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Bash: how to avoid duplicate result from random list?

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0

Suppose a script like this

#!/bin/bash

array[0]="123"

array[1]="333"

array[2]="5566"

array[3]="135"

array[4]="68"

array[5]="45"

array[6]="78"

array[7]="32"

array[8]="190"



number()

{

size=${#array[@]}

index=$(($RANDOM % $size))

sleep 0.5s

echo  Lucky number is ... ${array[$index]}

}



for i in {1..21}; do number; done

Is ok..I want to get random number in the interval,but show also a lot of dups

Lucky number is ... 135

Lucky number is ... 135

Lucky number is ... 5566

Lucky number is ... 78

Lucky number is ... 190

Lucky number is ... 333

Lucky number is ... 190

Lucky number is ... 135

Lucky number is ... 5566

Lucky number is ... 333

Lucky number is ... 45

Lucky number is ... 45

Lucky number is ... 68

Lucky number is ... 68

Lucky number is ... 333

Lucky number is ... 78

Lucky number is ... 78

Lucky number is ... 5566

Lucky number is ... 78

Lucky number is ... 333

Lucky number is ... 135

I want a result like this

Lucky number is ... 135

Lucky number is ... 5566

Lucky number is ... 78

Lucky number is ... 190

Lucky number is ... 333

Lucky number is ... 45

Lucky number is ... 68

So we can get the "Lucky number" without dups,unique
Someone know how to do?
Editing the script and reduced 21 to 9 don't change,I get even duplicates.

Lucky number is ... 135

Lucky number is ... 68

Lucky number is ... 45

Lucky number is ... 333

Lucky number is ... 78

Lucky number is ... 135

Lucky number is ... 135

Lucky number is ... 333

Lucky number is ... 32

bash random

share|improve this question

edited 1 hour ago

elbarna

asked 1 hour ago

elbarnaelbarna

4,4261212 gold badges4343 silver badges9696 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  you asked for 21 random numbers from a set of 9 numbers ... I'm not sure what you expected to happen there?
  
  – Jeff Schaller♦
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I have reduced 21 to 9 and is the same thing. Give me a 9 numbers but with some repeat for one or more times
  
  – elbarna
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  do an unset array[index] after you’ve echoed your Lucky number ? or piping your output to the uniq command ?
  
  – LL3
  1 hour ago
  

  
  
  
  

add a comment | 

0

Suppose a script like this

#!/bin/bash

array[0]="123"

array[1]="333"

array[2]="5566"

array[3]="135"

array[4]="68"

array[5]="45"

array[6]="78"

array[7]="32"

array[8]="190"



number()

{

size=${#array[@]}

index=$(($RANDOM % $size))

sleep 0.5s

echo  Lucky number is ... ${array[$index]}

}



for i in {1..21}; do number; done

Is ok..I want to get random number in the interval,but show also a lot of dups

Lucky number is ... 135

Lucky number is ... 135

Lucky number is ... 5566

Lucky number is ... 78

Lucky number is ... 190

Lucky number is ... 333

Lucky number is ... 190

Lucky number is ... 135

Lucky number is ... 5566

Lucky number is ... 333

Lucky number is ... 45

Lucky number is ... 45

Lucky number is ... 68

Lucky number is ... 68

Lucky number is ... 333

Lucky number is ... 78

Lucky number is ... 78

Lucky number is ... 5566

Lucky number is ... 78

Lucky number is ... 333

Lucky number is ... 135

I want a result like this

Lucky number is ... 135

Lucky number is ... 5566

Lucky number is ... 78

Lucky number is ... 190

Lucky number is ... 333

Lucky number is ... 45

Lucky number is ... 68

So we can get the "Lucky number" without dups,unique
Someone know how to do?
Editing the script and reduced 21 to 9 don't change,I get even duplicates.

Lucky number is ... 135

Lucky number is ... 68

Lucky number is ... 45

Lucky number is ... 333

Lucky number is ... 78

Lucky number is ... 135

Lucky number is ... 135

Lucky number is ... 333

Lucky number is ... 32

bash random

share|improve this question

edited 1 hour ago

elbarna

asked 1 hour ago

elbarnaelbarna

4,4261212 gold badges4343 silver badges9696 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  you asked for 21 random numbers from a set of 9 numbers ... I'm not sure what you expected to happen there?
  
  – Jeff Schaller♦
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I have reduced 21 to 9 and is the same thing. Give me a 9 numbers but with some repeat for one or more times
  
  – elbarna
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  do an unset array[index] after you’ve echoed your Lucky number ? or piping your output to the uniq command ?
  
  – LL3
  1 hour ago
  

  
  
  
  

add a comment | 

Suppose a script like this

#!/bin/bash

array[0]="123"

array[1]="333"

array[2]="5566"

array[3]="135"

array[4]="68"

array[5]="45"

array[6]="78"

array[7]="32"

array[8]="190"



number()

{

size=${#array[@]}

index=$(($RANDOM % $size))

sleep 0.5s

echo  Lucky number is ... ${array[$index]}

}



for i in {1..21}; do number; done

Is ok..I want to get random number in the interval,but show also a lot of dups

Lucky number is ... 135

Lucky number is ... 135

Lucky number is ... 5566

Lucky number is ... 78

Lucky number is ... 190

Lucky number is ... 333

Lucky number is ... 190

Lucky number is ... 135

Lucky number is ... 5566

Lucky number is ... 333

Lucky number is ... 45

Lucky number is ... 45

Lucky number is ... 68

Lucky number is ... 68

Lucky number is ... 333

Lucky number is ... 78

Lucky number is ... 78

Lucky number is ... 5566

Lucky number is ... 78

Lucky number is ... 333

Lucky number is ... 135

I want a result like this

Lucky number is ... 135

Lucky number is ... 5566

Lucky number is ... 78

Lucky number is ... 190

Lucky number is ... 333

Lucky number is ... 45

Lucky number is ... 68

So we can get the "Lucky number" without dups,unique
Someone know how to do?
Editing the script and reduced 21 to 9 don't change,I get even duplicates.

Lucky number is ... 135

Lucky number is ... 68

Lucky number is ... 45

Lucky number is ... 333

Lucky number is ... 78

Lucky number is ... 135

Lucky number is ... 135

Lucky number is ... 333

Lucky number is ... 32

bash random

share|improve this question

edited 1 hour ago

elbarna

asked 1 hour ago

elbarnaelbarna

4,4261212 gold badges4343 silver badges9696 bronze badges

#!/bin/bash

array[0]="123"

array[1]="333"

array[2]="5566"

array[3]="135"

array[4]="68"

array[5]="45"

array[6]="78"

array[7]="32"

array[8]="190"



number()

{

size=${#array[@]}

index=$(($RANDOM % $size))

sleep 0.5s

echo  Lucky number is ... ${array[$index]}

}



for i in {1..21}; do number; done

Lucky number is ... 135

Lucky number is ... 135

Lucky number is ... 5566

Lucky number is ... 78

Lucky number is ... 190

Lucky number is ... 333

Lucky number is ... 190

Lucky number is ... 135

Lucky number is ... 5566

Lucky number is ... 333

Lucky number is ... 45

Lucky number is ... 45

Lucky number is ... 68

Lucky number is ... 68

Lucky number is ... 333

Lucky number is ... 78

Lucky number is ... 78

Lucky number is ... 5566

Lucky number is ... 78

Lucky number is ... 333

Lucky number is ... 135

Lucky number is ... 135

Lucky number is ... 5566

Lucky number is ... 78

Lucky number is ... 190

Lucky number is ... 333

Lucky number is ... 45

Lucky number is ... 68

Lucky number is ... 135

Lucky number is ... 68

Lucky number is ... 45

Lucky number is ... 333

Lucky number is ... 78

Lucky number is ... 135

Lucky number is ... 135

Lucky number is ... 333

Lucky number is ... 32

share|improve this question

edited 1 hour ago

elbarna

asked 1 hour ago

elbarnaelbarna

4,4261212 gold badges4343 silver badges9696 bronze badges

share|improve this question

edited 1 hour ago

elbarna

asked 1 hour ago

elbarnaelbarna

4,4261212 gold badges4343 silver badges9696 bronze badges

asked 1 hour ago

elbarnaelbarna

4,4261212 gold badges4343 silver badges9696 bronze badges

asked 1 hour ago

elbarnaelbarna

4,4261212 gold badges4343 silver badges9696 bronze badges

1 Answer
1

active

oldest

votes

2

Treat the list like a deck of cards. Shuffle the numbers, then read them out one at a time.

See Simple method to shuffle the elements of an array in BASH shell?.

#!/bin/bash

array=("123" "333" "5566" "135" "68" "45" "78" "32" "190")



shuffle() {

    array=($(shuf -e "${array[@]}"))

    index=0

}



number() {

   size=${#array[@]}

   sleep 0.5s



   if (( index >= size )) ; then

      shuffle

   fi



   echo Lucky number is ... ${array[$index]}

   index=$((index + 1))

}



for i in {1..21}; do number; done

share|improve this answer

edited 1 hour ago

answered 1 hour ago

xiotaxiota

12588 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @steeldriver Advantage of shuffling the indices is the original array is left alone. But you'd still have to create an array of indices and track position within it to prevent duplicates. Otherwise, the shuf command may return the same index multiple times.
  
  – xiota
  1 hour ago
  

  
  
  
  

add a comment | 

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2

Treat the list like a deck of cards. Shuffle the numbers, then read them out one at a time.

See Simple method to shuffle the elements of an array in BASH shell?.

#!/bin/bash

array=("123" "333" "5566" "135" "68" "45" "78" "32" "190")



shuffle() {

    array=($(shuf -e "${array[@]}"))

    index=0

}



number() {

   size=${#array[@]}

   sleep 0.5s



   if (( index >= size )) ; then

      shuffle

   fi



   echo Lucky number is ... ${array[$index]}

   index=$((index + 1))

}



for i in {1..21}; do number; done

share|improve this answer

edited 1 hour ago

answered 1 hour ago

xiotaxiota

12588 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @steeldriver Advantage of shuffling the indices is the original array is left alone. But you'd still have to create an array of indices and track position within it to prevent duplicates. Otherwise, the shuf command may return the same index multiple times.
  
  – xiota
  1 hour ago
  

  
  
  
  

add a comment | 

2

Treat the list like a deck of cards. Shuffle the numbers, then read them out one at a time.

See Simple method to shuffle the elements of an array in BASH shell?.

#!/bin/bash

array=("123" "333" "5566" "135" "68" "45" "78" "32" "190")



shuffle() {

    array=($(shuf -e "${array[@]}"))

    index=0

}



number() {

   size=${#array[@]}

   sleep 0.5s



   if (( index >= size )) ; then

      shuffle

   fi



   echo Lucky number is ... ${array[$index]}

   index=$((index + 1))

}



for i in {1..21}; do number; done

share|improve this answer

edited 1 hour ago

answered 1 hour ago

xiotaxiota

12588 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @steeldriver Advantage of shuffling the indices is the original array is left alone. But you'd still have to create an array of indices and track position within it to prevent duplicates. Otherwise, the shuf command may return the same index multiple times.
  
  – xiota
  1 hour ago
  

  
  
  
  

add a comment | 

Treat the list like a deck of cards. Shuffle the numbers, then read them out one at a time.

See Simple method to shuffle the elements of an array in BASH shell?.

#!/bin/bash

array=("123" "333" "5566" "135" "68" "45" "78" "32" "190")



shuffle() {

    array=($(shuf -e "${array[@]}"))

    index=0

}



number() {

   size=${#array[@]}

   sleep 0.5s



   if (( index >= size )) ; then

      shuffle

   fi



   echo Lucky number is ... ${array[$index]}

   index=$((index + 1))

}



for i in {1..21}; do number; done

share|improve this answer

edited 1 hour ago

answered 1 hour ago

xiotaxiota

12588 bronze badges

#!/bin/bash

array=("123" "333" "5566" "135" "68" "45" "78" "32" "190")



shuffle() {

    array=($(shuf -e "${array[@]}"))

    index=0

}



number() {

   size=${#array[@]}

   sleep 0.5s



   if (( index >= size )) ; then

      shuffle

   fi



   echo Lucky number is ... ${array[$index]}

   index=$((index + 1))

}



for i in {1..21}; do number; done

share|improve this answer

edited 1 hour ago

answered 1 hour ago

xiotaxiota

12588 bronze badges

answered 1 hour ago

xiotaxiota

12588 bronze badges

answered 1 hour ago

xiotaxiota

12588 bronze badges