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有关随机数的shell例子

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发表于 2005-2-15 20:12:55 | 显示全部楼层 |阅读模式
[code:1]#!/bin/bash

# $RANDOM returns a different random integer at each invocation.
# Nominal range: 0 - 32767 (signed 16-bit integer).

MAXCOUNT=10
count=1

echo
echo "$MAXCOUNT random numbers:"
echo "-----------------"
while [ "$count" -le $MAXCOUNT ]      # Generate 10 ($MAXCOUNT) random integers.
do
  number=$RANDOM
  echo $number
  let "count += 1"  # Increment count.
done
echo "-----------------"

# If you need a random int within a certain range, use the 'modulo' operator.
# This returns the remainder of a division operation.

RANGE=500

echo

number=$RANDOM
let "number %= $RANGE"
echo "Random number less than $RANGE  ---  $number"

echo

# If you need a random int greater than a lower bound,
# then set up a test to discard all numbers below that.

FLOOR=200

number=0   #initialize
while [ "$number" -le $FLOOR ]
do
  number=$RANDOM
done
echo "Random number greater than $FLOOR ---  $number"
echo


# May combine above two techniques to retrieve random number between two limits.
number=0   #initialize
while [ "$number" -le $FLOOR ]
do
  number=$RANDOM
  let "number %= $RANGE"  # Scales $number down within $RANGE.
done
echo "Random number between $FLOOR and $RANGE ---  $number"
echo


# Generate binary choice, that is, "true" or "false" value.
BINARY=2
number=$RANDOM
T=1

let "number %= $BINARY"
#  Note that    let "number >>= 14"    gives a better random distribution
#+ (right shifts out everything except last binary digit).
if [ "$number" -eq $T ]
then
  echo "TRUE"
else
  echo "FALSE"
fi  

echo


# Generate toss of the dice.
SPOTS=6   # Modulo 6 gives range 0 - 5.
          # Incrementing by 1 gives desired range of 1 - 6.
          # Thanks, Paulo Marcel Coelho Aragao, for the simplification.
die1=0
die2=0

# Tosses each die separately, and so gives correct odds.

    let "die1 = $RANDOM % $SPOTS +1" # Roll first one.
    let "die2 = $RANDOM % $SPOTS +1" # Roll second one.

let "throw = $die1 + $die2"
echo "Throw of the dice = $throw"
echo


exit 0[/code:1]
很多地方 都用到的
这是个例子
 楼主| 发表于 2005-2-15 20:24:46 | 显示全部楼层
#!/bin/bash
# random-between.sh
# Random number between two specified values.
# Script by Bill Gradwohl, with minor modifications by the document author.
# Used with permission.


randomBetween() {
   #  Generates a positive or negative random number
   #+ between $min and $max
   #+ and divisible by $divisibleBy.
   #  Gives a "reasonably random" distribution of return values.
   #
   #  Bill Gradwohl - Oct 1, 2003

   syntax() {
   # Function embedded within function.
      echo
      echo    "Syntax: randomBetween [min] [max] [multiple]"
      echo
      echo    "Expects up to 3 passed parameters, but all are completely optional."
      echo    "min is the minimum value"
      echo    "max is the maximum value"
      echo    "multiple specifies that the answer must be a multiple of this value."
      echo    "    i.e. answer must be evenly divisible by this number."
      echo   
      echo    "If any value is missing, defaults area supplied as: 0 32767 1"
      echo    "Successful completion returns 0, unsuccessful completion returns"
      echo    "function syntax and 1."
      echo    "The answer is returned in the global variable randomBetweenAnswer"
      echo    "Negative values for any passed parameter are handled correctly."
   }

   local min=${1:-0}
   local max=${2:-32767}
   local divisibleBy=${3:-1}
   # Default values assigned, in case parameters not passed to function.

   local x
   local spread

   # Let's make sure the divisibleBy value is positive.
   [ ${divisibleBy} -lt 0 ] && divisibleBy=$((0-divisibleBy))

   # Sanity check.
   if [ $# -gt 3 -o ${divisibleBy} -eq 0 -o  ${min} -eq ${max} ]; then
      syntax
      return 1
   fi

   # See if the min and max are reversed.
   if [ ${min} -gt ${max} ]; then
      # Swap them.
      x=${min}
      min=${max}
      max=${x}
   fi

   #  If min is itself not evenly divisible by $divisibleBy,
   #+ then fix the min to be within range.
   if [ $((min/divisibleBy*divisibleBy)) -ne ${min} ]; then
      if [ ${min} -lt 0 ]; then
         min=$((min/divisibleBy*divisibleBy))
      else
         min=$((((min/divisibleBy)+1)*divisibleBy))
      fi
   fi

   #  If max is itself not evenly divisible by $divisibleBy,
   #+ then fix the max to be within range.
   if [ $((max/divisibleBy*divisibleBy)) -ne ${max} ]; then
      if [ ${max} -lt 0 ]; then
         max=$((((max/divisibleBy)-1)*divisibleBy))
      else
         max=$((max/divisibleBy*divisibleBy))
      fi
   fi

   #  ---------------------------------------------------------------------
   #  Now do the real work.

   #  Note that to get a proper distribution for the end points, the
   #+ range of random values has to be allowed to go between 0 and
   #+ abs(max-min)+divisibleBy, not just abs(max-min)+1.

   #  The slight increase will produce the proper distribution for the
   #+ end points.

   #  Changing the formula to use abs(max-min)+1 will still produce
   #+ correct answers, but the randomness of those answers is faulty in
   #+ that the number of times the end points ($min and $max) are returned
   #+ is considerably lower than when the correct formula is used.
   #  ---------------------------------------------------------------------

   spread=$((max-min))
   [ ${spread} -lt 0 ] && spread=$((0-spread))
   let spread+=divisibleBy
   randomBetweenAnswer=$(((RANDOM%spread)/divisibleBy*divisibleBy+min))   

   return 0

   #  However, Paulo Marcel Coelho Aragao points out that
   #+ when $max and $min are not divisible by $divisibleBy,
   #+ the formula fails.
   #
   #  He suggests instead the following formula:
   #    rnumber = $(((RANDOM%(max-min+1)+min)/divisibleBy*divisibleBy))

}

# Let's test the function.
min=-14
max=20
divisibleBy=3


#  Generate an array of expected answers and check to make sure we get
#+ at least one of each answer if we loop long enough.

declare -a answer
minimum=${min}
maximum=${max}
   if [ $((minimum/divisibleBy*divisibleBy)) -ne ${minimum} ]; then
      if [ ${minimum} -lt 0 ]; then
         minimum=$((minimum/divisibleBy*divisibleBy))
      else
         minimum=$((((minimum/divisibleBy)+1)*divisibleBy))
      fi
   fi


   #  If max is itself not evenly divisible by $divisibleBy,
   #+ then fix the max to be within range.

   if [ $((maximum/divisibleBy*divisibleBy)) -ne ${maximum} ]; then
      if [ ${maximum} -lt 0 ]; then
         maximum=$((((maximum/divisibleBy)-1)*divisibleBy))
      else
         maximum=$((maximum/divisibleBy*divisibleBy))
      fi
   fi


#  We need to generate only positive array subscripts,
#+ so we need a displacement that that will guarantee
#+ positive results.

displacement=$((0-minimum))
for ((i=${minimum}; i<=${maximum}; i+=divisibleBy)); do
   answer[i+displacement]=0
done


# Now loop a large number of times to see what we get.
loopIt=1000   #  The script author suggests 100000,
              #+ but that takes a good long while.

for ((i=0; i<${loopIt}; ++i)); do

   #  Note that we are specifying min and max in reversed order here to
   #+ make the function correct for this case.

   randomBetween ${max} ${min} ${divisibleBy}

   # Report an error if an answer is unexpected.
   [ ${randomBetweenAnswer} -lt ${min} -o ${randomBetweenAnswer} -gt ${max} ] && echo MIN or MAX error - ${randomBetweenAnswer}!
   [ $((randomBetweenAnswer%${divisibleBy})) -ne 0 ] && echo DIVISIBLE BY error - ${randomBetweenAnswer}!

   # Store the answer away statistically.
   answer[randomBetweenAnswer+displacement]=$((answer[randomBetweenAnswer+displacement]+1))
done



# Let's check the results

for ((i=${minimum}; i<=${maximum}; i+=divisibleBy)); do
   [ ${answer[i+displacement]} -eq 0 ] && echo "We never got an answer of $i." || echo "${i} occurred ${answer[i+displacement]} times."
done


exit 0
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