I have a file with below contents. I need to print each line side by side
hello
1223
man
2332
xyz
abc
Output desired:
hello 1223
man 2332
xyz abc
Is there any other alternative than paste command?
You can use this awk:
awk '{ORS = (NR%2 ? FS : RS)} 1' file
hello 1223
man 2332
xyz abc
This sets ORS (output record separator) equal to input field separator (FS) for odd numbered lines, for even numbered lines it will be set to input record separator (RS).
To get tabular data use column -t:
awk '{ORS = (NR%2 ? FS : RS)} 1' file | column -t
hello 1223
man 2332
xyz abc
awk/gawk solution:
$ gawk 'BEGIN{ OFS="\t"} { COL1=$1; getline; COL2=$1; print(COL1,COL2)}' file
hello 1223
man 2332
xyz abc
Bash solution (no paste command):
$ echo $(cat file) | while read col1 col2; do printf "%s\t%s\n" $col1 $col2; done
hello 1223
man 2332
xyz abc
Related
I want to use awk/sed to deal with two files(a.txt and b.txt) below and get the result
cat a.txt
a UK
b Japan
c China
d Korea
e US
And cat b.txt results
c Russia
e Canada
The result that I want is as below:
a UK
b Japan
c Russia
d Korea
e Canada
With awk:
First fill aray/hash a with complete row ($0) and use first column ($1) from this row as index. Finally, print all elements of array/hash a with a loop.
awk '{a[$1]=$0} END{for(i in a) print a[i]}' file1 file2
Output:
a UK
b Japan
c Russia
d Korea
e Canada
try:
awk 'FNR==NR{A[$1]=$NF;next} {printf("%s %s\n",$1,$1 in A?A[$1]:$NF)}' b.txt a.txt
Checking here condition FNR==NR which will be TRUE only when first file(b.txt) is being read. Then creating an array named A whose index is $1 and have the value last column. Then using printf for printing 2 strings where first string is $1 and another is if $1 of a.txt is present in array A then print array A's value whose index is $1 else print last column of a.tzt itself.
EDIT: as OP had carriage characters into Input_files so please remove them by following too.
tr -d '\r' < b.txt > temp_b.txt && mv temp_b.txt b.txt
You can use the below one-liner:
join -a 1 -a 2 a.txt <( awk '{print $1, "--", $0, "--"}' < b.txt ) | sed 's/ --$//' | awk -F ' -- ' '{print $NF}'
We use awk to prefix each line in b.txt with a key and -- to give us a split point later:
<( awk '{print $1, "--", $0, "--"}' < b.txt )
Use the join command to join the files on common keys. The -a 1 option tells the command to
join -a 1 -a 2 a.txt <( awk '{print $1, "--", $0, "--"}' < b.txt )
Use sed to remove the -- parts that are on some end of lines:
sed 's/ --$//'
Use awk to print the last item on each line:
awk -F ' -- ' '{print $NF}'
$ awk 'NR==FNR{b[$1]=$2;next} {print $1, ($1 in b ? b[$1] : $2)}' b.txt a.txt
a UK
b Japan
c Russia
d Korea
e Canada
I am having trouble matching specific column with grep command. I have a test file (test.txt) like this..
Bra001325 835 T 13 c$c$c$c$c$cccccCcc !!!!!68886676
Bra001325 836 C 8 ,,,,,.,, 68886676
Bra001325 841 A 6 ,$,.,,. BJJJJE
Bra001325 866 C 2 ,. HJ
And i want to extract all those lines which has a number 866 in the second column. When i use grep command i am getting all the lines that contains the number that number
grep "866" test.txt
Bra001325 835 T 13 c$c$c$c$c$cccccCcc !!!!!68886676
Bra001325 836 C 8 ,,,,,.,, 68886676
Bra001325 866 C 2 ,. HJ
How can i match specific column with grep command?
Try doing this :
$ awk '$2 == 866' test.txt
No need to add {print}, the default behaviour of awk is to print on a true condition.
with grep :
$ grep -P '^\S+\s+866\b' *
But awk can print filenames too & is quite more robust than grep here :
$ awk '$2 == 866{print FILENAME":"$0; nextfile}' *
In my case, the field separator is not space but comma. So I would have to add this, otherwise it won't work for me (On ubuntu 18.04.1).
awk -F ', ' '$2 == 866' test.txt
Given input such as:
1
1a
1.1b
2.0c
How to extract the integer/decimal number at beginning of each input line, using only Linux/Unix command line utilities?
Using awk, you could say:
awk '{print $0+0}'
Awk is available in Linux, BSD, and many other Unix-like operating systems. It helps in this way:
echo "1" | awk '{a+=$0; print a}' # output 1
echo "1a" | awk '{a+=$0; print a}' # output 1
echo "1.1b" | awk '{a+=$0; print a}' # output 1.1
echo "2.0c" | awk '{a+=$0; print a}' # output 2
Some more awk
For extracting only digits
$ awk 'gsub(/[[:alpha:]].*/,x,$1) + 1' << EOF
1
1a
1.1b
2.0c
EOF
1
1
1.1
2.0
For integer
$ awk '{print int($0)}' << EOF
1
1a
1.1b
2.0c
EOF
1
1
1
2
---edit---
If there is any blank line in file, you can avoid printing zero from following
$ awk 'NF{$0+=0}1' << EOF
1
1a
1.1b
2foot4c
2
EOF
1
1
1.1
2
2
Here is a way to do this with sed:
echo "12.3abc" | sed -n 's/^\([0-9.][0-9.]*\).*/\1/p'
Output:
12.3
The block in parentheses matches all numbers or periods '.' that occur at the beginning of the line. Everything after that is match by the '.*'.
The \1 says to replace the entire line with just the portion that was matched in the parentheses.
Assuming your version of grep supports -o:
grep -o '^[0-9.]\+' data.in
NB: This will match any sequence of digits and decimal points at the start of the line.
I want to find the last appearance of a string in a text file with linux commands. For example
1 a 1
2 a 2
3 a 3
1 b 1
2 b 2
3 b 3
1 c 1
2 c 2
3 c 3
In such a text file, i want to find the line number of the last appearance of b which is 6.
I can find the first appearance with
awk '/ b / {print NR;exit}' textFile.txt
but I have no idea how to do it for the last occurrence.
cat -n textfile.txt | grep " b " | tail -1 | cut -f 1
cat -n prints the file to STDOUT prepending line numbers.
grep greps out all lines containing "b" (you can use egrep for more advanced patterns or fgrep for faster grep of fixed strings)
tail -1 prints last line of those lines containing "b"
cut -f 1 prints first column, which is line # from cat -n
Or you can use Perl if you wish (It's very similar to what you'd do in awk, but frankly, I personally don't ever use awk if I have Perl handy - Perl supports 100% of what awk can do, by design, as 1-liners - YMMV):
perl -ne '{$n=$. if / b /} END {print "$n\n"}' textfile.txt
This can work:
$ awk '{if ($2~"b") a=NR} END{print a}' your_file
We check every second file being "b" and we record the number of line. It is appended, so by the time we finish reading the file, it will be the last one.
Test:
$ awk '{if ($2~"b") a=NR} END{print a}' your_file
6
Update based on sudo_O advise:
$ awk '{if ($2=="b") a=NR} END{print a}' your_file
to avoid having some abc in 2nd field.
It is also valid this one (shorter, I keep the one above because it is the one I thought :D):
$ awk '$2=="b" {a=NR} END{print a}' your_file
Another approach if $2 is always grouped (may be more efficient then waiting until the end):
awk 'NR==1||$2=="b",$2=="b"{next} {print NR-1; exit}' file
or
awk '$2=="b"{f=1} f==1 && $2!="b" {print NR-1; exit}' file
I have a file test.txt looking like this:
2092 Mary
103 Tom
1239 Mary
204 Mark
1294 Tom
1092 Mary
I am trying to create a shell script that will
Read each line and put the data in two columns into variable var1 and var2
If var2 in each line is the same, then add the var1 in those lines.
output the file into a text file.
The result should be unique values in the var2 column. Here's what I have so far:
#!/bin/sh
#!/usr/bin/sh
cat test.txt| while read line;
do
$var1=$(echo $line| awk -F\; '{print $1}')
$var2=$(echo $line| awk -F\; '{print $2}')
How can I reference the variable in each line and then compare them?
The expected output would be:
4423 Mary
1397 Tom
204 Mark
Using awk it is easy:
awk '{sum[$2] += $1} END {for (i in sum) printf "%4d %s\n", sum[i], i; }'
If you want to do it with bash 4.x (not 3.x), then:
declare -A sum
while read number name
do
((sum[$name] += $number))
done
for name in "${!sum[#]}"
do
echo ${sum[$name]} $name
done
The structure here is essentially isomorphic with the awk script, but a little less notationally convenient. It will read from standard input, using the names as indexes into the associative array sum. The ${!sum[#]} notation is described in the Shell Parameter Expansion section of the manual, and not even hinted at in the section on Arrays. The information is there if you know where to look.
If you want to process an arbitrary number of input files (like the awk script would) then you need to use cat to collect the data:
cat "$#" |
{
declare -A sum
while read number name
do
((sum[$name] += $number))
done
for name in "${!sum[#]}"
do
echo ${sum[$name]} $name
done
}
This is not UUOC because it handles no arguments (read standard input), one argument or many arguments.
For all the scripts, if you want to sort the output in number or name order, apply an appropriate sort to the output of the script:
script file1 file2 file3 | sort -k 1,1n # By sum increasing order
script file1 file2 file3 | sort -k 1,1nr # By sum decreasing order
script file1 file2 file3 | sort -k 2,2 # By name increasing order
script file1 file2 file3 | sort -k 2,2r # By name decreasing order