How to get "random" number based on a string in Bash?

Question

I know there are plenty of random number generators out there, but I am looking for something that may be a little more predictable. Is there a way to get a "random" number (one that would be the same in every instance) given a string? I would like to do this in bash. I am looking for something a little more advanced than the count of characters in the string, but not as advanced as a full checksum of it.

The end goal is to get a decimal value, so this could be ran against a string multiple times repeating the result.

Why not use a hash function like md5sum or sha1sum? echo -n "Hello World" | md5sum. — Kerrek SB
– Kerrek SB, Commented Sep 1, 2011 at 0:28
I agree, a hash is probably easiest; I use one for a similar purpose myself. — Kevin
– Kevin, Commented Sep 1, 2011 at 0:35
You said "not as advanced as a full checksum". Why? sha1sum is easy enough to use. — Keith Thompson
– Keith Thompson, Commented Sep 1, 2011 at 0:45
So a checksum is the best way to do this? I was hoping to get a decimal value, so couldn't the checksum value get turned into a decimal? — Kyle Hotchkiss
– Kyle Hotchkiss, Commented Sep 1, 2011 at 0:56

Mu Qiao · Accepted Answer · 2011-09-01 00:58:14Z

7

You need a random number, but you don't want a full checksum, it's contradiction. I think md5sum and sha1sum is really easy to use and should fit your needs:

md5sum <<< "$your_str"

or

sha1sum <<< "$your_str"

Update:

If you need decimal numbers, just:

n=$(md5sum <<< "$your_str")
echo $((0x${n%% *}))

answered Sep 1, 2011 at 0:58

Mu Qiao

7,1271 gold badge34 silver badges35 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Kyle Hotchkiss Over a year ago

Just what I'm looking for. Thanks Mu.

gene_wood Over a year ago

md5sum <<< "$your_str" uses the <<< herestring bash functionality. ${n%% *} extracts the first of the 2 fields returned by md5sum, which is the md5 hash. It does this by deleting the longest match of the substring " *" from the end of the variable. $((0x________)) converts the hexadecimal md5sum from hex (base 16) to decimal (base 10)

Ivan Ogai · Accepted Answer · 2016-12-14 19:30:02Z

7

To hash the string abc to 3 decimals:

echo $((0x$(echo abc | md5sum | cut -f 1 -d " " | cut -c 1-3))) | cut -c 1-3

To get e.g. 4 decimals instead of 3, replace the two occurrences of 3 by 4:

echo $((0x$(echo abc | md5sum | cut -f 1 -d " " | cut -c 1-4))) | cut -c 1-4

$(echo abc | md5sum | cut -f 1 -d " " | cut -c 1-4) gives an hexadecimal of 4 digits.
$((0xHEX)) converts an hexadecimal (HEX) to decimal, which may have more than 4 digits even when HEX has 4 digits, e.g. ffff is 65535.
cut -c 1-4 gets the first 4 digits.

answered Dec 14, 2016 at 19:30

Ivan Ogai

1,51615 silver badges10 bronze badges

Comments

Yasunori Kuji · Accepted Answer · 2016-10-27 06:43:19Z

2

$ echo abc | md5sum | grep -Eo "[[:digit:]]{3}"  | head -n1
Result -> 248
$ echo xyz | md5sum | grep -Eo "[[:digit:]]{3}"  | head -n1
Result -> 627

Combined other answers!

For HEX colors

echo abc | sha1sum | grep -Eo "[a-f0-9]{6}"  | head -n1
03cfd7
echo abc | sha256sum | grep -Eo "[a-f0-9]{6}"  | head -n1
edeaaf
echo abc | sha512sum | grep -Eo "[a-f0-9]{6}"  | head -n1
4f285d

edited Oct 27, 2016 at 6:43

answered Sep 8, 2016 at 3:04

Yasunori Kuji

6,3071 gold badge31 silver badges39 bronze badges

1 Comment

Ivan Ogai Over a year ago

The first solution for 3 digits fails in (rare) cases where the hash has no 3 consecutive decimals. See my solution for an approach that always returns 3 decimals.

Collectives™ on Stack Overflow

How to get "random" number based on a string in Bash?

3 Answers 3

2 Comments

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related