replace the last third line of a file using awk and sed - sed

I need to find the last third line of a file and if its "}," I should replace it with "}"
I was thinking of awk and sed
This finds the last third line of file
PATTERN=$(awk '{v[c++]=$0}END{print v[c-3]}' $file)
I wanted to use
sed -i 's/$PATTERN/}/' $file
But this is wrong
Don't know how to proceed


I need to find the last third line of a file and if its "}," I should
replace it with "}"
Using tac and sed:
tac filename | sed '3s/^};$/}/' | tac
On MacOS, you could say:
sed '1!G;h;$!d' filename | sed '3s/^};$/}/' | sed '1!G;h;$!d'


This awk should do it (last third line more is cleaned)
cat file
one
two },
three
four
more }, test }, hi
five },
yes
awk '{v[++c]=$0}END {for (i=1;i<=NR;i++) {if (v[c-3]~"},") gsub(/},/,"}",v[c-3]);print v[i]}}' t
one
two },
three
four
more }, test }, hi
five },
yes

Related

sed or awk: delete/comment n lines following a pattern before 3 lines

To delete/comment 3 lines befor a pattern (including the line with the pattern):
how can i achive it through sed command
Ref:
sed or awk: delete n lines following a pattern
the above ref blog help to achive the this with after a pattern match but i need to know before match
define host{
use xxx;
host_name pattern;
alias yyy;
address zzz;
}
the below sed command will comment the '#' after the pattern match for example
sed -e '/pattern/,+3 s/^/#/' file.cfg
define host{
use xxx;
#host_name pattern;
#alias yyy;
#address zzz;
#}
like this how can i do this for the before pattern?
can any one help me to resolve this

If tac is allowed :
tac|sed -e '/pattern/,+3 s/^/#/'|tac
If tac isn't allowed :
sed -e '1!G;h;$!d'|sed -e '/pattern/,+3 s/^/#/'|sed -e '1!G;h;$!d'
(source : http://sed.sourceforge.net/sed1line.txt)

Reverse the file, comment the 3 lines after, then re-reverse the file.
tac file | sed '/pattern/ {s/^/#/; N; N; s/\n/&#/g;}' | tac
#define host{
#use xxx;
#host_name pattern;
alias yyy;
address zzz;
}
Although I think awk is a little easier to read:
tac file | awk '/pattern/ {c=3} c-- > 0 {$0 = "#" $0} 1' | tac

This might work for you (GNU sed):
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/s/^/#/mg;P;D' file
Gather up 4 lines in the pattern space and if the last line contains pattern insert # at the beginning of each line in the pattern space.
To delete those 4 lines, use:
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/d;P;D' file
To delete the 3 lines before pattern but not the line containing pattern use:
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/s/.*\n//;P;D'

Merging Lines using sed

I have text file that consists of 45999 lines. Each line has a word (unigram). I want to create two-sequential words (bigrams). For example:
apple
pie
red
vine
I want 'apple pie', 'pie red', 'red vine'. I tried with sed 'N;s/\n/ /' but it creates just 'apple pie' and 'red vine'. How can I solve this problem? Thank you..

Could you please try following if you are ok with awk.
awk -v RS="" '
BEGIN{
OFS=","
s1="\047"
}
{
for(i=2;i<=NF;i++){
print s1 $(i-1) s1, s1 $i s1
}
}' Input_file
Output will be as follows.
'apple','pie'
'pie','red'
'red','vine'
2nd solution: since output of OP is not clear so adding this one too.
awk -v RS="" '
BEGIN{
OFS=","
s1="\047"
}
{
for(i=2;i<=NF;i++){
val=(val?val OFS:"")s1 $(i-1) s1 OFS s1 $i s1
}
}
END{
print val
}' Input_file
Output will be as follows.
'apple','pie','pie','red','red','vine'

This might work for you (GNU sed):
sed -nE 'N;s/\n(.*)/ \1&/;P;D' file
Append the next line to the current line, then replace the newline by a space and append the second line again. Print/delete the first line and repeat.
N.B. This does not print the last line as it is not a pair, if the last line is needed use:
sed -E 'N;s/\n(.*)/ \1&/;P;D' file
If the output is to be printed as a single line with each pair surrounded by single quotes and separated by a comma, use:
sed -E ':a;$!N;s/(\S+)\n(.*)/'\''\1 \2'\'', \2/;ta;s/ (\S+)$/ '\''\1'\''/' file
Or:
sed -E ':a;$!N;s/(\S+)\n(.*)/'\''\1 \2'\'', \2/;ta;s/, \S+$/' file

insert semi colon after 10 digit number

I have lines that start like this: 2141058222 11/22/2017 and I want to append a ; at the end of the ten digit number like this: 2141058222; 11/22/2017.
I've tried sed with sed -i 's/^[0-9]\{10\}\\$/;&/g' which does nothing.
What am I missing?

Try this:
echo "2141058222 11/22/2017" | sed -r 's/^([0-9]{10})/&;/'

echo "2141058222 11/22/2017" | sed 's/ /; /'
Output:
2141058222; 11/22/2017

If the input is always in the format specified, GNU cut works, and might even be more efficient than sed:
cut -c -10,11- --output-delimiter ';' <<< "2141058222 11/22/2017"
Output:
2141058222; 11/22/2017
For an input file that'd be:
cut -c -10,11- --output-delimiter ';' file

bash SED command explanation with semicolon

What is this sed command doing? and is there any online utility that kind of explains sed a little bit, like regex?
sed -i '1s/$/|,a Type,b Type,c Type/;/./!b;1!s/$/|,,,/' textflile.txt
I think in the beginning it is adding csv a type, b type, c type at the end of the line but what does the rest of the command too

I don't know of any such utility, but let me explain using a text editor:
sed -i '1s/$/|,a Type,b Type,c Type/;/./!b;1!s/$/|,,,/' textflile.txt
^ ^ ^ ^ ^^ ^^ ^
| | | | || || |
modify | End Non-empty || || input
the | of lines || |Negation, file
file | line only || |i.e. lines 2,3,...
in | || |
place | || First
First line Negation, i.e.| line
empty lines only|
Branch to
script end,
i.e. skip the rest
In other words, it adds |,a type, b Type,c Type to the first line, doesn't change empty lines, and adds |,,, to all the remaining lines.

sed -i '1s/$/|,a Type,b Type,c Type/;/./!b;1!s/$/|,,,/' textflile.txt
can be written as
sed -i '
1 s/$/|,a Type,b Type,c Type/
/./! b
1! s/$/|,,,/
' textflile.txt
on line 1 only, add some text to the end of the line
if the line is empty ("matches 1 character, not"), goto next "cycle" (i.e., print current line and go to next line)
on every line except line 1, add "|,,," to the end of the line
So, it looks like you're adding some blank fields to a CSV file.
info sed contains the complete sed manual.

This doesn't answer your question but it's important for people to know and requires more space and formatting than a comment so: FYI to do what #choroba says that sed script does, i.e.
it adds |,a type, b Type,c Type to the first line,
doesn't change empty lines,
and adds |,,, to all the remaining lines.
is just this in awk:
awk '
NR==1 { print $0 "|,a type, b Type,c Type"; next }
!NF { print }
NF { print $0 "|,,," }
'
or if you're familiar with ternary expressions and want to remove the redundant code:
awk '{
sfx = "|," (NR==1 ? "a type, b Type,c Type" : ",,")
print $0 (NF ? sfx : "")
}'

Print pattern on a string with special character

How to print only string figure with the following line :
\begin{figure}[h!]
I tried :
firstLine='\begin{figure}[h!]'
echo $firstLine | sed -n 's/\\begin{\(.*\)}/\1/p'
but returns :
figure[h!] instead of figure
It seems that issue comes from [] or ! character.

firstLine='\begin{figure}[h!]'
echo "$firstLine" | sed 's/.*{\(.*\)}.*/\1/'
Output:
figure
With your code (add .*):
echo $firstLine | sed -n 's/\\begin{\(.*\)}.*/\1/p'

This might work for you (GNU sed):
sed 's/.*{\(.*\)}.*/\1/' file
This assumes there is only one {...} expression and one line.
A more rigorous solution would be:
sed -n 's/.*\\begin{\([^}]*\)}.*/\1/p' file
However nothing would be output if no match was found.