Awk pass variable containing columns to be printed - awk

I want to pass a variable to awk containing which columns to print from a file?
In this trivial case file.txt contains a single line
11 22 33 44 55
This is what I've tried:
awk -v a='2/4' -v b='$2/$4' '{print a"\n"$a"\n"b"\n"$b}' file.txt
output:
2/4
22
$2/$4
11 22 33 44 55
desired output:
0.5
Is there any way to do this type of "eval" of variable as a command?


Here is one method for dividing columns specified in variables:
$ awk -v num=2 -v denom=4 '{print $num/$denom}' file.txt
0.5
If you trust the person who creates the shell variable b, then here is a method that offers flexibility:
$ b='$2/$4'; awk "{print $b}" file.txt
0.5
$ b='$1*$2'; awk "{print $b}" file.txt
242
$ b='$2,$2/$4,$5'; awk "{print $b}" file.txt
22 0.5 55
The flexibility here is due to the fact that b can contain any awk code. This approach requires that you trust the creator of b.

Related

awk Can not Select Column with empty value

i am trying to select a column with its missing value
here is my input file separated by tab
1 2 3
4 5
6
7 8
9
i am trying to select the first column in which output will look like
1
4
7
and the length of my column would be 5 in this case
I have tried
awk '$1!=""{print $1}' ./demo.txt
but it returns
1
4
6
7
9
can anybody help with this I am new in AWK

You can use cut:
$ cut -f 1 file # the default delimiter is a tab
Or with sed:
$ sed 's/[[:blank:]].*$//' file
Or awk:
$ awk '{sub(/[[:blank:]].*$/,"")}1' file
Or:
$ awk 'BEGIN{FS=OFS="\t"} {print $1}' file
All those print the first column and all five lines (blank or not)
Prints:
1
4
7

Tell awk to use a tab (\t) as the input field delimiter (-F):
$ awk -F'\t' '{ print $1 }' demo.txt
1
4
7
If you want to print multiple columns, maintaining the same delimiter for output, another approach using the FS and OFS variables:
$ awk 'BEGIN { FS=OFS="\t" } { print $1,$3 }' demo.txt
1 3
4 5
7
9

With sed something like:
sed 's/^\([^[:blank:]]*\).*/\1/' demo.txt

Using FIELDWIDTHS in gnu-awk you can do this for fixed width separated data:
awk 'BEGIN {FIELDWIDTHS = "4 4 *"} {print $1}' file
1
4
7
For demo purpose:
awk 'BEGIN {FIELDWIDTHS = "4 4 *"} {print NR ":", $1}' file
1: 1
2: 4
3:
4: 7
5:

if they're all single digits in 1st column :
echo \
'1 2 3
4 5
6
7 8
9' |
mawk NF=1 FS=
gcat -n
1 1
2 4
3
4 7
5
that's literally all you need. To play it safe, then do
nawk NF=1 FS='[[:space:]]' # overly-verbose so-called
# "proper" posix form
gawk NF=1 FS='[ \t]' # suffices unless the input
# happens to have uncommon bytes
# like \013 \v or \014 \f
or a very fringe way of fudging NF :
mawk 'NF ^= FS="[ \t]"'

Sed replace nth column of multiple tsv files without header

Here are multiple tsv files, where I want to add 'XX' characters only in the second column (everywhere except in the header) and save it to this same file.
Input:
$ls
file1.tsv file2.tsv file3.tsv
$head -n 4 file1.tsv
a b c
James England 25
Brian France 41
Maria France 18
Ouptut wanted:
a b c
James X1_England 25
Brian X1_France 41
Maria X1_France 18
I tried this, but the result is not kept in the file, and a simple redirection won't work:
# this works, but doesn't save the changes
i=1
for f in *tsv
do awk '{if (NR!=1) print $2}’ $f | sed "s|^|X${i}_|"
i=$((i+1))
done
# adding '-i' option to sed: this throws an error but would be perfect (sed no input files error)
i=1
for f in *tsv
do awk '{if (NR!=1) print $2}’ $f | sed -i "s|^|T${i}_|"
i=$((i+1))
done
Some help would be appreciated.

The second column is particularly easy because you simply replace the first occurrence of the separator.
for file in *.tsv; do
sed -i '2,$s/\t/\tX1_/' "$file"
done
If your sed doesn't recognize the symbol \t, use a literal tab (in many shells, you type it with ctrlv tab.) On *BSD (and hence MacOS) you need -i ''

AWK solution:
awk -i inplace 'BEGIN { FS=OFS="\t" } NR!=1 { $2 = "X1_" $2 } 1' file1.tsv
Input:
a b c
James England 25
Brian France 41
Maria France 18
Output:
a b c
James X1_England 25
Brian X1_France 41
Maria X1_France 18

awk + How do I find duplicates in a column?

How do I find duplicates in a column?
$ head countries_lat_long_int_code3.csv | cat -n
1 country,latitude,longitude,name,code
2 AD,42.546245,1.601554,Andorra,376
3 AE,23.424076,53.847818,United Arab Emirates,971
4 AF,33.93911,67.709953,Afghanistan,93
5 AG,17.060816,-61.796428,Antigua and Barbuda,1
6 AI,18.220554,-63.068615,Anguilla,1
7 AL,41.153332,20.168331,Albania,355
8 AM,40.069099,45.038189,Armenia,374
9 AN,12.226079,-69.060087,Netherlands Antilles,599
10 AO,-11.202692,17.873887,Angola,244
For instance this has duplicates in the 5th column.
5 AG,17.060816,-61.796428,Antigua and Barbuda,1
6 AI,18.220554,-63.068615,Anguilla,1
How do I view all the others in this file?
I know I can do this:
awk -F, 'NR>1{print $5}' countries_lat_long_int_code3.csv | sort
And I can eyeball and see if there is any duplicates, but is there a better way?
Or I can do this:
Find out how may are there completely
$ awk -F, 'NR>1{print $5}' countries_lat_long_int_code3.csv | sort | wc -l
210
Find out how many unique values are there
$ awk -F, 'NR>1{print $5}' countries_lat_long_int_code3.csv | sort | uniq | wc -l
183
Therefore there are at most 27 (210-183) duplicates.
EDIT1
My desired output would be something as follows, basically all the columns but just showing the rows that are duplicates:
5 AG,17.060816,-61.796428,Antigua and Barbuda,1
6 AI,18.220554,-63.068615,Anguilla,1

This will give you the duplicated codes
awk -F, 'a[$5]++{print $5}'
if you're only interested in count of duplicate codes
awk -F, 'a[$5]++{count++} END{print count}'
To print duplicated rows try this
awk -F, '$5 in a{print a[$5]; print} {a[$5]=$0}'
This will print the whole row with duplicates found in col $5:
awk -F, 'a[$5]++{print $0}'

This is the less memory aggressive i can guess:
$ cat infile
country,latitude,longitude,name,code
AD,42.546245,1.601554,Andorra,376
AE,23.424076,53.847818,United Arab Emirates,971
AF,33.93911,67.709953,Afghanistan,93
AG,17.060816,-61.796428,Antigua and Barbuda,1
AI,18.220554,-63.068615,Anguilla,1
AL,41.153332,20.168331,Albania,355
AM,40.069099,45.038189,Armenia,374
AN,12.226079,-69.060087,Netherlands Antilles,599
AO,-11.202692,17.873887,Angola,355
$ awk -F\, '$NF in a{if (a[$NF]!=0){print a[$NF];a[$NF]=0}print;next}{a[$NF]=$0}' infile
AG,17.060816,-61.796428,Antigua and Barbuda,1
AI,18.220554,-63.068615,Anguilla,1
AL,41.153332,20.168331,Albania,355
AO,-11.202692,17.873887,Angola,355
NOTE: I have included another duplicate for testing purposes.

If you just want to print out a unique value that repeat over the same file just add at the end of the awk:
awk ... ... | sort | uniq -u
That will print the unique values only on alphabetic order

awk and sprintf to zero fill

Using awk and sprintf how can I zero fill both before and after a decimal point
input
11
12.2
9.6
output
110
122
096
I can get either using these, but not both
sprintf("%.1f", $1)
output
110
122
96
sprintf("%03d", $1)
output
011
012
096

x = sprintf("%06.3f", 1.23)
Output:
$ awk 'BEGIN{x = sprintf("%06.3f", 1.23); print x}'
01.230
$

I really can't tell from your question but maybe one of these does whatever it is you want:
$ cat file
11
12.2
9.6
$ awk '{ x=sprintf("%03d",$0*10); print x }' file
110
122
096
$ awk '{ x=sprintf("%04.1f",$0); print x }' file
11.0
12.2
09.6
Obviously you could just use printf with no intermediate variable but you asked for sprintf().

Reading a file from line 4 to the end

I want to read a file from the line 4 to the very end is there anyway to this with awk or something?

This sed command will do:
sed -n '4,$p' file.txt
Or using awk:
awk 'NR>=4' file.txt
Or using tail:
tail +4 file.txt

awk 'NR >= 4 {print $0}'
For example
$> seq 101 110 | awk 'NR >= 4 {print $0}'
104
105
106
107
108
109
110

tail +4 filename ll serve ur purpose.
more on tail

heres a method (that can depend on the type of shell you use, bash should work):
tmpvar=`cat a_file | wc -l `; tail -$((tmpvar-4)) a_file
heres another method that should work in more shells:
cat a_file -n | awk '{if($1>4) print $2}'