I am learning shell script and trying to write a program which will generate random numbers with index and this will be written to a file. For example following is the sample content which should be written to a file
0, 45
1, 65
2, 1898
3, 42
and so on
I used the following for loop but it only writes 10, random_number and not index from 0 to 9. Can someone help me on this
for i in `seq 1 10`
do
echo $i, $RANDOM > inputFile
done
Related
awk newbie here! I am asking for help to solve a simple specific task.
Here is file.txt
1
2
3
5
6
7
8
9
As you can see a single number (the number 4) is missing. I would like to print on the console the number 4 that is missing. My idea was to compare the current line number with the entry and whenever they don't match I would print the line number and exit. I tried
cat file.txt | awk '{ if ($NR != $1) {print $NR; exit 1} }'
But it prints only a newline.
I am trying to learn awk via this small exercice. I am therefore mainly interested in solutions using awk. I also welcome an explanation for why my code does not do what I would expect.
Try this -
awk '{ if (NR != $1) {print NR; exit 1} }' file.txt
4
since you have a solution already, here is another approach, comparing with previous values.
awk '$1!=p+1{print p+1} {p=$1}' file
you positional comparison won't work if you have more than one missing value.
Maybe this will help:
seq $(tail -1 file)|diff - file|grep -Po '.*(?=d)'
4
Since I am learning awk as well
awk 'BEGIN{i=0}{i++;if(i!=$1){print i;i=$1}}' file
4
`awk` explanation read each number from `$1` into array `i` and increment that number list line by line with `i++`, if the number is not sequential, then print it.
cat file
1
2
3
5
6
7
8
9
11
12
13
15
awk 'BEGIN{i=0}{i++;if(i!=$1){print i;i=$1}}' file
4
10
14
I have a file with a range of inputs varying from 0-100. I want to generate the percentage of numbers below a range of values.
Is it possible to do it in awk? Can someone show me an example of it.
Eg input file: 1, 4, 6, 7, 7, 7, 8, 9, 2, 4
output:
below 2: 20%
below 4: 40%
below 6: 50%
below 8: 90%
** included below or equal
Assuming data is in a.txt file:
awk -v limit=16 '$1 < limit {count++} END {print 100*count/NR}' a.txt
Where limit is the cutoff value, it just counts the elements smaller than limit ($1 < limit), and after all lines are read it prints the count divided by number of records (NR).
I have a file that I need to extract segments from based on a character range given in another file. I would like to do it using an awk command.
File one would look like this ( a single line):
AATTGTGAAGGTAGATGGCTCGCTCCGCGGCGGGGCGCGCGCGCGCGCGCGGGCTCGCTATATAGAGATATATGCGCGCGGCGCGCGGCGCGCGCGGCGCGCGCGTATATATATAGGCGCGCGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGAAAAAAAAAAAAAAAAAAAAAAAAATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGCCCCCCCCCCCCCC
The second file would look like follows:
5 10
13 20
22 24
and the output would be:
GTGAAG
AGATGGCT
GCT
This one-liner will solve your problem:
awk 'BEGIN{getline sequence < "first_file"} {print substr(sequence, $1, $2 - $1 + 1) }' second_file
Explanation: This script reads string sequence from file named first_file(adjust it to the actual file name) using getline function. Then for each line of second file(which contains ranges for processing) it extracts necessary substring using substr function. substr accepts three parameters: string(sequence), position($1), and length($2 - $1 + 1).
Nya gave you the awk solution, here's one based on coreutils.
string
AATTGTGAAGGTAGATGGCTCGCTCCGCGGCGGGGCGCGCGCGCGCGCGCGGGCTCGCTATATAGAGATATATGCGCGCGGCGCGCGGCGCGCGCGGCGCGCGCGTATATATATAGGCGCGCGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGAAAAAAAAAAAAAAAAAAAAAAAAATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGCCCCCCCCCCCCCC
offlen
5 10
13 20
22 24
You can get the output you want with:
while read off len; do cut -c${off}-${len} string; done < offlen
Output:
GTGAAG
AGATGGCT
GCT
Hi I need to parse a number from sql output :
COUNT(*)
----------
924
140
173
583
940
77
6 rows selected.
if the the fisrt line is less then 10 I want to create a empty file,
The problem is I dont know how to parse it, the numbers are still changing (from 0 to ca. 10 000 ) .
Question is very unclear so I'll make some assumptions. You'll get the output above from sql either to file or stdout and you would like to test of the first line containing digits is less than 10. Correct?
This is one way to do it.
sed -n '3p' log | awk '{ print ($1 < 10) ? "true" : "false" }'
sed is used to print the 3rd line from your example
this is then piped into awk which makes to comparison.
...or putting it together in bash
#!/bin/bash
while read variable;
do
if [[ "$variable" =~ ^[0-9]+$ ]]
then
break
fi
done < input
if [ "$variable" -lt 10 ]
then
echo 'less than 10'
# add your code here, eg
# touch /path/to/file/to/be/created
fi
What's the easiest way to split a file and add a header to each section?
The unix split command does everything that I need minus being able to add a header.
Any easy way to do it with existing tools before I script it up?
It is probably easiest to do this in either awk or perl. If you aren't processing much data, then using a simple shell script to post-process the output of split is probably fine. However, this will traverse the input more than once which can be a bottleneck if you are doing this for any sort of online data processing task. Something like the following should work:
bash$ cat input-file | awk '
BEGIN {
fnum = 1
print "HEADER" > fnum
}
{ if ((NR % 10) == 0) {
close(fnum)
fnum++
print "HEADER" > fnum
}
print >> fnum
}
'
bash$ wc -l input-file
239 input-file
bash$ ls
1 19 6
10 2 7
11 20 8
12 21 9
13 22 input-file
14 23
15 24
16 3
17 4
18 5
bash$