How do I specify chance when setting a variable to random item in array?
Currently I do something like this to set a variable to a random item in an array:
array=("foo" "bar" "baz")
var=${array[$RANDOM % ${#array[@]} ]}
How would I do if I want to set $var to one of these values, but specify the chance that the variable will be set to the value of each item respectively? Say I want a 73,3% chance for foo, 26,6% chance for bar, and 0,1% chance for baz.
command-line variable array random
edited yesterday
terdon♦
130k32254432
asked yesterday
DisplayNameDisplayName
4,48094580
add a comment |
Currently I do something like this to set a variable to a random item in an array:
array=("foo" "bar" "baz")
var=${array[$RANDOM % ${#array[@]} ]}
How would I do if I want to set $var to one of these values, but specify the chance that the variable will be set to the value of each item respectively? Say I want a 73,3% chance for foo, 26,6% chance for bar, and 0,1% chance for baz.
command-line variable array random
edited yesterday
terdon♦
130k32254432
asked yesterday
DisplayNameDisplayName
4,48094580
Are you allowed to use only bash builtins and no external programs?
– Mark Plotnick
yesterday
@MarkPlotnick I would prefer bash, but anything that I can run from the command line works
– DisplayName
yesterday
add a comment |
Currently I do something like this to set a variable to a random item in an array:
array=("foo" "bar" "baz")
var=${array[$RANDOM % ${#array[@]} ]}
How would I do if I want to set $var to one of these values, but specify the chance that the variable will be set to the value of each item respectively? Say I want a 73,3% chance for foo, 26,6% chance for bar, and 0,1% chance for baz.
command-line variable array random
edited yesterday
terdon♦
130k32254432
asked yesterday
DisplayNameDisplayName
4,48094580
Currently I do something like this to set a variable to a random item in an array:
array=("foo" "bar" "baz")
var=${array[$RANDOM % ${#array[@]} ]}
How would I do if I want to set $var to one of these values, but specify the chance that the variable will be set to the value of each item respectively? Say I want a 73,3% chance for foo, 26,6% chance for bar, and 0,1% chance for baz.
command-line variable array random
command-line variable array random
edited yesterday
terdon♦
130k32254432
asked yesterday
DisplayNameDisplayName
4,48094580
edited yesterday
terdon♦
130k32254432
asked yesterday
DisplayNameDisplayName
4,48094580
edited yesterday
terdon♦
130k32254432
edited yesterday
terdon♦
130k32254432
edited yesterday
terdon♦
130k32254432
130k32254432
asked yesterday
DisplayNameDisplayName
4,48094580
asked yesterday
DisplayNameDisplayName
4,48094580
asked yesterday
DisplayNameDisplayName
4,48094580
4,48094580
Are you allowed to use only bash builtins and no external programs?
– Mark Plotnick
yesterday
@MarkPlotnick I would prefer bash, but anything that I can run from the command line works
– DisplayName
yesterday
add a comment |
Are you allowed to use only bash builtins and no external programs?
– Mark Plotnick
yesterday
@MarkPlotnick I would prefer bash, but anything that I can run from the command line works
– DisplayName
yesterday
Are you allowed to use only bash builtins and no external programs?
– Mark Plotnick
yesterday
Are you allowed to use only bash builtins and no external programs?
– Mark Plotnick
yesterday
@MarkPlotnick I would prefer bash, but anything that I can run from the command line works
– DisplayName
yesterday
@MarkPlotnick I would prefer bash, but anything that I can run from the command line works
– DisplayName
yesterday
add a comment |
2 Answers
2
active
oldest
votes
One way: set up a parallel array with the corresponding percentage chances; below, I've scaled them to 1000. Then, choose a random number between 1 and 1000 and iterate through the array until you've run out of chances:
#!/bin/bash
array=( "foo" "bar" "baz")
chances=(733 266 1)
choice=$((1 + (RANDOM % 1000)))
value=
for((index=0; index < ${#array[@]}; index++))
do
choice=$((choice - ${chances[index]}))
if [[ choice -le 0 ]]
then
value=${array[index]}
break
fi
done
[[ index -eq ${#array[@]} ]] && value=${array[index]}
printf '%sn' "$value"
answered yesterday
Jeff SchallerJeff Schaller
40.9k1056130
I've also used this method before. It works well. However: it might be worth to calculate the sum dynamically. If it's not actually 1000, you get the wrong results. Also, if RANDOM is limited to 32768, there might be some random bias in the selection.
– frostschutz
yesterday
Great points, frostschutz; I didn't do any error-checking of the sums, since I assumed they were all assigned manually.
– Jeff Schaller
yesterday
add a comment |
The shell can't do floating point math, but if we just move the decimal point, we can use $RANDOM and integer math:
#!/usr/local/bin/bash
array=("foo" "bar" "baz")
dieroll=$(($RANDOM % 1000))
if [[ "$dieroll" -lt 1 ]]; then
printf "%sn" "${array[2]}"
elif [[ "$dieroll" -lt 266 ]]; then
printf "%sn" "${array[1]}"
else
printf "%sn" "${array[0]}"
fi
This has the advantages of not having to blow the array up to 1000 entries or needing any for loops.
answered yesterday
DopeGhotiDopeGhoti
45.2k55988
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
One way: set up a parallel array with the corresponding percentage chances; below, I've scaled them to 1000. Then, choose a random number between 1 and 1000 and iterate through the array until you've run out of chances:
#!/bin/bash
array=( "foo" "bar" "baz")
chances=(733 266 1)
choice=$((1 + (RANDOM % 1000)))
value=
for((index=0; index < ${#array[@]}; index++))
do
choice=$((choice - ${chances[index]}))
if [[ choice -le 0 ]]
then
value=${array[index]}
break
fi
done
[[ index -eq ${#array[@]} ]] && value=${array[index]}
printf '%sn' "$value"
answered yesterday
Jeff SchallerJeff Schaller
40.9k1056130
I've also used this method before. It works well. However: it might be worth to calculate the sum dynamically. If it's not actually 1000, you get the wrong results. Also, if RANDOM is limited to 32768, there might be some random bias in the selection.
– frostschutz
yesterday
Great points, frostschutz; I didn't do any error-checking of the sums, since I assumed they were all assigned manually.
– Jeff Schaller
yesterday
add a comment |
One way: set up a parallel array with the corresponding percentage chances; below, I've scaled them to 1000. Then, choose a random number between 1 and 1000 and iterate through the array until you've run out of chances:
#!/bin/bash
array=( "foo" "bar" "baz")
chances=(733 266 1)
choice=$((1 + (RANDOM % 1000)))
value=
for((index=0; index < ${#array[@]}; index++))
do
choice=$((choice - ${chances[index]}))
if [[ choice -le 0 ]]
then
value=${array[index]}
break
fi
done
[[ index -eq ${#array[@]} ]] && value=${array[index]}
printf '%sn' "$value"
answered yesterday
Jeff SchallerJeff Schaller
40.9k1056130
I've also used this method before. It works well. However: it might be worth to calculate the sum dynamically. If it's not actually 1000, you get the wrong results. Also, if RANDOM is limited to 32768, there might be some random bias in the selection.
– frostschutz
yesterday
Great points, frostschutz; I didn't do any error-checking of the sums, since I assumed they were all assigned manually.
– Jeff Schaller
yesterday
add a comment |
One way: set up a parallel array with the corresponding percentage chances; below, I've scaled them to 1000. Then, choose a random number between 1 and 1000 and iterate through the array until you've run out of chances:
#!/bin/bash
array=( "foo" "bar" "baz")
chances=(733 266 1)
choice=$((1 + (RANDOM % 1000)))
value=
for((index=0; index < ${#array[@]}; index++))
do
choice=$((choice - ${chances[index]}))
if [[ choice -le 0 ]]
then
value=${array[index]}
break
fi
done
[[ index -eq ${#array[@]} ]] && value=${array[index]}
printf '%sn' "$value"
answered yesterday
Jeff SchallerJeff Schaller
40.9k1056130
One way: set up a parallel array with the corresponding percentage chances; below, I've scaled them to 1000. Then, choose a random number between 1 and 1000 and iterate through the array until you've run out of chances:
#!/bin/bash
array=( "foo" "bar" "baz")
chances=(733 266 1)
choice=$((1 + (RANDOM % 1000)))
value=
for((index=0; index < ${#array[@]}; index++))
do
choice=$((choice - ${chances[index]}))
if [[ choice -le 0 ]]
then
value=${array[index]}
break
fi
done
[[ index -eq ${#array[@]} ]] && value=${array[index]}
printf '%sn' "$value"
answered yesterday
Jeff SchallerJeff Schaller
40.9k1056130
answered yesterday
Jeff SchallerJeff Schaller
40.9k1056130
answered yesterday
Jeff SchallerJeff Schaller
40.9k1056130
answered yesterday
Jeff SchallerJeff Schaller
40.9k1056130
40.9k1056130
I've also used this method before. It works well. However: it might be worth to calculate the sum dynamically. If it's not actually 1000, you get the wrong results. Also, if RANDOM is limited to 32768, there might be some random bias in the selection.
– frostschutz
yesterday
Great points, frostschutz; I didn't do any error-checking of the sums, since I assumed they were all assigned manually.
– Jeff Schaller
yesterday
add a comment |
I've also used this method before. It works well. However: it might be worth to calculate the sum dynamically. If it's not actually 1000, you get the wrong results. Also, if RANDOM is limited to 32768, there might be some random bias in the selection.
– frostschutz
yesterday
Great points, frostschutz; I didn't do any error-checking of the sums, since I assumed they were all assigned manually.
– Jeff Schaller
yesterday
I've also used this method before. It works well. However: it might be worth to calculate the sum dynamically. If it's not actually 1000, you get the wrong results. Also, if RANDOM is limited to 32768, there might be some random bias in the selection.
– frostschutz
yesterday
I've also used this method before. It works well. However: it might be worth to calculate the sum dynamically. If it's not actually 1000, you get the wrong results. Also, if RANDOM is limited to 32768, there might be some random bias in the selection.
– frostschutz
yesterday
Great points, frostschutz; I didn't do any error-checking of the sums, since I assumed they were all assigned manually.
– Jeff Schaller
yesterday
Great points, frostschutz; I didn't do any error-checking of the sums, since I assumed they were all assigned manually.
– Jeff Schaller
yesterday
add a comment |
The shell can't do floating point math, but if we just move the decimal point, we can use $RANDOM and integer math:
#!/usr/local/bin/bash
array=("foo" "bar" "baz")
dieroll=$(($RANDOM % 1000))
if [[ "$dieroll" -lt 1 ]]; then
printf "%sn" "${array[2]}"
elif [[ "$dieroll" -lt 266 ]]; then
printf "%sn" "${array[1]}"
else
printf "%sn" "${array[0]}"
fi
This has the advantages of not having to blow the array up to 1000 entries or needing any for loops.
answered yesterday
DopeGhotiDopeGhoti
45.2k55988
add a comment |
The shell can't do floating point math, but if we just move the decimal point, we can use $RANDOM and integer math:
#!/usr/local/bin/bash
array=("foo" "bar" "baz")
dieroll=$(($RANDOM % 1000))
if [[ "$dieroll" -lt 1 ]]; then
printf "%sn" "${array[2]}"
elif [[ "$dieroll" -lt 266 ]]; then
printf "%sn" "${array[1]}"
else
printf "%sn" "${array[0]}"
fi
This has the advantages of not having to blow the array up to 1000 entries or needing any for loops.
answered yesterday
DopeGhotiDopeGhoti
45.2k55988
add a comment |
The shell can't do floating point math, but if we just move the decimal point, we can use $RANDOM and integer math:
#!/usr/local/bin/bash
array=("foo" "bar" "baz")
dieroll=$(($RANDOM % 1000))
if [[ "$dieroll" -lt 1 ]]; then
printf "%sn" "${array[2]}"
elif [[ "$dieroll" -lt 266 ]]; then
printf "%sn" "${array[1]}"
else
printf "%sn" "${array[0]}"
fi
This has the advantages of not having to blow the array up to 1000 entries or needing any for loops.
answered yesterday
DopeGhotiDopeGhoti
45.2k55988
The shell can't do floating point math, but if we just move the decimal point, we can use $RANDOM and integer math:
#!/usr/local/bin/bash
array=("foo" "bar" "baz")
dieroll=$(($RANDOM % 1000))
if [[ "$dieroll" -lt 1 ]]; then
printf "%sn" "${array[2]}"
elif [[ "$dieroll" -lt 266 ]]; then
printf "%sn" "${array[1]}"
else
printf "%sn" "${array[0]}"
fi
This has the advantages of not having to blow the array up to 1000 entries or needing any for loops.
answered yesterday
DopeGhotiDopeGhoti
45.2k55988
edited yesterday
answered yesterday
DopeGhotiDopeGhoti
45.2k55988
answered yesterday
DopeGhotiDopeGhoti
45.2k55988
answered yesterday
DopeGhotiDopeGhoti
45.2k55988
45.2k55988
add a comment |
add a comment |
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Are you allowed to use only bash builtins and no external programs?
– Mark Plotnick
yesterday
@MarkPlotnick I would prefer bash, but anything that I can run from the command line works
– DisplayName
yesterday