13TeV.00276262.DAOD_FTAG1.root
ntuple.data15_13TeV.00276329.DAOD_FTAG2.root
ntuple.data15_13TeV.00276336.DAOD_FTAG3.root
etc...

how can I use regexp to achieve this goal? Maybe we can use this syntax:

for f in `ls`;do if [....];then echo $f;fi;done  > log.list

Answer 1

In regexp land, many roads lead to rome. :)

ls | egrep '^\w*\.\w*\.[0-9]*\.\w*\.root$'

^ marks the beginning of a line $ marks the end of a line \\w is a word character \\w* is many work characters . is a literal '.' character, an unmasked '.' in the regurlar expression stands for "any character" [0-9] is any of the numbers between 0 and 9

And for your specific example:

for f in `ls`;do echo $f | egrep '^\w*\.\w*\.[0-9]*\.\w*\.root$';done

And now including the if statement:

for f in `ls`; do if [[ $f =~ '\w*\.\w*\.[0-9]*\.\w*\.root' ]]; then echo $f; fi; done

In this case, I had to remove the line beginning and end (^...$) for it to match. Not sure why. In general, =~ will check for regular expressions.

Answer 2

ls | grep '..*[.]..*[.][0-9][0-9]*[.]..*[.]root > log.list 

应该做的工作

Answer 3

It doesn't have to be complicated like that. You just need to list out the files which match a certain pattern - wildcards are basically enough, there's no need for regexes.

ls -1 ntuple.data*.*.*.root > log.list