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对数组中给定的元素切片求和（bash）

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对数组中给定的元素切片求和（bash）

好的，这是我的代码，您可能可以了解我想要做的事情的要点。我不太擅长 Linux bash。

#random numbers
MAXCOUNT=100
count=1


while [ "$count" -le $MAXCOUNT ]; do
    number[$count]=$RANDOM
    let "count += 1"
done

#adding function
function sum()
{
    echo $(($1+$2))
}

#main section
first="${count[1-20]}"
echo "The sum of the first 20 elements are: $first"
last="${count[1-20]}"
echo "The sum of the last 20 elements is: $last"

if [ $first -gt $last ]; then
    echo "The sum of the first 20 numbers is greater."
else
    echo "The sum of the last 20 numbers is greater."
fi

我使用这个脚本的目标：

  • 获取并回显包含随机数的数组的前 20 个数字的总和。

  • 获取并回显包含随机数的数组的最后 20 个数字的总和。

  • 回显第一个和是否大于第二个和。

任何帮助都会很棒！请猛击。

答案1

让我们从求和函数开始。我们实际上想让它更通用一些——让它添加所有参数，这样我们就可以摆脱几个循环做类似的事情reduce func array。

# Since you are using bash, let's use declare to make things easier.
# Don't use those evil `function foo` or `function foo()` stuffs -- obsolete bourne thing.
sum(){ declare -i acc; for i; do acc+=i; done; echo $acc; }

剩下的就很容易了。

MAXCOUNT=100 num=()
# Let's use the less evil native 0-based indices.
for ((i=0; i<MAXCOUNT; i++)); do nums+=($RANDOM); done

# https://gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion
# set f to the sum of the 20 elements of nums starting from elem 0
f=$(sum "${nums[@]:0:20}"); echo f20=$f
# set l to the sum of the LAST 20 elems of nums, mind the space
l=$(sum "${nums[@]: -20}"); echo l20=$l
if ((f > l)); then echo f20g; else echo l20g; fi

答案2

unset  num[@] sum; c=-1
while  num[c+=1]=$RANDOM
do     case $c  in      ([2-7]?) ;;
       ($((sum+=num[c])):|99) ! eval '
               printf "$@$sum" "$'"$((sum<$3?2:4))\" greater";;
       (19)    set "The sum of the %-5s 20 elements is:\t%s\n" \
                    first "$sum" last "";  sum=0
       esac||break
done

The sum of the first 20 elements is:    308347
The sum of the last  20 elements is:    306596
The sum of the first 20 elements is:    greater

答案3

另一种可能的解决方案：

#!/bin/bash

        max=100 rng=20   ### Problem conditions

texta="The sum of the %.3sst %d elements is: %d\n"      ### Output
textb="The first sum is %ser than the last sum\n"       ### Output

unset   num                                             ### used vars
for     (( s1=s2=c=0 ;  c<max ; c++ ))
do      num[c]=$RANDOM
        (( c<rng      )) && (( s1+=num[c] ))            ### first sum.
        (( c>=max-rng )) && (( s2+=num[c] ))            ### last sum.
done

    compare=small; (( s1 > s2 )) && compare=bigg

    printf "$texta" "first" "$rng" "$s1"
    printf "$texta" " last" "$rng" "$s2"
    printf "$textb" "$compare"

The sum of the first 20 elements is: 348899
The sum of the  last 20 elements is: 336364
The first sum is bigger than the last sum

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