I am working on trying to duplicate characters on certain words, but the sed script I wrote is not quite doing what I want it to. For example, I am trying to duplicate each character in a words like so:
FILE into FFIILLEE
I know how to remove the duplicate letters with :
sed s/\(.\)\1/\1/g' file.txt
This is the sed script I wrote, but it just ends up duplicating the whole word and I just get:
FILE FILE
This is what I wrote:
sed 's/[A-Z]*/& &/g' file.txt
How can I grab each letter and duplicate just the letter?
A slight variation on your first script should work:
sed 's/\(.\)/\1\1/g' file.txt
Translation: For every character seen, replace it by itself followed by itself.
sed 's/[[:alpha:]]/&&/g' file.txt
[:alpha:]class is the whole scope of letter available, you could extend with [:alnum:]including digit
& in replacement pattern is the whole search pattern matching. In this case 1 letter
g for each possible occurence
Your probleme was to use the * in search pattern that mean all occurence of previous pattern so the pattern is the whole word at once and not every letter of this word
Related
I have text file input.txt which has
april,december,month.gmail.com
lion,tiger,animal.gmail.com
Using sed change first and second columns to uppercase? Is there a way to do it?
With GNU sed:
sed 's/^[a-z]*,[a-z]*,/\U&/' file
s: substitute command
[a-z]*,: search for zero ore more lowercase letter followed by a ,. The pattern is repeated for second field
the \U sequence turns the replacement to uppercase
\U is applied to & which reference the matched string
or if there is only three comma separated fields:
sed 's/^[a-z].*,/\U&/' file
output:
APRIL,DECEMBER,month.gmail.com
LION,TIGER,animal.gmail.com
As #Sundeep suggests, the second sed can be shortened to:
s/^.*,/\U&/
which converts all characters until last , is found
For more on GNU sed substitution command, see this article
I want to insert a range of lines from a file, say something like 210,221r before the first occurrence of a pattern in a bunch of other files.
As I am clearly not a GNU sed expert, I cannot figure how to do this.
I tried
sed '0,/pattern/{210,221r file
}' bunch_of_files
But apparently file is read from line 210 to EOF.
Try this:
sed -r 's/(FIND_ME)/PUT_BEFORE\1/' test.text
-r enables extendend regular expressions
the string you are looking for ("FIND_ME") is inside parentheses, which creates a capture group
\1 puts the captured text into the replacement.
About your second question: You can read the replacement from a file like this*:
sed -r 's/(FIND_ME)/`cat REPLACEMENT.TXT`\1/' test.text
If replace special characters inside REPLACEMENT.TXT beforehand with sed you are golden.
*= this depends on your terminal emulator. It works in bash.
In https://stackoverflow.com/a/11246712/4328188 CodeGnome gave some "sed black magic" :
In order to insert text before a pattern, you need to swap the pattern space into the hold space before reading in the file. For example:
sed '/pattern/ {
h
r file
g
N
}' in
However, to read specific lines from file, one may have to use a two-calls solution similar to dummy's answer. I'd enjoy knowing of a one-call solution if it is possible though.
I want to remove last 2 words in the string which is in a file.
I am using this command first to delete the last word. But I couldn't do it. can someone help me
sed 's/\w*$//' <file name>
my strings are like this
Input:
asbc/jahsf/jhdsflk/jsfh/ -0.001 (exam)
I want to remove both numerical value and the one in brackets.
Output:
asbc/jahsf/jhdsflk/jsfh/
Using GNU sed:
$ sed -r 's/([[:space:]]+[-+.()[:alnum:]]+){2}$//' file
asbc/jahsf/jhdsflk/jsfh/
How it works
[[:space:]]+ matches one or more spaces.
[-+.()[:alnum:]]+ matches the 'words' which are allowed to contain any number of plus or minus signs, periods, parens, or any alphanumeric characters.
Note that, when a period is inside square brackets, [.], it is just a period, not a wildcard: it does not need to be escaped.
([[:space:]]+[-+.()[:alnum:]]+) matches one or more spaces followed by a word.
([[:space:]]+[-+.()[:alnum:]]+){2}$ matches two words and the spaces which precede them.
Note the use of character classes like [:space:] and [:alnum:]. Unlike the old-fashioned classes like [a-zA-Z0-9], these classes are unicode safe.
OSX (BSD) sed
The above was tested on GNU sed. For BSD sed, try:
sed -E 's/([[:space:]][[:space:]]*[-+.()[:alnum:][:alnum:]]*){2}$//' file
To remove everything that follows a number with decimal places
This looks for a decimal number with optional sign and removes it, the spaces which precede it, and everything which follows it:
$ sed -r 's/[[:space:]]+[-+]?[[:digit:]]+[.][[:digit:]]+[[:space:]].*//' file
asbc/jahsf/jhdsflk/jsfh/
How it works:
[[:space:]]+ matches one or more spaces
[-+]? matches zero or one signs.
[[:digit:]]+ matches any number of digits.
[.] matches a decimal point (period).
[[:digit:]]+ matches one or more digits following the decimal point.
[[:space:]] matches a space following the number.
.* matches anything which follows.
It looks like there is a tab between what you want to keep and what you want to get rid of. I don't have linux in front of me but try this.
sed 's/\t.*//'
This is assuming your strings are always formatted similarily which is what I take from your comment.
This might work for you (GNU sed):
sed -r 's/\s+\S+\s+\S+\s*$//' file
or if you prefer:
sed -r 's/(\s+\S+){2}\s*$//' file
This matches and removes: one or more whitespaces followed by one or more non-whitespaces twice followed by zero or more whitespaces at the end of the line.
I want to insert a newline after the following pattern
lcl|NC_005966.1_gene_750
While the last number(in this case the 750) changes. The numbers are in a range of 1-3407.
How can I tell sed to keep this pattern together and not split them after the first number?
So far i found
sed 's/lcl|NC_005966.1_gene_[[:digit:]]/&\n/g' file
But this breaks off, after the first digit.
Try:
sed 's/lcl|NC_005966.1_gene_[[:digit:]]*/&\n/g' file
(note the *)
Alternatively, you could say:
sed '/lcl|NC_005966.1_gene_[[:digit:]]/G' file
which would add a newline after the specified pattern is encountered.
sed 's/lcl|NC_005966\.1_gene_[[:digit:]][[:digit:]]*/&\
/g' file
You need to escape . as it's an RE metacharacter, and you need [[:digit:]][[:digit:]]* to represent 1-or-more digits and you need to use \ followed by a literal newline for portability across seds.
This is a simple question, I'm not sure if i'm able to do this with sed/awk
How can I make sed search for these 3 lines and replace with a line with a determined string?
<Blarg>
<Bllarg>
<Blllarg>
replace with
<test>
I tried with sed "s/<Blarg>\n<Bllarg>\n<Blllarg>/<test>/g" But it just don't seem to find these lines. Probably something with my break line character (?) \n. Am I missing something?
Because sed usually handles only one line at a time, your pattern will never match. Try this:
sed '1N;$!N;s/<Blarg>\n<Bllarg>\n<Blllarg>/<test>/;P;D' filename
This might work for you:
sed '/<Blarg>/ {N;N;s/<Blarg>\n<Bllarg>\n<Blllarg>/<test>/}' <filename>
It works as follows:
Search the file till <Blarg> is found
Then append the two following lines to the current pattern space using N;N;
Check if the current pattern space matches <Blarg>\n<Bllarg>\n<Blllarg>
If so, then substitute it with <test>
You can use range addresses with regular expressions an the c command, which does exactly what you are asking for:
sed '/<Blarg>/,/<Blllarg>/c<test>' filename