I need to generate 12 digit Hex numbers in KSH on Solaris
Thanks
#!/bin/ksh
set -A hex 0 1 2 3 4 5 6 7 8 9 A B C D E F
for i in {1..12}
do
printf ${hex[$((RANDOM%16))]}
done
Start with this Python program, hex12.py.
hex12.py
#!/usr/bin/env python
import random
import hashlib
h= hashlib.sha1(str(random.random())).hexdigest()
print h[:12]
In your shell you can now use hex.py to create 12 hex digits on standard out.
Try this one:
DIGITS=`head -c 6 /dev/urandom | od -x | head -n 1 | sed -e 's/^0* //' -e 's/ //g'
As RANDOM variable generates a 15 bit number (from 0 to 32767) you can concatenate several RANDOM values.
You will need a 48 bit number as 12 hex digits are 12 * 4 = 48 bits.
Either:
$ printf '%x\n' $(( ((RANDOM<<15|RANDOM)<<15|RANDOM)<<3|RANDOM%8 ))
9142467b46d3
Or:
$ printf '%x' $((RANDOM%4096)) $((RANDOM%4096)) $((RANDOM%4096)) $((RANDOM%4096)); echo
808878c21e19
Related
I'm trying to sort the numbers on each line of a file individually. The numbers within one line are separated by tabs. (I used spaces but they're actually tabs.)
For example, for the following input
5 8 7 6
1 5 6 8
8 9 7 1
the desired output would be:
5 6 7 8
1 5 6 7
1 7 8 9
My attempt so far is:
let i=1
while read line
do
echo "$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
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)
Is it possible to format seq in a way that it will display the range desired but with N numbers per line?
Let say that I want seq 20 but with the following output:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
My next guess would be a nested loop but I'm not sure how...
Any help would be appreciated :)
Use can use awk to format it as per your needs.
$ seq 20 | awk '{ORS=NR%5?FS:RS}1'
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
ORS is awk's built-in variable which stands for Output Record Separator and has a default value of \n. NR is awk's built-in variable which holds the line number. FS is built-in variable that stands for Field Separator and has the default value of space. RS is built-in variable that stands for Record Separator and has the default value of \n.
Our action which is a ternary operator, to check if NR%5 is true. When it NR%5 is not 0 (hence true) it uses FS as Output Record Separator. When it is false we use RS which is newline as Output Record Separator.
1 at the end triggers awk default action that is to print the line.
You can use xargs to limit the sequence displayed per line.
$ seq 20 | xargs -n 5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
The parameter -n 5 tells xargs to only display 5 sequence numbers.
If you have bash you can use the builtin sequence.
echo {1..20} | xargs -n 5
Using Bash:
while read num; do
((num % 5)) && printf "$line " || echo "$line"
done < <(seq 20)
Or:
for i in {1..20}; do
s+="$i "
if ! ((i % 5)); then
echo $s
s=""
fi
done
I have a txt file with some lines such as:
a
b
c
f
e
f
1
2
3
4
5
6
now I want to random lines and print it to another txt file for example:
f
6
e
1
and so on...
could any body help me?
I am new in bash scripting
You could use shuf (a part of GNU coreutils).
shuf inputfile > outfile
For example:
$ seq 10 | shuf
7
5
8
3
9
4
10
1
6
2
There is an option for that
sort -R /your/file.txt
Expanation
-R, --random-sort
sort by random hash of keys
Iterate over the file, outputting each line with a certain probability (in this example, with roughly a 10% chance for each line:
while read line; do
if (( RANDOM % 10 == 0 )); then
echo "$line"
fi
done < file.txt
(I say "roughly", because the value of RANDOM ranges between 0 and 32767. As such, there are slightly more values that will produce a remainder of 0-7 than there are that will produce a remainder of 8 or 9 when divided by 10. Other probabilities are have similar problems; you can fine-tune the expression to be more precise, but I leave that as an exercise to the reader.)
For less fortunates systems without GNU utils like BSD/OSX you can use this code:
for ((i=0; i<10; i++)); do
n=$((RANDOM%10))
sed $n'q;d' file
done
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.