With the sed command, is it possible to do internal string commands? in this case the actual lines are:
s/9G /9F6 09999F7 09999F8 09999F9 09999G /g
s/0G /0F6 09999F7 09999F8 09999F9 09999G /g
The number can be set using [09] but I didn't know if I could retrieve it from, say, & and use it before the F6 in something like the following:
s/[09]G /(&:0:1)F6 09999F7 09999F8 09999F9 09999G /g
This actual code does not work, by the way.
You are looking for a so called sub expression in the form of \(SUB_PATTERN\):
sed 's/\([09]\)G /\1F6 09999F7 09999F8 09999F9 09999G /g' file
From man sed:
s/regexp/replacement/
Attempt to match regexp against the pattern space. If successful, replace that portion matched with replacement.
The replacement may contain the special character & to refer to that portion of the pattern space which matched, and
the special escapes \1 through \9 to refer to the corresponding matching sub-expressions in the regexp.
Related
I want to replace a line, that represents a part of mathematical equation:
f(x,z,time,temp)=-(2.0)/(exp(128*((x-2.5*time)*(x-2.5*time)+(z-0.2)*(z-0.2))))+(
with a new one similar to the above. Both new and old lines are saved in bash variables.
Main problem is that mathematical equation is full with special characters that do not allow proper search and replace in bash mode, even when I used as delimiter special character that is not used in equation.
I used
sed -n "s|$OLD|$NEW|g" restart.k
and
sed -i "s|$OLD|$NEW|g" restart.k
but all times I get wrong results.
Any idea to solve this?
There is only * in your pattern here that is special for sed, so escape it and do replacement as usual:
sed "s:$(sed 's:[*]:\\&:g' <<<"$old"):$new:" infile
if there are more special characters in your real sample, then you will need to add them inside bracket []; there are some exceptions like:
if ^ character: it can be place anywhere in [] but not first character, because ^ character at first negates the characters within its bracket expression.
if ] character: it should be the first character, because this character is also used to end the bracket expression.
if - character: it should be the first or last character, because this character is also can be used for defining the range of characters too.
I wanted to replace underscores with hyphens in all places where the character('_') is preceded and following by uppercase letters e.g. QWQW_IOIO, OP_FD_GF_JK, TRT_JKJ, etc. The replacement is needed throughout one document.
I tried to replace this in vim using:
:%s/[A-Z]_[A-Z]/[A-Z]-[A-Z]/g
But that resulted in QWQW_IOIO with QWQ[A-Z]-[A-Z]OIO :(
I tried using a sed command:
sed -i '/[A-Z]_[A-Z]/ s/_/-/g' ./file_name
This resulted in replacement over the whole line. e.g.
QWQW_IOIO variable may contain '_' or '-' line was replaced by
QWQW-IOIO variable may contain '-' or '-'
You had the right idea with your first vim approach. But you need to use a capturing group to remember what character was found in the [A-Z] section. Those are nicely explained here and under :h /\1. As a side note, I would recommend using \u instead of [A-Z], since it is both shorter and faster. That means the solution you want is:
:%s/\(\u\)_\(\u\)/\1-\2/g
Or, if you would like to use the magic setting to make it more readable:
:%s/\v(\u)_(\u)/\1-\2/g
Another option would be to limit the part of the search that gets replaced with the \zs and \ze atoms:
:%s/\u\zs_\ze\u/-/g
This is the shortest solution I'm aware of.
This should do what you want, assuming GNU sed.
sed -i -r -e 's/([A-Z]+)_([A-Z]+)/\1-\2/g' ./file_name
Explanation:
-r flag enables extended regex
[A-Z]+ is "one or more uppercase letters"
() groups a pattern together and creates a numbered memorized match
\1, \2 put those memorized matches in the replacement.
So basically this finds a chunk of uppercase letters followed by an underscore, followed by another chunk of uppercase letters, memorizes only the letter chunks as 2 groups,
([A-Z]+)_([A-Z]+)
Then it replays those groups, but with a hyphen in between instead of an underscore.
\1-\2
The g flag at the end says to do this even if the pattern shows up multiple times on one line.
Note that this falls apart a little in this case:
QWQW_IOIO_ABAB
Because it matches the first time, but not the second; the second part won't match because IOIO was consumed by the first match. So that would result in
QWQW-IOIO_ABAB
This version drops the + so it only matches one uppercase letter, and won't break in the same way:
sed -i -r -e 's/([A-Z])_([A-Z])/\1-\2/g'
It still has a small flaw, if you have a string like this:
A_B_C
Same issue as before, just one letter now instead of multiple.
I'm partially successfully using sed to replace variables in a text file. I'm stuck on an exception.
A script reads input from a list - say the $roll_symbol is C20.
sed replaces C20, GC20, and KC20 (because C20 matches part of the string).
I searched the web and tried the variations I found - no success.
I tried these variations without success:
escape the reserved character $
escape braces
escape both
use double quotes instead of single quotes.
*the best version so far (but only partially):
sed -i 's/'${roll_symbol}'/'${roll_symbol}\,${contract_month}'/g' $OUTPUT_DIRECTORY/$OUTPUT_FILE;
You need to tell sed what characters are legal before the start of your match to limit where it can match. To only match at start-of-word boundaries try \<.
sed -i "s/\<${roll_symbol}/${roll_symbol},${contract_month}/g" "$OUTPUT_DIRECTORY/$OUTPUT_FILE";
I have a line in a source file: [12 13 15]. In vim, I type:
:%s/\([0-90-9]\) /\0, /g
wanting to add a coma after 12 and 13. It works, but not quite, as it inserts an extraspace [12 , 13 , 15].
How can I achieve the desired effect?
Use \1 in the replacement expression, not \0.
\1 is the text captured by the first \(...\). If there were any more pairs of escaped parens in your pattern, \2 would match the text capture between the pair starting at the second \(, \3 at the third \(, and so on.
\0 is the entire text matched by the whole pattern, whether in parentheses or not. In your case this includes the space at the end of your pattern.
Also note that [0-90-9] is the same as [0-9]: each [...] collection matches just one character. It happens to work anyway, because in your data ‘a digit followed by a space’ matches in the same places as ‘2 digits followed by a space’. (If you actually needed to only insert commas after 2 digits, you could write [0-9][0-9].)
"I have a line in a source file:..."
then you type :%s/... this will do the substitution on all lines, if it matched. or that is the single line in your file?
If it is the single line, you don't have to group, or [0-9], just :%s/ \+/,/g will do the job.
The fine answers already point interesting solutions, but here's another one,
making use of the \zs, which marks the start of the match. In this pattern:
/[0-9]\zs /
The searched text is /[0-9] /, but only the space counts as a match. Note
that you can use the class \d to simplify the digit character class, so the
following command shall work for your needs:
:s/\d\d\zs /, /g ; matches only the space, replace by `, '
You said you have multiple lines and these changes are only to certain lines.
You can either visually select the lines to be changed or use the :global
command, which searches for lines matching a pattern and applies a command to
them. Now you'd need to build an expression to match the lines to be changed
in a less precise as possible way. If the lines that begins with optional
spaces, a [ and two digits are the only lines to be matched and no other
ones, then this would work for you:
:g/\s*[\d\d/s/\d\d\zs /, /g
Check the help for pattern.txt for \ze and similar and
:global.
Homework: use the help to understand \zs and see how this works:
:s/\d\d\zs\ze /,/g
Is it possible to search a string that is a regex, without escaping all the fancy characters?
For example, I want to find this string in my source file: ^[\d\| *]$, without escaping \, $, etc.
I would like to just copy and paste the regex and get the result.
What you want is a grep that searches for matching strings, rather than attempting to match a regular expression. With GNU grep, you can invoke the command with the
-F or --fixed-strings flags, or just invoke the command as fgrep instead. The following are all equivalent:
grep -F '^[\d\| *]$'
grep --fixed-strings '^[\d\| *]$'
fgrep '^[\d\| *]$'
Fixed-string searches are exactly what you need when you want to match code that represents a regular expression, or when you want a faster grep that doesn't need the advanced matching capability of a regular expression engine.
There is an easy way to avoid special treatment of the characters
in Vim regular expressions. The \V specifier allows to switch
interpretation of the rest of the pattern to the “very nomagic” mode,
which means that every character but the backslash is understood
literally.
Therefore, one can set the last search register accordingly:
:let #/ = '\V' . escape('^[\d\| *]$', '\')
and use n and N for searching immediately.
You can use the command line tool fgrep (“fast grep”),
Supposing you have yanked the regexp to search for to the default register (#"), you can do this:
/\V<C-R><C-R>=escape(#", '/\')<CR><CR>
The \V starts a "very nomagic" search, where only atoms starting with a backslash have special, non-literal meaning. The escape() renders all those contained backslashes ineffective (and escapes / which would otherwise end the search pattern), so that this is a purely literal search. The text is inserted via Ctrl+R= into the search command line.