For example, my file has the following data:
$ cat sample.txt
19999119999,string1,dddddd
18888135790,string2,dddddd
15555555500,string3,dddddd
This is a sample data. How can we remove ONLY first digit from each row? My output should be:
$ cat output.txt
9999119999,string1,dddddd
8888135790,string2,dddddd
5555555500,string3,dddddd
Is there any way to parse each line character wise using grep or sed?
Or any other way to get the desired output?
You just need to print from the second character on:
$ cut -c2- file
9999119999,string1,dddddd
8888135790,string2,dddddd
5555555500,string3,dddddd
Or, using sed, remove the first char:
$ sed 's/^.//' file
9999119999,string1,dddddd
8888135790,string2,dddddd
5555555500,string3,dddddd
Try this:
$ sed -r 's/^[0-9](.*)/\1/' sample.txt
Output:
9999119999,string1,dddddd
8888135790,string2,dddddd
5555555500,string3,dddddd
^[0-9] - The first digit of each line
(.*) - The content of each line except the first digit
\1 - Denote the content of (.*)
Sorry for my bad English.
Grep can solve this with a look behind. For that you need -P option :
grep -Po '(?<=^\d)(.+)' file
or in shorthand :
grep -Po '^\d\K.+' file
The (?<=^\d)/^\d\K part is the look behind that matches the first digit.
Related
I have a file which contains the following records
+aaaa
+bbbb
cccc-123
-dddd
eeee+789
-fff+456
ggg
Now I want to keep only the records if the first character is a "+" or "-" sign
so the (new) file should look like this
+aaaa
+bbbb
-dddd
-fff+456
Can this be done via a grep or sed command ?
Try this:
grep '^\[+-\]' myfile.txt
or
grep '^[+-]' myfile.txt
Depending on your flavour of grep
This might work for you (GNU sed):
sed '/^[+-]/!d' file
or:
sed -n '/^[+-]/p' file
In the first solution: if the first character of the line is not + or - delete the line.
In the second solution: if the first character of the line is + or - print it.
27211;18:05:03479;20161025;0;0;0;0;10991;0;10991;000;0;0;000;1000000;0;0;000;0;0;0;82
Second string after ; is time. gg:mm:sssss:. I just want to be gg:mm:ss:
Like so:
27211;18:05:03;20161025;0;0;0;0;10991;0;10991;000;0;0;000;1000000;0;0;000;0;0;0;82
I tried with cut but it deletes everything after n'th occurance of character, and for now I am stuck, please help.
give this one liner a try:
awk -F';' -v OFS=";" 'sub(/...$/,"",$2)+1' file
It removes the last 3 chars from column 2.
update with sed one liner
If you are a fan of sed:
sed -r 's/(;[^;]*)...;/\1;/' file
With sed:
sed -r 's/^([^;]+;[^;]+)...;/\1;/' file
(Or)
sed -r 's/^([^;]+;[0-9]{2}:[0-9]{2}:[0-9]{2})...;/\1;/' file
It also can be something like sed 's/(.*)([0-9]{2}\:){2}([0-9]{3})[0-9]*\;(.*)/\1\2\3\4/g'
It is not very clean, but at least is more clear for me.
Regards
I'd use perl for this:
perl -pe 's/(?<=:\d\d)\d+(?=;)//' file
That removes any digits between "colon-digit-digit" and the semicolon (first match only, not globally in the line).
If you want to edit the file in-place: perl -i -pe ...
With sed:
sed -E 's/(:[0-9]{2})[0-9]{3}/\1/' file
or perl:
perl -pe's/:\d\d\B\K...//' file
I am trying to grep the line from file and then from $1 I am trying to change the character.
eg
cat file1.txt
Surjit
Shilpa
cchiku
end of file
I tried and grepped the line which start with s.
grep -e "S"
Then I want to replace the 4th character to x for all grepped result in the file1.txt
I tried
sed -i "s/./x/4" file1.txt
How can I do this only for grepped results?
You can use the sed '/pattern/s/find/replace/' file syntax:
sed '/^S/s/./x/4' file
# ^^ ^^^^^^^
# | replace the 4th character with x
# |
# on lines starting with S
With your file:
$ sed '/^S/s/./x/4' file
Surxit
Shixpa
cchiku
end of file
Note I am using /^S/ as a pattern to match lines starting with S, because if you just say /S/ it will match any line containing S. The anchor ^ indicates the beginning of the line.
An alternative to fedorqui's answer is to include the starting with S condition into the pattern itself:
sed 's/^\(S..\)./\1x/' file
The command matches lines starting with S and puts the S and the following two characters into a matching group. In the replacement part the content of the matching group will get reused and next character after it will get replaced by x.
awk -v FS="" -v OFS="" '/^S/{$4="x"}1' infile
Surxit
Shixpa
cchiku
end of file
I know how to remove all trailing spaces from a file, e.g :
sed -i 's/ *$//' file
Is there a way to do it, but not in lines containing only spaces?
Something in the spirit of :
sed -i 's/[a-zA-Z0-9;}{] *$/[a-zA-Z0-9;}{]/' file
^ keep the original characters
Preferably, but not necessariliy, with sed.
Any linux supported solution will do.
Thanks
Just make sure some other character appears before:
sed -r 's/([^\s])\s+$/\1/' file
This checks if a non-space character (\s) appears followed by any amount of spaces. If so, just print this non-space character back, so that the trailing spaces are removed.
Test
Using cat -vet to see the markers:
$ cat -vet a
hello $
$
bye $
$ sed -r 's/([^\s])\s+$/\1/' a | cat -vet -
hello$
$
bye$
I am looking for I one liner hopefully, that can trim the first and last character of a line, on multiple lines e.g. test.txt
Before:
xyyyyyyyyyyyyyyyyyyyx
pyyyyyyyyyyyyyyyyyyyz
After:
yyyyyyyyyyyyyyyyyyy
yyyyyyyyyyyyyyyyyyy
$ cat /tmp/txt
xyyyyyyyyyyyyyyyyyyyx
pyyyyyyyyyyyyyyyyyyyz
$ sed 's/^.\(.*\).$/\1/' /tmp/txt
yyyyyyyyyyyyyyyyyyy
yyyyyyyyyyyyyyyyyyy
There is little trick :)
sed 's/^.(.*).$/\1/' file > file1 ; rm file ; echo file1 > file ; rm file1
sed -ne 's,^.\(.*\).$,\1,p'
This command will delete all lines that have less than two characters, since one cannot really strip the first and last character from them.