I have a text file that looks similar to below.
Code 1 (3)
5 10 10
6 10 10
7 10 10
Code 2 (2)
9 11 11
10 8 8
Code 3 (1)
12 10 9
Code 4 (2)
14 8 10
15 8 10
I am only interested in the first and last numbers, in the first column. I would like to extract the first, last, and difference (1+last-first) to a new text file with a column for each first, last, and difference. The result should look like below. Technically, the difference column could be the number between the parentheses, as this number would always be the 1+difference between the last and first numbers between each string. Note, the last row in the input text file does not have a string below it.
5 7 3
9 10 2
12 12 1
14 15 2
Trying awk '/Code/{flag=1;next}/Code/{flag=0}flag' gives me all the lines and columns between each string. Trying awk '$1 ~ /Code/{flag=1;next},$1 ~ 1 /Code/{flag=0}flag' results in a syntax error at ,.
You may use this awk:
awk -v OFS='\t' '/^Code/ {
if (NR > 1)
print first, prev, (prev-first+1)
first = prev = ""
next
}
(first == "") {
first = $1
}
{
prev = $1
}
END {
print first, prev, (prev-first+1)
}' file
5 7 3
9 10 2
12 12 1
14 15 2
Related
Guys I have a file like this
NR column
1 1
2 1
3 0
4 0
5 0
6 1
7 1
8 1
9 1
10 0
11 0
12 0
13 1
14 1
What I need is to find the NR what will tell me where there are 1.
so my ideal output should tell me from NR=1 - 2 (there are 1s, then), NR=6 - 9, NR=13 - 14
or
1
2
6
9
13
14
Since, I think is easier not consider in the output the first row and the last. I expect that the output is
2
6
9
13
I've been trying a way to use getline but unsuccessfully.
I am sure there is an easy way to do this, help?
Thanks
Assuming your output above was incorrect (and it should really be the line number where the 0/1 or 1/0 transition happens - so the lines would be: "1, 3, 6, 10, 13"), then an awk oneliner is:
awk 'prev!=$0{print NR};{prev=$0}' file
which says:
for every line that doesn't match the prev line, print the line number, and
for every line, save the prev line
$ awk 'NR>1 && $0!=prev{print NR} {prev=$0}' file
3
6
10
13
or for your updated requirements:
$ awk '$1!=prev{print NR-prev} {prev=$1} END{if (prev) print NR}' file
1
2
6
9
13
14
awk to the rescue!
$ awk '!p&&$2==1{p=$1}
p&&!$2{print p"-"($1-1);p=0}
END{if(p) print p"-"$1}' file
1-2
6-9
13-14
{
if (NR > 1 && last != $0) {
print NR;
}
last = $0;
}
Another way
awk '$2!=x{x=$2;print NR-!($2)}END{if(x)print NR}' file
1
2
6
9
13
14
My data is a large text file that consists of 12 rows repeating. It looks something like this:
{
1
2
3
4
5
6
7
8
9
10
}
repeating over and over. I want to turn every 12 rows into columns. so the data would look like this:
{ 1 2 3 4 5 6 7 8 9 10 }
{ 1 2 3 4 5 6 7 8 9 10 }
{ 1 2 3 4 5 6 7 8 9 10 }
I have found some examples of how to convert all the rows to columns using awk: awk '{printf("%s ", $0)}', but no examples of how to convert every 12 rows into columns and then repeat the process.
Here is an idiomatic way (read golfed down version of Tom Fenech's answer) of doing it with awk:
$ awk '{ORS=(NR%12?FS:RS)}1' file
{ 1 2 3 4 5 6 7 8 9 10 }
{ 1 2 3 4 5 6 7 8 9 10 }
{ 1 2 3 4 5 6 7 8 9 10 }
ORS stands for Output Record Separator. We set the ORS to FS which by default is space for every line except the 12th line where we set it to RS which is a newline by default.
You could use something like this:
awk '{printf "%s%s", $0, (NR%12?OFS:RS)}' file
NR%12 evaluates to true except when the record number is exactly divisible by 0. When it is true, the output field separator is used (which defaults to a space). When it is false, the record separator is used (by default, a newline).
Testing it out:
$ awk '{printf "%s%s", $0, (NR%12?OFS:RS)}' file
{ 1 2 3 4 5 6 7 8 9 10 }
I want to read first colum in a matrix, and then print columns of this matrix using this first colum as reference. And example:
mat.txt
2 10 6 12 3
4 11 1 22 6
5 15 3 18 9
Using first column as reference, I would like to get columns 2, 4 and 5, and also put the value of first colum at the begining.
2 10 12 3
4 11 22 6
5 15 18 9
I try this, but doesn't work well:
awk 'FNR==NR{c++;cols[c]=$1;end}
{for(i=1;i<=c;i++) printf("%s%s",$(cols[i]+1),i<c ? OFS : "\n")}' mat.txt mat.txt
This may do:
awk 'FNR==NR {a[NR]=$1;next} {printf "%s ",a[FNR];for (i in a) printf "%s ",$(a[i]);print ""}' mat.txt{,}
2 10 12 3
4 11 22 6
5 15 18 9
The {,} make the file be used two times.
I am looking for a way to select a column (e. g. eighth column) of a data file and write the first five numbers of that column in a row, the next five numbers in second row, and so on.
I have been testing with awk and printf without success.
The awk way to do this is to switch from using OFS and ORS to separate the output using the modulus function:
$ seq 1 20 | awk '{printf "%s", $1 (NR % 5 ? OFS : ORS)}'
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
Change $1 to $8 for the eigth column for example and NR % 5 to NR % 10 for rows of 10 instead of 5. The seq command just generate a single column of digits from 1 to 20 used for demonstration.
I also find using xargs useful for this kind of thing:
$ seq 1 20 | awk '{print $1}' | xargs -n5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
The awk isn't necessary for the example as seq only produces a single column however for your question change $1 to $8 to select only the eighth column from your input. With this approach you could also switch out awk with cut.
This will also produce the format requested
seq 1 20 | awk '{printf("%s ", $1); if (NR % 5 == 0) printf("\n")}'
where $1 indicates de column number which could be changed when passing an archive to the awk line.
I have a file like this: (data.dat)
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 7
5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 9
6 6 6 6 6 6 6 6 6 6 6 7 6 7
7 9 7 7 7 7 7 7 7 7 7 8 7 9
8 10 8 9 8 9 8 8 8 8 8 9
9 11 9 10 9 9 9 9 9 10
10 12 10 11 10 10 10 11
The odd columns are simple line counters (NR), the even columns are simple values. I would like to get those values, in which the second (or even) colum values are the same in all even columns, i.e. I should get this output:
1
2
3
9
I have already tried to make this line, but something is wrong:
awk '{arr1[$1]=$2;arr2[$3]=$4;arr3[$5]=$6;arr4[$7]=$8;arr5[$9]=$10;arr6[$11]=$12;arr7[$13]=$14;arr8[$15]=$16;}END{for(x in arr1) if(x in arr2 && x in arr3 && x in arr4 && x in arr5 && x in arr6 && x in arr7 && x in arr8) print arr1[x];}' data.dat | sort -n
Is there a better way, by the way?
UPDATE: The real problem is that the array indices are different. So, the arr[...] method does not work... :(
This would work -
awk '
BEGIN{x=0}
{if (x<NF) x=NF;for (i=2;i<=NF;i+=2) a[$i]++}
END{x=x/2;for (y in a) if (x==a[y]) print y}' INPUT_FILE
Explanation:
We set a variable x=0 in the BEGIN statement.
We use this variable to get to find out maximum number of fields (This is useful later).
We store value of every second column to an array and get their number of occurrences.
We divide the variable x by 2 to verify maximum number a value can occur in every second column.
If the occurrences of numbers in an array matches this variable it means they are present in every second column.
Test: with your sample file
[jaypal:~/Temp] awk '
BEGIN{x=0}
{if (x<NF) x=NF;for (i=2;i<=NF;i+=2) a[$i]++}
END{x=x/2;for (y in a) if (x==a[y]) print y}' file
2
3
9
1
You can either pipe the output to sort -n to get it in order or use this -
awk '
BEGIN{x=0}
{if (x<NF) x=NF;for (i=2;i<=NF;i+=2) a[$i]++}
END{x=x/2;for (i=1;i<=length(a);i++) if (x==a[i]) print i}' INPUT_FILE
Your example works with just a simple;
awk '{if($2==$4 && $2==$6 && $2==$8 && $2==$10 && $2==$12 && $2==$14 && $2==$16) print $1}' test.txt | sort -n
Any other requirements I'm missing?
EDIT: Apparently with the missing columns you added :) Try
awk '{if(NF>1) { found=1; for(i=4; i<NF+1; i+=2) { if($2!=$i) { found=0; } } } if(found) print $1}' test.txt | sort -n
In your input data row # 9 doesn't have all even columns same so not sure how you show 9 in your desired output. You can try following awk command to print 1st col for your task:
awk '{same=0; prev=-1; for(i=2;i<=NF;i+=2) {if (prev != -1 && prev != $i) {same=1; break;} else prev=$i;} if (same==0) print $1;}' awk '{same=0; prev=-1; for(i=2;i<=NF;i+=2) {if (prev != -1 && prev != $i) {same=1; break;} else prev=$i;} if (same==0) print $1;}'