I am trying to get more familiar with awk statements, especially ones that can be done with just one line. I have a file that looks like this
9 5 0 2
8 7 4 3
4 8 2 1
I want the output to look like
0
3
1
Is there a way I can do this with just a one liner using awk? Thank you.
Using awk:
awk '{min=$1; for (i=2; i<=NF; i++) if ($i < min) min=$i; print min}' file
0
3
1
The are languages with built-in "min" functions:
ruby -ane 'puts $F.min' file
Or available libraries
perl -MList::Util=min -lane 'print min #F' file
Limiting to shell:
min() { printf "%s\n" "$#" | sort -n | head -1; }
while read -a nums; do
echo $(min "${nums[#]}")
done < file
GNU awk, which you'll find in most Linux distributions, has a built-in sort function, asort.
echo -e "9 5 0 2\n8 7 4 3\n4 8 2 1" |
awk '{ split($0,a); asort(a); print a[1]; }'
0
3
1
Related
Im trying to get the columsums (exept for the first one) of a tab delimited containing numbers.
To find out the number of columns an store it in a variable I use:
cols=$(awk '{print NF}' file.txt | sort -nu | tail -n 1
next I want to calculate the sum of all numbers in that column and store this in a variable again in a for loop:
for c in 2:$col
do
num=$(cat file.txt | awk '{sum+$2 ; print $0} END{print sum}'| tail -n 1
done
this
num=$(cat file.txt | awk '{sum+$($c) ; print $0} END{print sum}'| tail -n 1
on itself with a fixed numer and without variable input works find but i cannot get it to accept the for-loop variable.
Thanks for the support
p.s. It would also be fine if i could sum all columns (expept the first one) at once without the loop-trouble.
Assuming you want the sums of the individual columns,
$ cat file
1 2 3 4
5 6 7 8
9 10 11 12
$ awk '
{for (i=2; i<=NF; i++) sum[i] += $i}
END {for (i=2; i<=NF; i++) printf "%d%s", sum[i], OFS; print ""}
' file
18 21 24
In case you're not bound to awk, there's a nice tool for "command-line statistical operations" on textual files called GNU datamash.
With datamash, summing (probably the simplest operation of all) a 2nd column is as easy as:
$ datamash sum 2 < table
9
Assuming the table file holds tab-separated data like:
$ cat table
1 2 3 4
2 3 4 5
3 4 5 6
To sum all columns from 2 to n use column ranges (available in datamash 1.2):
$ n=4
$ datamash sum 2-$n < table
9 12 15
To include headers, see the --headers-out option
I have a text file that has a single column of numbers, like this:
1
2
3
4
5
6
I want to convert it into two columns, in the left to right order this way:
1 2
3 4
5 6
I can do it with:
awk '{print>"line-"NR%2}' file
paste line-0 line-1 >newfile
But I think the reliance on two intermediate files will make it fragile in a script.
I'd like to use something like cat file | mystery-zip-command >newfile
You can use paste to do this:
paste -d " " - - < file > newfile
You can also use pr:
pr -ats" " -2 file > newfile
-a - use round robin order
-t - suppress header and trailer
-s " " - use single space as the delimiter
-2 - two column output
See also:
Convert a text file into columns
another alternative
$ seq 6 | xargs -n2
1 2
3 4
5 6
or with awk
$ seq 6 | awk '{ORS=NR%2?FS:RS}1'
1 2
3 4
5 6
if you want the output terminate with a new line in case of odd number of input lines..
$ seq 7 | awk '{ORS=NR%2?FS:RS}1; END{ORS=NR%2?RS:FS; print ""}'
1 2
3 4
5 6
7
awk 'NR % 2 == 1 { printf("%s", $1) }
NR % 2 == 0 { printf(" %s\n", $1) }
END { if (NR % 2 == 1) print "" }' file
The odd lines are printed with no newline after them, to print the first column. The even lines are printed with a space first and a newline after, to print the second column. At the end, if there were an odd number of lines, we print a newline so we don't end in the middle of the line.
With bash:
while IFS= read -r odd; do IFS= read -r even; echo "$odd $even"; done < file
Output:
1 2
3 4
5 6
$ seq 6 | awk '{ORS=(NR%2?FS:RS); print} END{if (ORS==FS) printf RS}'
1 2
3 4
5 6
$
$ seq 7 | awk '{ORS=(NR%2?FS:RS); print} END{if (ORS==FS) printf RS}'
1 2
3 4
5 6
7
$
Note that it always adds a terminating newline - that is important as future commands might depend on it, e.g.:
$ seq 6 | awk '{ORS=(NR%2?FS:RS); print}' | wc -l
3
$ seq 7 | awk '{ORS=(NR%2?FS:RS); print}' | wc -l
3
$ seq 7 | awk '{ORS=(NR%2?FS:RS); print} END{if (ORS==FS) printf RS}' | wc -l
4
Just change the single occurrence of 2 to 3 or however many columns you want if your requirements change:
$ seq 6 | awk '{ORS=(NR%3?FS:RS); print} END{if (ORS==FS) printf RS}'
1 2 3
4 5 6
$ seq 7 | awk '{ORS=(NR%3?FS:RS); print} END{if (ORS==FS) printf RS}'
1 2 3
4 5 6
7
$ seq 8 | awk '{ORS=(NR%3?FS:RS); print} END{if (ORS==FS) printf RS}'
1 2 3
4 5 6
7 8
$ seq 9 | awk '{ORS=(NR%3?FS:RS); print} END{if (ORS==FS) printf RS}'
1 2 3
4 5 6
7 8 9
$
Short awk approach:
awk '{print ( ((getline nl) > 0)? $0" "nl : $0 )}' file
The output:
1 2
3 4
5 6
(getline nl)>0 - getline will get the next record and assign it to variable nl. The getline command returns 1 if it finds a record and 0 if it encounters the end of the file
Short GNU sed approach:
sed 'N;s/\n/ /' file
N - add a newline to the pattern space, then append the next line of input to the pattern space
s/\n/ / - replace newline with whitespace within captured pattern space
seq 6 | tr '\n' ' ' | sed -r 's/([^ ]* [^ ]* )/\1\n/g'
I'm trying to sort the numbers on each line of a file individually. The numbers within one line are separated by tabs. (I used spaces but they're actually tabs.)
For example, for the following input
5 8 7 6
1 5 6 8
8 9 7 1
the desired output would be:
5 6 7 8
1 5 6 7
1 7 8 9
My attempt so far is:
let i=1
while read line
do
echo "$line" | tr " " "\n" | sort -g
cut -f $i fileName | paste -s >> tempFile$$
((++i))
done < fileName
This is the best I got - I'm sure it can be done in 6 characters with awk/sed/perl:
while read line
do
echo $(printf "%d\n" $line | sort -n) | tr ' ' \\t >> another-file.txt
done < my-input-file.txt
Using a few features that are specific to GNU awk:
$ awk 'BEGIN{ PROCINFO["sorted_in"] = "#ind_num_asc" }
{ delete(a); n = 0; for (i=1;i<=NF;++i) a[$i];
for (i in a) printf "%s%s", i, (++n<NF?FS:RS) }' file
5 6 7 8
1 5 6 8
1 7 8 9
Each field is set as a key in the array a. In GNU awk it is possible to specify the order in which the for (i in a) loop traverses the array - here, I've set it to do so in ascending numerical order.
Here is a bash script that can do it. It takes a filename argument or reads stdin, was tested on CentOS and assumes IFS=$' \t\n'.
#!/bin/bash
if [ "$1" ] ; then exec < "$1" ; fi
cat - | while read line
do
set $line
echo $(for var in "$#"; do echo $var; done | sort -n) | tr " " "\t"
done
If you want to put the output in another file run it as:
cat input_file | sorting_script > another_file
or
sorting_script input_file > another file
Consider using perl for this:
perl -ape '#F=sort #F;$_="#F\n"' input.txt
Here -a turns on automatic field splitting (like awk does) into the array #F, -p makes it execute the script for each line and print $_ each time, and -e specifies the script directly on the command line.
Not quite 6 characters, I'm afraid, Sean.
This should have been simple in awk, but it doen't quite have the features needed. If there had been an array $# corresponding to the fields $1, $2, etc., then the solution would have been awk '{asort $#}' input.txt, but sadly no such array exits. The loops required to move the fields into an array and out of it again make it longer than the bash version:
awk '{for(i=1;i<=NF;i++)a[i]=$i;asort(a);for(i=1;i<=NF;i++)printf("%s ",a[i]);printf("\n")}' input.txt
So awk isn't the right tool for the job here. It's also a bit odd that sort itself doesn't have a switch to control its sorting direction.
Using awk
$ cat file
5 8 7 6
1 5 6 8
8 9 7 1
$ awk '{c=1;while(c!=""){c=""; for(i=1;i<NF;i++){n=i+1; if($i>$n){c=$i;$i=$n;$n=c}}}}1' file
5 6 7 8
1 5 6 8
1 7 8 9
Better Readable version
awk '{
c=1
while(c!="")
{
c=""
for(i=1;i<NF;i++)
{
n=i+1
if($i>$n)
{
c=$i
$i=$n
$n=c
}
}
}
}1
' file
If you have ksh, you may try this
#!/usr/bin/env ksh
while read line ; do
set -s +A cols $line
echo ${cols[*]}
done < "input_file"
Test
[akshay#localhost tmp]$ cat test.ksh
#!/usr/bin/env ksh
cat <<EOF | while read line ; do set -s +A cols $line; echo ${cols[*]};done
5 8 7 6
1 5 6 8
8 9 7 1
EOF
[akshay#localhost tmp]$ ksh test.ksh
5 6 7 8
1 5 6 8
1 7 8 9
using awk command I tried to print the upper triangle of a matrix
awk '{for (i=1;i<=NF;i++) if (i>=NR) printf $i FS "\n"}' matrix
but the output is shown as a single row
Consider this sample matrix:
$ cat matrix
1 2 3
4 5 6
7 8 9
To print the upper-right triangle:
$ awk '{for (i=1;i<=NF;i++) printf "%s%s",(i>=NR)?$i:" ",FS; print""}' matrix
1 2 3
5 6
9
Or:
$ awk '{for (i=1;i<=NF;i++) printf "%2s",(i>=NR)?$i:" "; print""}' matrix
1 2 3
5 6
9
To print the upper-left triangle:
$ awk '{for (i=1;i<=NF+1-NR;i++) printf "%s%s",$i,FS; print""}' matrix
1 2 3
4 5
7
Or:
$ awk '{for (i=1;i<=NF+1-NR;i++) printf "%2s",$i; print""}' matrix
1 2 3
4 5
7
This might work for you (GNU sed):
sed -r ':a;n;H;G;s/\n//;:b;s/^\S+\s*(.*)\n.*/\1/;tb;$!ba' file
Use the hold space as a counter for those lines that have been processed and for each current line remove those many fields from the front of the current line.
N.B. The counter is set following the printing of the current line otherwise the first line would be minus the first field.
On reflection an alternative/more elegant solution is:
sed -r '1!G;h;:a;s/^\S+\s*(.*)\n.*/\1/;ta' file
And to print the upper-left triangle:
sed -r '1!G;h;:a;s/^([^\n]*)\S+[^\n]*(.*)\n.*/\1\2/;ta' file
$ awk '{for (i=NR;i<=NF;i++) printf "%s%s",$i,(i<NF?FS:RS)}' file
1 2 3
5 6
9
How can I select the lines from the second line to the line before the last line of a file by using head and tail in unix?
For example if my file has 15 lines I want to select lines from 2 to 14.
tail -n +2 /path/to/file | head -n -1
perl -ne 'print if($.!=1 and !(eof))' your_file
tested below:
> cat temp
1
2
3
4
5
6
7
> perl -ne 'print if($.!=1 and !(eof))' temp
2
3
4
5
6
>
alternatively in awk you can use below:
awk '{a[count++]=$0}END{for(i=1;i<count-1;i++) print a[i]}' your_file
To print all lines but first and last ones you can use this awk as well:
awk 'NR==1 {next} {if (f) print f; f=$0}'
This always prints the previous line. To prevent the first one from being printed, we skip the line when NR is 1. Then, the last one won't be printed because when reading it we are printing the penultimate!
Test
$ seq 10 | awk 'NR==1 {next} {if (f) print f; f=$0}'
2
3
4
5
6
7
8
9