Suppose I have a vi file as the following:
cat file1
abc 123 pqr
lmn 234 rst
jkl 100 mon
I want to take the 2nd field of each line (viz, in this case is 123, 234 and 100) and append it to the end of that same line.
How will I do that?
The output should look like the following:
abc 123 pqr 123
lmn 234 rst 234
jkl 100 mon 100
With awk:
$ awk '{NF=NF+1; $NF=$2}1' file
abc 123 pqr 123
lmn 234 rst 234
jkl 100 mon 100
It increments the number of field in one and sets the last one as the 2nd. Then 1 is a true condition, which is evaluated as the default awk behaviour: {print $0}.
Or also
awk '{print $0, $2}' file
It prints the full line plus the second field.
Or even shorter, thanks Håkon Hægland!:
awk '{$(NF+1)=$2}1' file
You have many ways to do that in Vi(m). This is the simplest that comes to my mind:
:%norm 0f<space>yaw$p
Explanation:
:{range}norm command executes normal mode command on each line in {range}
% is a shortcut range meaning "all lines in the buffer" so we will execute what follows on every line in the buffer
0 puts the cursor on the first column on the current line (not strictly necessary but good practice)
f<space> jumps the cursor on the first <space> after the cursor on the current line
yaw yanks the word and the <space> under the cursor
$ jumps to the end of the line
p pastes the previously yanked text
prompt with mark, you can do it in vi
:%s/\( [^ ]*\)\(.*\)/\1\2\1/
Another way, Using sed
sed -r 's/( [^ ]*)(.*)/\1\2\1/' file
Related
I have a file with the lines below
123
456
123
789
abc
efg
xyz
I need to search with abc and replace immediate above 123 with 111. This is the requirement, abc is only one occurrence in the file but 123 can be multiple occurrences and 123 can be at any position above abc.
Please help me.
I have tried with below sed command
sed -i.bak "/abc/!{x;1!p;d;};x;s/123/1111" filename
With the above command, it is only replacing 123, if 123 is just above abc, if 123 is 2 lines above abc then replace is failing.
There's more than on way to do it. Here's one:
sed -i.bak '1{h;d;};/123/{x;p;d;};/abc/{x;s/123/111/;p;d;};H;${x;p;};d' filename
ed comes in handy for complex editing of files in scripts:
ed -s file <<EOF
/^abc$/;?^123$?;.c
111
.
w
EOF
This: Sets the current line to the first one matching abc (/^abc$/;). Then changes the first line before that point that matches 123 to 111 (?XXX? searches backwards for a matching regular expression, and ?^123$?;. selects that single line for c to change) and finally saves the modified file.
This is a classic case where you keep track of your previous line and change stuff depeinding on conditions satisfying the current line. Genearlly, an awk program looks like this:
awk '(FNR==1){prev=$0; next}
(condition_on_$0) { action_on_prev }
{ print prev; prev = $0 }
END { print $0 }'
So in the case of the OP, this would read:
awk '(FNR==1){prev=$0; next}
$0 == "abc" { if (prev == "123") prev = "111" }
{ print prev; prev = $0 }
END { print $0 }'
This might work for you (GNU sed):
sed -Ez 's/(.*)(\n123.*\nabc)/\1\n111\2/' file
This slurps the file into memory and inserts 111 in front of the last occurrence of 123 before abc.
A less memory intensive solution:
sed -E '/^123$/{:a;N;/\n123$/{h;s///p;g;s/.*\n//;ba};/\nabc$/!ba;s/^/111\n/}' file
This gathers up lines following a line containing 123. If another line containing 123 is encountered it offloads all lines before it and begins gathering lines again. If it finds a line containing abc it inserts 111 at the front of the lines gathered so far.
Another alternative:
sed '/abc/{x;/./{s/^/111\n/p;z};x;b};/123/{x;/./p;x;h;$!d;b};x;/./{x;H;$!d};x' file
$ tac file | awk 'f && sub(/123/,"111"){f=0} /abc/{f=1} 1' | tac
123
456
111
789
abc
efg
xyz
I have a file (input.txt) with the following structure:
>day_1
ABC
DEF
GHI
>day_2
JKL
MNO
PQR
>day_3
STU
VWX
YZA
>month_1
BCD
EFG
HIJ
>month_2
KLM
NOP
QRS
...
I would like to split this file into multiple files (day.txt; month.txt; ...). Each new text file would contain all "header" lines (the one starting with >) and their content (lines between two header lines).
day.txt would therefore be:
>day_1
ABC
DEF
GHI
>day_2
JKL
MNO
PQR
>day_3
STU
VWX
YZA
and month.txt:
>month_1
BCD
EFG
HIJ
>month_2
KLM
NOP
QRS
I cannot use split -l in this case because the amount of lines is not the same for each category (day, month, etc.). However, each sub-category has the same number of lines (=3).
EDIT: As per OP adding 1 more solution now.
awk -F'[>_]' '/^>/{file=$2".txt"} {print > file}' Input_file
Explanation:
awk -F'[>_]' ' ##Creating field separator as > or _ in current lines.
/^>/{ file=$2".txt" } ##Searching a line which starts with > if yes then creating a variable named file whose value is 2nd field".txt"
{ print > file } ##Printing current line to variable file(which will create file name of variable file's value).
' Input_file ##Mentioning Input_file name here.
Following awk may help you on same.
awk '/^>day/{file="day.txt"} /^>month/{file="month.txt"} {print > file}' Input_file
You can set the record separator to > and then just set the file name based on the category given by $1.
$ awk -v RS=">" 'NF {f=$1; sub(/_.*$/, ".txt", f); printf ">%s", $0 > f}' input.txt
$ cat day.txt
>day_1
ABC
DEF
GHI
>day_2
JKL
MNO
PQR
>day_3
STU
VWX
YZA
$ cat month.txt
>month_1
BCD
EFG
HIJ
>month_2
KLM
NOP
QRS
Here's a generic solution for >name_number format
$ awk 'match($0, /^>[^_]+_/){k = substr($0, RSTART+1, RLENGTH-2);
if(!(k in a)){close(op); a[k]; op=k".txt"}}
{print > op}' ip.txt
match($0, /^>[^_]+_/) if line matches >name_ at start of line
k = substr($0, RSTART+1, RLENGTH-2) save the name portion
if(!(k in a)) if the key is not found in array
a[k] add key to array
op=k".txt" output file name
close(op) in case there are too many files to write
print > op print input record to filename saved in op
Since each subcategory is composed of the same amount of lines, you can use grep's -A / --after flag to specify that number of lines to match after a header.
So if you know in advance the list of categories, you just have to grep the headers of their subcategories to redirect them with their content to the correct file :
lines_by_subcategory=3 # number of lines *after* a subcategory's header
for category in "month" "day"; do
grep ">$category" -A $lines_by_subcategory input.txt >> "$category.txt"
done
You can try it here.
Note that this isn't the most efficient solution as it must browse the input once for each category. Other solutions could instead browse the content and redirect each subcategory to their respective file in a single pass.
I am given a file. If a line has "xxx" as its third word then I need to replace it with "yyy". My final output must have all the original lines with the modified lines.
The input file is-
abc xyz mno
xxx xyz abc
abc xyz xxx
abc xxx xxx xxx
The required output file should be-
abc xyz mno
xxx xyz abc
abc xyz yyy
abc xxx yyy xxx
I have tried-
grep "\bxxx\b" file.txt | awk '{if ($3=="xxx") print $0;}' | sed -e 's/[^ ]*[^ ]/yyy/3'
but this gives the output as-
abc xyz yyy
abc xxx yyy xxx
Following simple awk may help you in same.
awk '$3=="xxx"{$3="yyy"} 1' Input_file
Output will be as follows.
abc xyz mno
xxx xyz abc
abc xyz yyy
abc xxx yyy xxx
Explanation: Checking condition here if $3 3rd field is equal to string xxx then setting $3's value to string yyy. Then mentioning 1 there, since awk works on method of condition then action. I am making condition TRUE here by mentioning 1 here and NOT mentioning any action here so be default print of current line will happen(either with changed 3rd field or with new 3rd field).
sed solution:
sed -E 's/^(([^[:space:]]+[[:space:]]+){2})apathy\>/\1empathy/' file
The output:
abc xyz mno
apathy xyz abc
abc xyz empathy
abc apathy empathy apathy
To modify the file inplace add -i option: sed -Ei ....
In general the awk command may look like
awk '{command set 1}condition{command set 2}' file
The command set 1 would be executed for every line while command set 2 will be executed if the condition preceding that is true.
My final output must have all the original lines with the modified
lines
In your case
awk 'BEGIN{print "Original File";i=1}
{print}
$3=="xxx"{$3="yyy"}
{rec[i++]=$0}
END{print "Modified File";for(i=1;i<=NR;i++)print rec[i]}'file
should solve that.
Explanation
$3 is the the third space-delimited field in awk. If it matches "xxx", then it is replaced. Print the unmodified lines first while storing the modified lines in an array. At the end, print the modified lines. BEGIN and END blocks are executed only at the beginning and the end respectively. NR is the awk built-in variable which denotes that number of records processed till the moment. Since it is used in the END block it should give us the total number of records.
All good :-)
Ravinder has already provided you with the shortest awk solution possible.
In sed, the following would work:
sed -E 's/(([^ ]+ ){2})xxx/\1yyy/'
Or if your sed doesn't include -E, you can use the more painful BRE notation:
sed 's/\(\([^ ][^ ]* \)\{2\}\)xxx/\1yyy/'
And if you're in the mood to handle this in bash alone, something like this might work:
while read -r line; do
read -r -a a <<<"$line"
[[ "${a[2]}" == "xxx" ]] && a[2]="yyy"
printf '%s ' "${a[#]}"
printf '\n'
done < input.txt
I have two files:
abc
ghi
and the second (aka database file)
abc 123
def 456
ghi 789
and I want to query the database file to print the second column into the second column of the first file if there is a match
So my output would be
abc 123
ghi 789
logically, I understand what I have to do, but I lack the commands in bash for it...
my attempt was to use join with the -1 but I do not understand how to implement it...
what's wrong with join?
$ cat 1
abc
ghi
$ cat 2
abc 123
def 456
ghi 789
$ join 1 2
abc 123
ghi 789
then if you want to store it somewhere just redirect the stdout.
join is a little overkill here (as it requires sorting) because file1 has just one column. Can you not use grep -f?
grep -Fwf file1 file2
-F treats the content of file1 as strings, not patterns
-w looks for the whole word to match
how to use awk on the following file named "awk.txt" and print all fields in proper length of space or tab length between.
# cat /root/awk.txt
abc hij klm
def pqr hij
mmm fgf hgt
yyt ghf jkw
I wanted to use awk on this and print in the following proper format.
abc hij klm
def pqr hij
mmm fgf hgt
yyt ghf jkw
Please help!!
Use the column command from coreutils:
column -t file
In this special case, where all entries have the same length, the following awk command would do the trick as well, however column can do the job even if the entries have different length:
awk '{$1=$1}1' OFS=' ' file
This line of awk will format the output using printf (documentation)
awk '{printf "%3s\t%3s\t%3s\n",$1,$2,$3}' awk.txt
If you want to strip the first line starting with #
awk '!/^#/{printf "%3s\t%3s\t%3s\n",$1,$2,$3}'