I've got an input file like this:
line 1
line 2
line 3
line 4
line 5
line 6
I'd like to use awk to insert a blank line every few lines; for example, every two:
line 1
line 2
line 3
line 4
line 5
line 6
How can I get awk to put a blank line into my file every n lines?
A more "awk-ish" way to write smcameron's answer:
awk -v n=5 '1; NR % n == 0 {print ""}'
The "1;" is a condition that is always true, and will trigger the default action which is to print the current line.
awk '{ if ((NR % 5) == 0) printf("\n"); print; }'
for n == 5, of course. Substitute whatever your idea of n is.
More awkishness:
awk 'ORS=NR%5?RS:RS RS'
For example:
$ printf "%s\n" {1..12} | awk 'ORS=NR%5?RS:RS RS'
1
2
3
4
5
6
7
8
9
10
11
12
awk '{print; if (FNR % 5 == 0 ) printf "\n";}' your_file
I guess 'print' should be before 'printf', and FNR is more accurate for your task.
$ awk -v n=5 '$0=(!(NR%n))?"\n"$0:$0'
If you want to change 'n', please set the parameter 'n' by awk's -v option.
Related
I am trying this
awk '{B=$(NF-1);A=$NF; $NF=$(NF-2); $(NF-1) = $(NF-3); $(NF-2)=A; $(NF-3) = B; print;}' input_text.txt
but I get the error:
awk: cmd. line:1: (FILENAME=cazzo.txt FNR=2) fatal: attempt to access field -1
Sample input:
$ cat input_text.txt
1 7 9 11 0 5 2
The same happens if I replace the spaces with tabs in the input_text.txt file.
Expected output:
1 7 9 5 2 11 0
I am running with Cygwin on Windows 10.
You can try this awk for swapping values:
awk 'NF > 3 {a=$NF; b=$(NF-1); $NF=$(NF-2); $(NF-1)=$(NF-3); $(NF-3)=b; $(NF-2)=a} 1' file
1 7 9 5 2 11 0
If there are DOS line breaks then use:
awk -v RS='\r?\n' 'NF > 3 {a=$NF; b=$(NF-1); $NF=$(NF-2); $(NF-1)=$(NF-3); $(NF-3)=b; $(NF-2)=a} 1' file
If you have gnu awk then you can use this regex based approach:
awk -v RS='\r?\n' 'NF > 3 {
$0 = gensub(/(\S+\s+\S+)(\s+)(\S+\s+\S+)$/, "\\3\\2\\1", "1")} 1' file
1 7 9 5 2 11 0
To swap the last n fields with the n fields before them:
$ awk -v n=2 'NF>=(2*n){ for (i=NF-(n-1); i<=NF; i++) {t=$i; $i=$(i-n); $(i-n)=t} } 1' file
1 7 9 5 2 11 0
$ awk -v n=3 'NF>=(2*n){ for (i=NF-(n-1); i<=NF; i++) {t=$i; $i=$(i-n); $(i-n)=t} } 1' file
1 0 5 2 7 9 11
With your shown samples, please try following code. This is a Generic code, where you have 2 awk variables named fromFields and toFields. So you need to give their values like: let's say you want to substitute 4th field value with 6th field AND 5th field value with 7th field, so you will set it like: fromFields="4,5" and toFields="6,7". I am assuming user will understand that values which are given are feasible with respect to Input_file.
awk -v fromFields="4,5" -v toFields="6,7" '
BEGIN{
num1=split(fromFields,arr1,",")
num2=split(toFields,arr2,",")
}
{
tmp=""
for(i=1;i<=num1;i++){
tmp=$arr1[i]
$arr1[i]=$arr2[i]
$arr2[i]=tmp
}
}
1
' Input_file
In the awk below I am trying to move the last line only, to the one above it. The problem with the below is that since my input file varies (not always 4 lines like in the below), I can not use i=3 everytime and can not seem to fix it. Thank you :).
file
this is line 1
this is line 2
this is line 3
this is line 4
desired output
this is line 1
this is line 2
this is line 4
this is line 3
awk (seems like the last line is being moved, but to i=2)
awk '
{lines[NR]=$0}
END{
print lines[1], lines[NR];
for (i=3; i<NR; i++) {print lines[i]}
}
' OFS=$'\n' file
this is line 1
this is line 2
this is line 4
this is line 3
$ seq 4 | awk 'NR>2{print p2} {p2=p1; p1=$0} END{print p1 ORS p2}'
1
2
4
3
$ seq 7 | awk 'NR>2{print p2} {p2=p1; p1=$0} END{print p1 ORS p2}'
1
2
3
4
5
7
6
try following awk once:
awk '{a[FNR]=$0} END{for(i=1;i<=FNR-2;i++){print a[i]};print a[FNR] ORS a[FNR-1]}' Input_file
Explanation: Creating an array named a with index FNR(current line's number) and keeping it's value to current line's value. Now in END section of awk, starting a for loop from i=1 to i<=FNR-2 why till FNR-2 because you need to swap only last 2 lines here. Once it prints all the lines then simply printing a[FNR](which is last line) and then printing a[FNR-1] with ORS(to print new line).
Solution 2nd: By counting the number of lines in a Input_file and putting them into a awk variable.
awk -v lines=$(wc -l < Input_file) 'FNR==(lines-1){val=$0;next} FNR==lines{print $0 ORS val;next} 1' Input_file
You nearly had it. You just have to change the order.
awk '
{lines[NR]=$0}
END{
for (i=1; i<NR-1; i++) {print lines[i]}
print lines[NR];
print lines[NR-1];
}
' OFS=$'\n' file
I'd reverse the file, swap the first two lines, then re-reverse the file
tac file | awk 'NR==1 {getline line2; print line2} 1' | tac
awk newbie here! I am asking for help to solve a simple specific task.
Here is file.txt
1
2
3
5
6
7
8
9
As you can see a single number (the number 4) is missing. I would like to print on the console the number 4 that is missing. My idea was to compare the current line number with the entry and whenever they don't match I would print the line number and exit. I tried
cat file.txt | awk '{ if ($NR != $1) {print $NR; exit 1} }'
But it prints only a newline.
I am trying to learn awk via this small exercice. I am therefore mainly interested in solutions using awk. I also welcome an explanation for why my code does not do what I would expect.
Try this -
awk '{ if (NR != $1) {print NR; exit 1} }' file.txt
4
since you have a solution already, here is another approach, comparing with previous values.
awk '$1!=p+1{print p+1} {p=$1}' file
you positional comparison won't work if you have more than one missing value.
Maybe this will help:
seq $(tail -1 file)|diff - file|grep -Po '.*(?=d)'
4
Since I am learning awk as well
awk 'BEGIN{i=0}{i++;if(i!=$1){print i;i=$1}}' file
4
`awk` explanation read each number from `$1` into array `i` and increment that number list line by line with `i++`, if the number is not sequential, then print it.
cat file
1
2
3
5
6
7
8
9
11
12
13
15
awk 'BEGIN{i=0}{i++;if(i!=$1){print i;i=$1}}' file
4
10
14
Assume the following file
#zvview.exe
#begin Present/3
77191.0000 189.320100 0 0 3 0111110 16 1
-8.072430+6-8.072430+6 77190 0 1 37111110 16 2
37 2 111110 16 3
8.115068+6 0.000000+0 8.500000+6 6.390560-2 9.000000+6 6.803440-1111110 16 4
9.500000+6 1.685009+0 1.000000+7 2.582780+0 1.050000+7 3.260540+0111110 16 5
37 2 111110 16 18
What I would like to do, is print in two columns, the fields after line 6. This can be done using NR. The tricky part is the following : Every second field, should go in one column as well as adding an E before the sign, so that the output file will look like this
8.115068E+6 0.000000E+0
8.500000E+6 6.390560E-2
9.000000E+6 6.803440E-1
9.500000E+6 1.685009E+0
1.000000E+7 2.582780E+0
1.050000E+7 3.260540E+0
From the output file you see that I want to keep in $6 only length($6)=10 characters.
How is it possible to do it in awk?
can do all in awk but perhaps easier with the unix toolset
$ sed -n '6,7p' file | cut -c2-66 | tr ' ' '\n' | pr -2ats' '
8.115068+6 0.000000+0
8.500000+6 6.390560-2
9.000000+6 6.803440-1
9.500000+6 1.685009+0
1.000000+7 2.582780+0
1.050000+7 3.260540+0
Here is a awk only solution or comparison
$ awk 'NR>=6 && NR<=7{$6=substr($6,1,10);
for(i=1;i<=6;i+=2) {f[++c]=$i;s[c]=$(i+1)}}
END{for(i=1;i<=c;i++) print f[i],s[i]}' file
8.115068+6 0.000000+0
8.500000+6 6.390560-2
9.000000+6 6.803440-1
9.500000+6 1.685009+0
1.000000+7 2.582780+0
1.050000+7 3.260540+0
Perhaps shorter version,
$ awk 'NR>=6 && NR<=7{$6=substr($6,1,10);
for(i=1;i<=6;i+=2) print $i FS $(i+1)}' file
8.115068+6 0.000000+0
8.500000+6 6.390560-2
9.000000+6 6.803440-1
9.500000+6 1.685009+0
1.000000+7 2.582780+0
1.050000+7 3.260540+0
to convert format to standard scientific notation, you can pipe the result to
sed or embed something similar in awk script (using gsub).
... | sed 's/[+-]/E&/g'
8.115068E+6 0.000000E+0
8.500000E+6 6.390560E-2
9.000000E+6 6.803440E-1
9.500000E+6 1.685009E+0
1.000000E+7 2.582780E+0
1.050000E+7 3.260540E+0
With GNU awk for FIELDWIDTHS:
$ cat tst.awk
BEGIN { FIELDWIDTHS="9 2 9 2 9 2 9 2 9 2 9 2" }
NR>5 && NR<8 {
for (i=1;i<NF;i+=4) {
print $i "E" $(i+1), $(i+2) "E" $(i+3)
}
}
$ awk -f tst.awk file
8.115068E+6 0.000000E+0
8.500000E+6 6.390560E-2
9.000000E+6 6.803440E-1
9.500000E+6 1.685009E+0
1.000000E+7 2.582780E+0
1.050000E+7 3.260540E+0
If you really want to get rid of the leading blanks then there's various ways to do it (simplest being gsub(/ /,"",$<field number>) on the relevant fields) but I left them in because the above allows your output to line up properly if/when your numbers start with a -, like they do on line 4 of your sample input.
If you don't have GNU awk, get it as you're missing a LOT of extremely useful functionality.
I tried to combine #karafka 's answer using substr, so the following does the trick!
awk 'NR>=6 && NR<=7{$6=substr($6,1,10);for(i=1;i<=6;i+=2) print substr($i,1,8) "E" substr($i,9) FS substr($(i+1),1,8) "E" substr($(i+1),9)}' file
and the output is
8.115068E+6 0.000000E+0
8.500000E+6 6.390560E-2
9.000000E+6 6.803440E-1
9.500000E+6 1.685009E+0
1.000000E+7 2.582780E+0
1.050000E+7 3.260540E+0
I have got a text file with some structure like this:
2 2 4 5 6
1 9 7 6 2
1 5 2 8 5
I want to be able to divide any element of any row by an element of another row. For example if I wanted to divide the 3rd element of the 1st row by the 2nd element of the 3rd row that would give:
4/5 = 0.8
Couldn't figure out a smart way to do this with AWK. Suggestions?
This MAY be what you want but it's hard to tell without more details and the expected output:
$ awk -v num=1,5 -v den=3,3 '{for (i=1;i<=NF;i++) cell[NR","i]=$i} END{print (cell[den] ? cell[num]/cell[den] : "NaN")}' file
3
$ awk -v num=3,4 -v den=1,2 '{for (i=1;i<=NF;i++) cell[NR","i]=$i} END{print (cell[den] ? cell[num]/cell[den] : 0)}' file
4
If (i1, j1) and (i2, j2) are the coordinates of the numerator and the denominator, you can do this :
i1=1
j1=3
i2=3
j2=2
awk 'NR=='$i1'{a=$'$j1'} NR=='$i2' {b=$'$j2'} END {print a"/"b " = " a/b}' file