I've got like 230 directories of this kind (1367018589_name_nameb_namec_named) and would like to rename them into (Name Nameb Namec Named).
To be more precise:
Removing numbers
Replacing underscores with spaces (except the first understore which comes after the numbers)
First letter into capital letter
a easy one-liner is preferred since I'm quite a newbie regarding Linux and bash.
Bash script wouldn't be a problem either - just a small explanation how to use it would be very much appreciated.
Meaning that I can understand once I know the command, but having troubles coming up with in my own.
Much thanks in andvance
In one line (updated to capitalize first letter of each word -- missed that the first time):
$ for f in * ; do g=$(echo $f | sed s/[0-9_]*// | sed s/_/\ /g | sed "s/\b\(.\)/\u\1g") ; echo "mv \"$f\" to \"$g\"" ; done
Once you are happy that it is going to do what you want change
echo "mv \"$f\" to \"$g\""
to
mv -i "$f" "$g"
Note, the -i option is to avoid the case of accidentally overwriting a file (say if you had files 123_test and 345_test for instance)
Related
I have seen questions that are close to this but I have not seen the exact answer I need and can't seem to get my head wrapped around the regex, awk, sed, grep, rename that I would need to make it happen.
I have files in one directory sequentially named from multiple sub directories of a different directory created using find piped to xargs.
Command I used:
find `<dir1>` -name "*.png" | xargs cp -t `<dir2>`
This resulted in the second directory containing duplicate filenames sequentially named as follows:
<name>.png
<name>.png.~1~
<name>.png.~2~
...
<name>.png.~n~
What I would like to do is take all files ending in ~*~ and rename it as follows:
<name>.#.png where the '#" is the number between the "~"s at the end of the file name
Any help would be appreciated.
With Perl's rename (stand alone command):
rename -nv 's/^([^.]+)\.(.+)\.~([0-9]+)~/$1.$3.$2/' *
If everything looks fine remove option -n.
There might be an easier way to this, but here is a small shell script using grep and awk to achieve what you wanted
for i in $(ls|grep ".png."); do
name=$(echo $i|awk -F'png' '{print $1}');
n=$(echo $i|awk -F'~' '{print $2}');
mv $i $name$n.png;
done
Lets say my data looks like this
iqwertyuiop
and I want to replace all the letters i after column 3 with a Z.. so my output would look like this
iqwertyuZop
How can I do this with sed or awk?
It's not clear what you mean by "column" but maybe this is what you want using GNU awk for gensub():
$ echo iqwertyuiop | awk '{print substr($0,1,3) gensub(/i/,"Z","g",substr($0,4))}'
iqwertyuZop
Perl is handy for this: you can assign to a substring
$ echo "iiiiii" | perl -pe 'substr($_,3) =~ s/i/Z/g'
iiiZZZ
This would totally be ideal for the tr command, if only you didn't have the requirement that the first 3 characters remain untouched.
However, if you are okay using some bash tricks plus cut and paste, you can split the file into two parts and paste them back together afterwords:
paste -d'\0' <(cut -c-3 foo) <(cut -c4- foo | tr i Z)
The above uses paste to rejoin together the two parts of the file that get split with cut. The second section is piped through tr to translate i's to Z's.
(1) Here's a short-and-simple way to accomplish the task using GNU sed:
sed -r -e ':a;s/^(...)([^i]*)i/\1\2Z/g;ta'
This entails looping (t), and so would not be as efficient as non-looping approaches. The above can also be written using escaped parentheses instead of unescaped characters, and so there is no real need for the -r option. Other implementations of sed should (in principle) be up to the task as well, but your MMV.
(2) It's easy enough to use "old awk" as well:
awk '{s=substr($0,4);gsub(/i/,"Z",s); print substr($0,1,3) s}'
The most intuitive way would be to use awk:
awk 'BEGIN{FS="";OFS=FS}{for(i=4;i<=NF;i++){if($i=="i"){$i="Z"}}}1' file
FS="" splits the input string by characters into fields. We iterate trough character/field 4 to end and replace i by Z.
The final 1 evaluates to true and make awk print the modified input line.
With sed it looks not very intutive but still it is possible:
sed -r '
h # Backup the current line in hold buffer
s/.{3}// # Remove the first three characters
s/i/Z/g # Replace all i by Z
G # Append the contents of the hold buffer to the pattern buffer (this adds a newline between them)
s/(.*)\n(.{3}).*/\2\1/ # Remove that newline ^^^ and assemble the result
' file
I have the following code:
#!/bin/sh
while read line; do
printf "%s\n" $line
done < input.txt
Input.txt has the following lines:
one\two
eight\nine
The output is as follows
onetwo
eightnine
The "standard" solutions to retain the slashes would be to use read -r.
However, I have the following limitations:
must run under #!/bin/shfor reasons of portability/posix compliance.
not all systems
will support the -r switch to read under /sh
The input file format cannot be changed
Therefore, I am looking for another way to retain the backslash after reading in the line. I have come up with one working solution, which is to use sed to replace the \ with some other value (e.g.||) into a temporary file (thus bypassing my last requirement above) then, after reading them in use sed again to transform it back. Like so:
#!/bin/sh
sed -e 's/[\/&]/||/g' input.txt > tempfile.txt
while read line; do
printf "%s\n" $line | sed -e 's/||/\\/g'
done < tempfile.txt
I'm thinking there has to be a more "graceful" way of doing this.
Some ideas:
1) Use command substitution to store this into a variable instead of a file. Problem - I'm not sure command substitution will be portable here either and my attempts at using a variable instead of a file were unsuccessful. Regardless, file or variable the base solution is really the same (two substitutions).
2) Use IFS somehow? I've investigated a little, but not sure that can help in this issue.
3) ???
What are some better ways to handle this given my constraints?
Thanks
Your constraints seem a little strict. Here's a piece of code I jotted down(I'm not too sure of how valuable your while loop is for the other stuffs you would like to do, so I removed it off just for ease). I don't guarantee this code to be robustness. But anyway, the logic would give you hints in the direction you may wish to proceed. (temp.dat is the input file)
#!/bin/sh
var1="$(cut -d\\ -f1 temp.dat)"
var2="$(cut -d\\ -f2 temp.dat)"
iter=1
set -- $var2
for x in $var1;do
if [ "$iter" -eq 1 ];then
echo $x "\\" $1
else
echo $x "\\" $2
fi
iter=$((iter+1))
done
As Larry Wall once said, writing a portable shell is easier than writing a portable shell script.
perl -lne 'print $_' input.txt
The simplest possible Perl script is simpler still, but I imagine you'll want to do something with $_ before printing it.
I want to know in how to get the first 3 letters in a filename in very simple way. Thanks and regards
Depends on the shell you're using. In bash, you can just use the substring extraction, something like:
pax> fname=xyzzy.txt
pax> echo ${fname}
xyzzy.txt
pax> first3=${fname:0:3} ; echo ${first3}
xyz
If you're not using bash, another option is to use cut, which tends to be available on most systems. It's an external program, meaning it's not as efficient as the internal bash method above, but you'll probably only notice that if you're doing it thousands of times per second.
pax> first3=$(echo ${fname} | cut -c1-3) ; echo ${first3}
xyz
I went through several grep examples, but don't see how to do the following.
Say, i have a file with a line
! some test here and number -123.2345 text
i can get this line using
grep ! input.txt
but how do i get the number (possibly positive or negative) from this line and append it to the end of another file? Is it possible to apply grep to grep results?
If yes, then i could get the number via something like
grep -Eo "[0-9]{1,}|\-[0-9]{1,}"
p/s/ i am using OS-X
p/p/s/ i'm trying to fetch data from several files and put into a single file for later plotting.
The format with your commands would be:
grep ! input.txt | grep -Eo "[0-9]{1,}|\-[0-9]{1,}" >> output
To grep from grep we use the pipe operator | this lets us chain commands together. To append this output to a file we use the redirection operator >>.
However there are a couple of problems. You regexp is better written: grep -Eoe '-?[0-9.]+' this allows for the decimal and returns the single number instead of two and if you want lines that start with ! then grep ^! is better to avoid matches with lines what contain ! but don't start with it. Better to do:
grep '^!' input | grep -Eoe '-?[0-9.]+' >> output
perl -lne 'm/.*?([\d\.\-]+).*/g;print $1' your_file >>anotherfile_to_append
$foo="! some test here and number -123.2345 text"
$echo $foo | sed -e 's/[^0-9\.-]//g'
$-123.2345
Edit:-
for a file,
[ ]$ cat log
! some test here and number -123.2345 text
some blankline
some line without "the character" and with number 345.566
! again a number 34
[ ]$ sed -e '/^[^!]/d' -e 's/[^0-9.-]//g' log > op
[ ]$ cat op
-123.2345
34
Now lets see the toothpicks :) '/^[^!]/d' / start of pattern, ^ not (like multiply with false), [^!] anyline starting with ! and d delete. Second expression, [^0-9.-] not matching anything within 0 to 9, and . and -, (everything else) // replace with nothing (i.e. delete) and done :)