1 ajar 45000
2 Sunil 25000
3 varoom 50000
4 Amit 47000
5 tanru 15000
6 Deepak 23000
7 Sunil 13000
8 sattvic 80000
I did it using awk. I want using sed command
$ awk 'NR==3, NR==6 {print NR,$0}' employee.txt
sed -n '3,6p' employee.txt
-n tells sed to not print each line;
3,6 is an "address", it tells sed to only apply the following command to the given range of lines;
p tells sed to print the line.
Related
Here's my code : ls -lt | sed -n 'p;n'
That code makes me skip from a line to another when listing file names but doesn't start by skipping the first one, how to make that happen?
Here's an exemple without my code to skip to make it clear:
And here's an exemple of when I use the skip code:
You have to invert your sed command: it should be n;p instead of p;n:
Your code:
for x in {1..20}; do echo $x ; done | sed -n 'p;n'
1
3
5
7
9
11
13
15
17
19
The version with sed inverted:
for x in {1..20}; do echo $x ; done | sed -n 'n;p'
Output:
2
4
6
8
10
12
14
16
18
20
You can use sed's ~ operator: first~step
$ seq 1 10 | sed -n '1~2p'
1
3
5
7
9
$ seq 1 10 | sed -n '2~2p'
2
4
6
8
10
would you please help me how to convert every 4-sequantial rows into one tab-separated column?
convert:
A
1
2
3
3
3
4
1
to :
A 1 2 3
3 3 4 1
A simple way to do this is to use xargs:
$ xargs -n4 < file
A 1 2 3
3 3 4 1
With awk you would do:
$ awk '{printf "%s%s",$0,(NR%4?FS:RS)}' file
A 1 2 3
3 3 4 1
Another flexible approach is to use pr:
$ pr -tas' ' --columns 4 file
A 1 2 3
3 3 4 1
Both the awk and pr solution can be easily modified to change the output separator to a TAB:
$ pr -at --columns 4 file
A 1 2 3
3 3 4 1
$ awk '{printf "%s%s",$0,(NR%4?OFS:RS)}' OFS='\t' file
A 1 2 3
3 3 4 1
$ perl -pe 's{\n$}{\t} if $. % 4' old.file > new.file
or simply (thanks to mpapec's comment):
$ perl -pe 'tr_\n_\t_ if $. % 4' old.file > new.file
I would like to print $1 and $2 if $2<25from text files. I also need to get the total number of students with marks less than 25 from all files. How can I do this with awk or sed?
students marks
jerry 12
peter 35
john 5
jerry 15
john 10
Desired output
jerry 12
john 5
jerry 15
john 10
Total no:of students :- 4
In awk:
$ awk '$2<25 {print; i++} END{print "\nTotal number of students:- "i}' file
Output:
jerry 12
john 5
jerry 15
john 10
Total number of students:- 4
If you want the output sorted by grade (lowest to highest):
$ sort -n -k2,2 file | awk '$2<25 {print; i++} END{print "\nTotal number of students:- "i}'
Sorted Output:
john 5
john 10
jerry 12
jerry 15
Total number of students:- 4
-n numerical sort;
-k2,2 sort on the second field.
awk '$2<25{count++ ; print}END{print "Total No of Students :-",count}' your_file
tested below:
> awk '$2<25{count++ ; print}END{print "Total No of Students :-",count}' temp
jerry 12
john 5
jerry 15
john 10
Total No of Students :- 4
I have this so far:
sed -n '0,10p' yourfile > newfile
But it is not working, just outputs a blank file :(
Your question is ambiguous, so here is every permutation I can think of:
Print only the first 10 lines
head -n10 yourfile > newfile
Skip the first 10 lines
tail -n+10 yourfile > newfile
Print every 10th line
awk '!(NR%10)' yourfile > newfile
Delete every 10th line
awk 'NR%10' yourfile > newfile
(Since an ambiguous questions can only have an ambiguous answer...)
To print every tenth line (GNU sed):
$ seq 1 100 | sed -n '0~10p'
10
20
30
40
...
100
Alternatively (GNU sed):
$ seq 1 100 | sed '0~10!d'
10
20
30
40
...
100
To delete every tenth line (GNU sed):
$ seq 1 100 | sed '0~10d'
1
...
9
11
...
19
21
...
29
31
...
39
41
...
To print the first ten lines (POSIX):
$ seq 1 100 | sed '11,$d'
1
2
3
4
5
6
7
8
9
10
To delete the first ten lines (POSIX):
$ seq 1 100 | sed '1,10d'
11
12
13
14
...
100
python -c "import sys;sys.stdout.write(''.join(line for i, line in enumerate(open('yourfile')) if i%10 == 0 ))" >newfile
It is longer, but it is a single language - not different syntax and aprameters for each thing one tries to do.
With non-GNU sed, to print every 10th line use
sed '10,${p;n;n;n;n;n;n;n;n;n;}'
(GNU : sed -n '0~10p')
and to delete every 10th line use
sed 'n;n;n;n;n;n;n;n;n;d;'
(GNU : sed -n '0~10d')
I would like to extract some lines from a text file, I have started to tweak sed lately,
I have a file with the structure
88 3 3 0 0 1 101 111 4 3
89 3 3 0 0 1 3 4 112 102
90 3 3 0 0 1 102 112 113 103
91 3 3 0 0 2 103 113 114 104
What I would like to do is to extract the information according to the second column, I use sth like in my bash script(argument 2 is infile)
sed -n '/^[0-9]* [23456789]/ p' < $2 > out
however I have different entries other than the range [23456789], for instance 10, since it is composed of 1 and 0, to get that these two characters should be in the range I guess, however there are entries with '1'(for the second column) that I do not like to keep so how can write '10's but not '1's.
Best,
Umut
sed -rn '/^[0-9]* ([23456789]|10)/ p' < $2 > out
You need the extend-regexp support (-r) to have the | operator (or)
Another interesting way is:
sed -rn '/^[0-9]* ([23456789]|[0-9]{2,})/ p' < $2 > out
Which means [23456789] or 2 or more repetition of a digit.
The instant you see variable-sized columns in your data, you should start thinking about awk:
awk '$2 > 1 && $2 < 11 {print}{}'
will do the trick assuming your file format is correct.
sed -rn '/^[0-9]* (2|3|4|5|6|7|8|9|10)/p' < $2 > out