Bash:数组是否可以保存另一个数组的名称?

时间:2012-07-09 20:04:22

标签: arrays bash for-loop

我正在编写一个程序并试图分解存储在数组中的数据,以使其运行得更快。 我试图这样做:

    data_to_analyze=(1 2 3 4 5 6 7 8 9 10)
    #original array size
    dataSize=(${#data_to_analyze[@]})

    #half of that size
    let samSmall="$dataSize/2"

    #the other half
    let samSmall2=("$dataSize - $samSmall -1")

    #the first half
    smallArray=("${data_to_analyze[@]:0:$samSmall}")

    #the rest
    smallArray2=("${data_to_analyze[@]:$samSmall:$samSmall2}")

    #an array of names(which correspond to arrays)
    combArray=(smallArray smallArray2)
    sizeComb=(${#combArray[@]})

    #for the length of the new array
    for ((i=0; i<= $sizeComb ; i++)); do
        #through first set of data and then loop back around for the second arrays data? 
        for sample_name in ${combArray[i]}; do
            command
            wait
            command
            wait
        done

我认为这样做首先只给出for循环的第一个数据数组。第一个数组完成后,应该再次使用第二个数组。

这让我有两个问题。 combArray真的通过了两个较小的数组吗?还有更好的方法吗?

2 个答案:

答案 0 :(得分:5)

您可以创建一个看起来像数组引用的字符串,然后使用它来间接访问引用数组的元素。它甚至适用于包含空格的元素!

combArray=(smallArray smallArray2)
for array in "${combArray[@]}"
do
    indirect=$array[@]    # make a string that sort of looks like an array reference
    for element in "${!indirect}"
    do
        echo "Element: $element"
    done
done

答案 1 :(得分:0)

#!/bin/bash
data_to_analyze=(1 2 3 4 5 6 7 8 9 10)
dataSize=${#data_to_analyze[@]}
((samSmall=dataSize/2,samSmall2=dataSize-samSmall))
smallArray=("${data_to_analyze[@]:0:$samSmall}")
smallArray2=("${data_to_analyze[@]:$samSmall:$samSmall2}")
combArray=(smallArray smallArray2)
sizeComb=${#combArray[@]}
for ((i=0;i<$sizeComb;i++));do
    eval 'a=("${'${combArray[i]}'[@]}")'
    for sample_name in "${a[@]}";do
        ...
    done 
done

编辑:删除$ {combArray [i]}附近的双引号并替换&lt; = by&lt; in for