Combining columns within a single file using awk - awk

I am trying to reformat a large file. The first 6 columns of each line are OK but the rest of the columns in the line need to be combined in increments of 2 with a "/" character in between.
Example file (showing only a few columns but have many more in actual file):
1 1 0 0 1 2 A T A C
Into:
1 1 0 0 1 2 A/T A/C
So far I have been trying awk and this is where I am at...
awk '{print $1,$2,$3,$4,$5; for(i=7; i < NF; i=i+2) print $i+"/"+$i+1}' myfile.txt > mynewfile.txt


awk '{for(i=j=7; i < NF; i+=2) {$j = $i"/"$(i+1); j++} NF=j-1}1' input


Please try this:
awk '{print $1" "$2" "$3" "$4" "$5" "$6" "$7"/"$8" "$9"/"$10}' myfile.txt > mynewfile.txt


"+" is the arithmetic "and" operator, string concatenation is done by simply listing the strings adjacent to each other, i.e. to get the string "foobar" you'd write:
"foo" "bar"
not:
"foo" + "bar"
Anyway, try this:
awk -v ORS= '{print $1,$2,$3,$4,$5,$6; for(i=7;i<=NF;i++) print (i%2?OFS:"/") $i; print "\n"}' myfile.txt > mynewfile.txt

Related

awk conditional statement based on a value between colon

I was just introduced to awk and I'm trying to retrieve rows from my file based on the value on column 10.
I need to filter the data based on the value of the third value if ":" was used as a separator in column 10 (last column).
Here is an example data in column 10. 0/1:1,9:10:15:337,0,15.
I was able to extract the third value using this command awk '{print $10}' file.txt | awk -F ":" '/1/ {print $3}'
This returns the value 10 but how can I return other rows (not just the value in column 10) if this third value is less than or greater than a specific number?
I tried this awk '{if($10 -F ":" "/1/ ($3<10))" print $0;}' file.txt but it returns a syntax error.
Thanks!

Your code:
awk '{print $10}' file.txt | awk -F ":" '/1/ {print $3}'
should be just 1 awk script:
awk '$10 ~ /1/ { split($10,f,/:/); print f[3] }' file.txt
but I'm not sure that code is doing what you think it does. If you want to print the 3rd value of all $10s that contain :s, as it sounds like from your text, that'd be:
awk 'split($10,f,/:/) > 1 { print f[3] }' file.txt
and to print the rows where that value is less than 7 would be:
awk '(split($10,f,/:/) > 1) && (f[3] < 7)' file.txt

linux csv file concatenate columns into one column

I've been looking to do this with sed, awk, or cut. I am willing to use any other command-line program that I can pipe data through.
I have a large set of data that is comma delimited. The rows have between 14 and 20 columns. I need to recursively concatenate column 10 with column 11 per row such that every row has exactly 14 columns. In other words, this:
a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p
will become:
a,b,c,d,e,f,g,h,i,jkl,m,n,o,p
I can get the first 10 columns. I can get the last N columns. I can concatenate columns. I cannot think of how to do it in one line so I can pass a stream of endless data through it and end up with exactly 14 columns per row.
Examples (by request):
How many columns are in the row?
sed 's/[^,]//g' | wc -c
Get the first 10 columns:
cut -d, -f1-10
Get the last 4 columns:
rev | cut -d, -f1-4 | rev
Concatenate columns 10 and 11, showing columns 1-10 after that:
awk -F',' ' NF { print $1","$2","$3","$4","$5","$6","$7","$8","$9","$10$11}'

Awk solution:
awk 'BEGIN{ FS=OFS="," }
{
diff = NF - 14;
for (i=1; i <= NF; i++)
printf "%s%s", $i, (diff > 1 && i >= 10 && i < (10+diff)?
"": (i == NF? ORS : ","))
}' file
The output:
a,b,c,d,e,f,g,h,i,jkl,m,n,o,p

With GNU awk for the 3rd arg to match() and gensub():
$ cat tst.awk
BEGIN{ FS="," }
match($0,"(([^,]+,){9})(([^,]+,){"NF-14"})(.*)",a) {
$0 = a[1] gensub(/,/,"","g",a[3]) a[5]
}
{ print }
$ awk -f tst.awk file
a,b,c,d,e,f,g,h,i,jkl,m,n,o,p

If perl is okay - can be used just like awk for stream processing
$ cat ip.txt
a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p
1,2,3,4,5,6,3,4,2,4,3,4,3,2,5,2,3,4
1,2,3,4,5,6,3,4,2,4,a,s,f,e,3,4,3,2,5,2,3,4
$ awk -F, '{print NF}' ip.txt
16
18
22
$ perl -F, -lane '$n = $#F - 4;
print join ",", (#F[0..8], join("", #F[9..$n]), #F[$n+1..$#F])
' ip.txt
a,b,c,d,e,f,g,h,i,jkl,m,n,o,p
1,2,3,4,5,6,3,4,2,43432,5,2,3,4
1,2,3,4,5,6,3,4,2,4asfe3432,5,2,3,4
-F, -lane split on , results saved in #F array
$n = $#F - 4 magic number, to ensure output ends with 14 columns. $#F gives the index of last element of array (won't work if input line has less than 14 columns)
join helps to stitch array elements together with specified string
#F[0..8] array slice with first 9 elements
#F[9..$n] and #F[$n+1..$#F] the other slices as needed
Borrowing from Ed Morton's regex based solution
$ perl -F, -lape '$n=$#F-13; s/^([^,]*,){9}\K([^,]*,){$n}/$&=~tr|,||dr/e' ip.txt
a,b,c,d,e,f,g,h,i,jkl,m,n,o,p
1,2,3,4,5,6,3,4,2,43432,5,2,3,4
1,2,3,4,5,6,3,4,2,4asfe3432,5,2,3,4
$n=$#F-13 magic number
^([^,]*,){9}\K first 9 fields
([^,]*,){$n} fields to change
$&=~tr|,||dr use tr to delete the commas
e this modifier allows use of Perl code in replacement section
this solution also has the added advantage of working even if input field is less than 14

You can try this gnu sed
sed -E '
s/,/\n/9g
:A
s/([^\n]*\n)(.*)(\n)(([^\n]*\n){4})/\1\2\4/
tA
s/\n/,/g
' infile

First variant - with awk
awk -F, '
{
for(i = 1; i <= NF; i++) {
OFS = (i > 9 && i < NF - 4) ? "" : ","
if(i == NF) OFS = "\n"
printf "%s%s", $i, OFS
}
}' input.txt
Second variant - with sed
sed -r 's/,/#/10g; :l; s/#(.*)((#[^#]){4})/\1\2/; tl; s/#/,/g' input.txt
or, more straightforwardly (without loop) and probably faster.
sed -r 's/,(.),(.),(.),(.)$/#\1#\2#\3#\4/; s/,//10g; s/#/,/g' input.txt
Testing
Input
a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p
a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r
a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u
Output
a,b,c,d,e,f,g,h,i,jkl,m,n,o,p
a,b,c,d,e,f,g,h,i,jklmn,o,p,q,r
a,b,c,d,e,f,g,h,i,jklmnopq,r,s,t,u

Solved a similar problem using csvtool. Source file, copied from one of the other answers:
$ cat input.txt
a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p
1,2,3,4,5,6,3,4,2,4,3,4,3,2,5,2,3,4
1,2,3,4,5,6,3,4,2,4,a,s,f,e,3,4,3,2,5,2,3,4
Concatenating columns:
$ cat input.txt | csvtool format '%1,%2,%3,%4,%5,%6,%7,%8,%9,%10%11%12,%13,%14,%15,%16,%17,%18,%19,%20,%21,%22\n' -
a,b,c,d,e,f,g,h,i,jkl,m,n,o,p,,,,,,
1,2,3,4,5,6,3,4,2,434,3,2,5,2,3,4,,,,
1,2,3,4,5,6,3,4,2,4as,f,e,3,4,3,2,5,2,3,4
anatoly#anatoly-workstation:cbs$ cat input.txt

awk: print each column of a file into separate files

I have a file with 100 columns of data. I want to print the first column and i-th column in 99 separate files, I am trying to use
for i in {2..99}; do awk '{print $1" " $i }' input.txt > data${i}; done
But I am getting errors
awk: illegal field $(), name "i"
input record number 1, file input.txt
source line number 1
How to correctly use $i inside the {print }?

Following single awk may help you too here:
awk -v start=2 -v end=99 '{for(i=start;i<=end;i++){print $1,$i > "file"i;close("file"i)}}' Input_file

An all awk solution. First test data:
$ cat foo
11 12 13
21 22 23
Then the awk:
$ awk '{for(i=2;i<=NF;i++) print $1,$i > ("data" i)}' foo
and results:
$ ls data*
data2 data3
$ cat data2
11 12
21 22
The for iterates from 2 to the last field. If there are more fields that you desire to process, change the NF to the number you'd like. If, for some reason, a hundred open files would be a problem in your system, you'd need to put the print into a block and add a close call:
$ awk '{for(i=2;i<=NF;i++){f=("data" i); print $1,$i >> f; close(f)}}' foo

If you want to do what you try to accomplish :
for i in {2..99}; do
awk -v x=$i '{print $1" " $x }' input.txt > data${i}
done
Note
the -v switch of awk to pass variables
$x is the nth column defined in your variable x
Note2 : this is not the fastest solution, one awk call is fastest, but I just try to correct your logic. Ideally, take time to understand awk, it's never a wasted time

Add new lines up to specific row

Hoping somebody can help me out.
I have large number of files with different number of lines.
I would like to add new lines in to the files up to specific rows, say 6.
Infile.txt
text1
text2
text3
The out file I would like to have is
Outfile.txt
text1
text2
text3
\n
\n
\n

Short awk solution:
awk -v r=6 'END{ while((r--)-NR>0) print "" }1' file
-v r=6 - variable r indicating total/maximal number of rows

In awk's END block, the built-in variable NR will contain the row number of the last line of the file. From there it's easy to print the needed number of additional empty rows.
$ awk -v lines=6 '1; END {for (i=NR; i<lines; ++i) print ""}' file
text1
text2
text3
$ awk -v lines=6 '1; END {for (i=NR; i<lines; ++i) print ""}' file | wc -l
6

IMHO the clearest and most obvious way to handle this is to simply loop from the last line number plus 1 to the target number of lines:
$ seq 3 | awk -v n=6 '{print} END{for (i=NR+1; i<=n; i++) print ""}'
1
2
3
$
You can also count down if you want to save a variable:
$ seq 3 | awk -v n=6 '{print} END{while (n-- > NR) print ""}'
1
2
3
$
but IMHO that's sacrificing clarity in favor of brevity and not worthwhile.

AWK how to count patterns on the first column?

I was trying get the total number of "??", " M", "A" and "D" from this:
?? this is a sentence
M this is another one
A more text here
D more and more text
I have this sample line of code but doesn't work:
awk -v pattern="\?\?" '{$1 == pattern} END{print " "FNR}'

$ awk '{ print $1 }' file | sort | uniq -c
1 ??
1 A
1 D
1 M
If for some reason you want an awk-only solution:
awk '{ ++cnt[$1] } END { for (i in cnt) print cnt[i], i }' file
but I think that's needlessly complicated compared to using the built-in unix tools that already do most of the work.
If you just want to count one particular value:
awk -v value='??' '$1 == value' file | wc -l
If you want to count only a subset of values, you can use a regex:
$ awk -v pattern='A|D|(\\?\\?)' '$1 ~ pattern { print $1 }' file | sort | uniq -c
1 ??
1 A
1 D
Here you do need to send a \ in order that the ?s are escaped within the regular expression. And because the \ is itself a special character within the string being passed to awk, you need to escape it first (hence the double backslash).