Say I have a text file called "demo.txt" who looks like this:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Now I want to read a certain line, say line 2, with a command which will look something like this:
Line2 = read 2 "demo.txt"
So when I'll print it:
echo "$Line2"
I'll get:
5 6 7 8
I know how to use 'sed' command in order to print a n-th line from a file, but not how to read it. I also know the 'read' command but dont know how to use it in order a certain line.
Thanks in advance for the help.
Using head and tail
$ head -2 inputFile | tail -1
5 6 7 8
OR
a generalized version
$ line=2
$ head -"$line" input | tail -1
5 6 7 8
Using sed
$ sed -n '2 p' input
5 6 7 8
$ sed -n "$line p" input
5 6 7 8
What it does?
-n suppresses normal printing of pattern space.
'2 p' specifies the line number, 2 or ($line for more general), p commands to print the current patternspace
input input file
Edit
To get the output to some variable use some command substitution techniques.
$ content=`sed -n "$line p" input`
$ echo $content
5 6 7 8
OR
$ content=$(sed -n "$line p" input)
$ echo $content
5 6 7 8
To obtain the output to a bash array
$ content= ( $(sed -n "$line p" input) )
$ echo ${content[0]}
5
$ echo ${content[1]}
6
Using awk
Perhaps an awk solution might look like
$ awk -v line=$line 'NR==line' input
5 6 7 8
Thanks to Fredrik Pihl for the suggestion.
Perl has convenient support for this, too, and it's actually the most intuitive!
The flip-flop operator can be used with line numbers:
$ printf "0\n1\n2\n3\n4" | perl -ne 'printf if 2 .. 4'
1
2
3
Note that it's 1-based.
You can also mix regular expressions:
$ printf "0\n1\nfoo\n3\n4" | perl -ne 'printf if /foo/ .. -1'
foo
3
4
(-1 refers to the last line)
Related
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
I need to move the last 4 lines of a text file and move them to the second row in the text file.
I'm assuming that tail and sed are used but, I haven't much luck so far.
Here is a head and tail solution. Let us start with the same sample file as Glenn Jackman:
$ seq 10 >file
Apply these commands:
$ head -n1 file ; tail -n4 file; tail -n+2 file | head -n-4
1
7
8
9
10
2
3
4
5
6
Explanation:
head -n1 file
Print first line
tail -n4 file
Print last four lines
tail -n+2 file | head -n-4
Print the lines starting with line 2 and ending before the fourth-to-last line.
If I'm assuming correctly, ed can handle your task:
seq 10 > file
ed file <<'COMMANDS'
$-3,$m1
w
q
COMMANDS
cat file
1
7
8
9
10
2
3
4
5
6
lines 7,8,9,10 have been moved to the 2nd line
$-3,$m1 means, for the range of lines from "$-3" (3 lines before the last line) to "$" (the last line, move them ("m") below the first line ("1")
Note that the heredoc has been quoted so the shell does not try to interpret the strings $- and $m1 as variables
If you don't want to actually modify the file, but instead print to stdout:
ed -s file <<'COMMANDS'
$-3,$m1
%p
Q
COMMANDS
Here is an awk solution:
seq 10 > file
awk '{a[NR]=$0} END {for (i=1;i<=NR-4;i++) if (i==2) {for (j=NR-3;j<=NR;j++) print a[j];print a[i]} else print a[i]}' file
1
7
8
9
10
2
3
4
5
6
I have the data file below:
136110828724515000007700877
137110904734015000007700877
138110911724215000007700877
127110626724515000007700871
127110626726015000007700871
131110724724515000007700871
134110814725015000007700871
134110814734015000007700871
104110122726027000001810072
107110208724527000002900000
And I want to extract value of column 3 ie values of 6787714447.
I tried by using:-
awk "print $3" <filename>
but it didn't work. What should I use instead?
It is a better job for cut:
$ cut -c 3 < file
6
7
8
7
7
1
4
4
4
7
As per man cut:
-c, --characters=LIST
select only these characters
To make them appear all in the same line, pipe tr -d '\n':
$ cut -c 3 < file | tr -d '\n'
6787714447
Or even to sed to have the new line at the end:
$ cut -c 3 < file | tr -d '\n' | sed 's/$/\n/'
6787714447
With grep:
$ grep -oP "^..\K." file
6
7
8
7
7
1
4
4
4
7
with sed:
$ sed -r 's/..(.).*/\1/' file
6
7
8
7
7
1
4
4
4
7
with awk:
$ awk '{split ($0, a, ""); print a[3]}' file
6
7
8
7
7
1
4
4
4
7
Cut is probably the simpler/cleaner option, but here two alternatives:
AWK version:
awk '{print substr($1, 3, 1) }' <filename>
Python version:
python -c 'print "\n".join(map(lambda x: x[2], open("<filename>").readlines()))'
EDIT: Please see 1_CR's comments and disregard this option in favour of his.
Is there a Linux utility or a Bash command I can use to sort a space delimited string of numbers?
Here's a simple example to get you going:
echo "81 4 6 12 3 0" | tr " " "\n" | sort -g
tr translates the spaces delimiting the numbers, into carriage returns, because sort uses carriage returns as delimiters (ie it is for sorting lines of text). The -g option tells sort to sort by "general numerical value".
man sort for further details about sort.
This is a variation from #JamesMorris answer:
echo "81 4 6 12 3 0" | xargs -n1 | sort -g | xargs
Instead of tr, I use xargs -n1 to convert to new lines. The final xargs is to convert back, to a space separated sequence of numbers.
This is a variation on ghostdog74's answer that's too big to fit in a comment. It shows digits instead of names of numbers and both the original string and the result are in space-delimited strings (instead of an array which becomes a newline-delimited string).
$ s="3 2 11 15 8"
$ sorted=$(echo $(printf "%s\n" $s | sort -n))
$ echo $sorted
2 3 8 11 15
$ echo "$sorted"
2 3 8 11 15
If you didn't use the echo when setting the value of sorted, then the string has newlines in it. In that case echoing it without quotes puts it all on one line, but, as echoing it with quotes would show, each number would appear on its own line. This is the case whether the original is an array or a string.
# demo
$ s="3 2 11 15 8"
$ sorted=$(printf "%s\n" $s | sort -n)
$ echo $sorted
2 3 8 11 15
$ echo "$sorted"
2
3
8
11
15
$ s=(one two three four)
$ sorted=$(printf "%s\n" ${s[#]}|sort)
$ echo $sorted
four one three two
Using Bash parameter expansion (to replace spaces with newlines) we can do:
str="3 2 11 15 8"
sort -n <<< "${str// /$'\n'}"
# alternative
NL=$'\n'
str="3 2 11 15 8"
sort -n <<< "${str// /${NL}}"
If you actually have a space-delimited string of numbers, then one of the other answers provided would work fine. If your list is a bash array, then:
oldIFS="$IFS"
IFS=$'\n'
array=($(sort -g <<< "${array[*]}"))
IFS="$oldIFS"
might be a better solution. The newline delimiter would help if you want to generalize to sorting an array of strings instead of numbers.
Improving on Evan Krall's nice Bash "array sort" by limiting the scope of IFS to a single command:
printf "%q\n" "${IFS}"
array=(3 2 11 15 8)
array=($(IFS=$'\n' sort -n <<< "${array[*]}"))
echo "${array[#]}"
printf "%q\n" "${IFS}"
$ awk 'BEGIN{split(ARGV[1], numbers);for(i in numbers) {print numbers[i]} }' \
"6 7 4 1 2 3" | sort -n
I added this to my .zshrc (or .bashrc) file:
#sort a space-separated list of words (e.g. a list of HTML classes)
sortwords() {
echo $1 | xargs -n1 | sort -g | xargs
}
Call it from the terminal like this:
sortwords "banana date apple cherry"
# apple banana cherry date
Thanks to #FranMowinckel and others for inspiration.