I have a file and want to add numbers infront of it, below is an example:
I have a file with the following:
0.152
0.153
0.158
0.156
0.157
and I want to put increasing numbers infront of it with a space, like this:
1 0.152
2 0.153
3 0.158
4 0.156
5 0.157
ascendingnumber*space*numberinfile
I would be very greatful if anyone can help. I have a large amount of data so it would take me for ages to add the numbers in manually. Its Linux stuff.
Many Thanks
A struggling student :)!
cat file | awk '{ print NR " " $1 }'
use awk
awk '{print NR " " $0}' input.txt > output.txt
several ways
awk '{print NR,$0}' file
cat -n file
nl file
sed '=' file
ruby -ne 'print "#{$.} #{$_}"' file
Of course, just bash
c=1; while read -r line; do echo $((c++)) $line; done < file
If your system has the nl command:
$ cat numbers.txt
0.152
0.153
0.158
0.156
0.157
$ nl -w 1 -s ' ' numbers.txt
1 0.152
2 0.153
3 0.158
4 0.156
5 0.157
The -w 1 flag specifies the column width of the ascending number. The -s ' ' flag tells nl to use one space to separate the numbers.
Related
I am a newbie in linux. Need help for a command.
I have file in linux with following values:
2-1
2-10
2-11
2-12
2-2
2-3
1-1
1-10
1-11
1-2
1-3
1-9
Needed output needed is 23. Sum of maximum from 1- & 2- pattern i.e. 11 from 1-11 & 12 from 2-12
awk -F"-" 'BEGIN{a=0; b=0;} {if(int($1)==1 && int($2)>a){a=int($2)}; if(int($1)==2 && int($2)>b){b=int($2)}}END{print a+b}' file
output:
23
Another awk using ternary operator
awk -v FS='-' '{m1=($1==1?(m1>$2?m1:$2):m1);m2=($1==2?(m2>$2?m2:$2):m2)}END{print m1+m2}' file
sort + awk pipeline:
sort -t- -k2 -n file | awk -F'-' '{a[$1]=$2}END{ print a[1]+a[2] }'
The output:
23
$ awk -F'-' '{max[$1] = ($2 > max[$1] ? $2 : max[$1])} END{for (key in max) sum+=max[key]; print sum}' file
23
$ awk -F- 'a[$1]<$2{a[$1]=$2}END{for(i in a)s+=a[i]; print s}' infile
23
I want to sum all the numbers in a file (columns and lines) given by the first parameter, but my program shows sum=sum+$i instead of the numeric sum:
sum=0;
file=$1
for i in $file
do
sum=sum+$i;
done;
echo "The sum is: " $sum
Input file:
$cat file.txt
10 20 10
40
50
Expected output :
The sum is: 21
Maybe if there is an awk method to solve this?
Try this -
$cat file1.txt
10 20 10
40
50
$awk '{for(i=1;i<=NF;i++) {sum+=$i}} END {print sum}' file1.txt
130
OR
$xargs < file1.txt| tr ' ' + | bc
130
cat file.txt | xargs | sed -e 's/\ /+/g' | bc
You can also use a simple read and an array to sum the value relying on word splitting to separate the values into an array via the default IFS (Internal Field Separator), e.g.
#!/bin/bash
declare -i sum=0
fn="${1:-/dev/stdin}" ## read from file as 1st argument (default stdin)
while read -r line; do ## read each line
a=( $line ) ## separate values into array
for i in ${a[#]}; do ## for each value in array
((sum += i)) ## add to sum
done
done <"$fn"
echo "sum: $sum"
Example Input File
$ cat dat/numfile.txt
10 20 10
40
50
Example Use/Output
$ bash sumnumfile.sh dat/numfile.txt
sum: 130
Another for some awks (at least mawk and gawk):
$ awk -v RS="[^0-9]" '{s+=$1}END{print s}' file
130
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 two sets of text files. First set is in AA folder. Second set is in BB folder. The content of ff.txt file from first set(AA folder) is shown below.
Name number marks
john 1 60
maria 2 54
samuel 3 62
ben 4 63
I would like to print the second column(number) from this file if marks>60. The output will be 3,4. Next, read the ff.txt file in BB folder and delete the lines containing numbers 3,4. How can I do this with bash?
files in BB folder looks like this. second column is the number.
marks 1 11.824 24.015 41.220 1.00 13.65
marks 1 13.058 24.521 40.718 1.00 11.82
marks 3 12.120 13.472 46.317 1.00 10.62
marks 4 10.343 24.731 47.771 1.00 8.18
awk 'FNR == NR && $3 > 60 {array[$2] = 1; next} {if ($2 in array) next; print}' AA/ff.txt BB/filename
This works, but is not efficient (is that matter?)
gawk 'BEGIN {getline} $3>60{print $2}' AA/ff.txt | while read number; do gawk -v number=$number '$2 != number' BB/ff.txt > /tmp/ff.txt; mv /tmp/ff.txt BB/ff.txt; done
Of course, the second awk can be replaced with sed -i
For multi files:
ls -1 AA/*.txt | while read file
do
bn=`basename $file`
gawk 'BEGIN {getline} $3>60{print $2}' AA/$bn | while read number
do
gawk -v number=$number '$2 != number' BB/$bn > /tmp/$bn
mv /tmp/$bn BB/$bn
done
done
I didn't test it, so if there is a problem, please comment.