如何在 Bash 中对数组进行切片

How to slice an array in Bash

提问by Chen Levy

Looking the "Array" section in the bash(1) man page, I didn't find a way to slice an array.

查看 bash(1) 手册页中的“数组”部分，我没有找到对数组进行切片的方法。

So I came up with this overly complicated function:

所以我想出了这个过于复杂的函数：

#!/bin/bash

# @brief: slice a bash array
# @arg1:  output-name
# @arg2:  input-name
# @args:  seq args
# ----------------------------------------------
function slice() {
   local output=
   local input=
   shift 2
   local indexes=$(seq $*)

   local -i i
   local tmp=$(for i in $indexes 
                 do echo "$(eval echo \"${$input[$i]}\")" 
               done)

   local IFS=$'\n'
   eval $output="( $tmp )"
}

Used like this:

像这样使用：

$ A=( foo bar "a  b c" 42 )
$ slice B A 1 2
$ echo "${B[0]}"  # bar
$ echo "${B[1]}"  # a  b c

Is there a better way to do this?

有一个更好的方法吗？

回答by Paused until further notice.

See the Parameter Expansionsection in the Bash manpage. A[@]returns the contents of the array, :1:2takes a slice of length 2, starting at index 1.

请参阅Bash页面中的参数扩展部分man。A[@]返回数组的内容，:1:2取一个长度为 2 的切片，从索引 1 开始。

A=( foo bar "a  b c" 42 )
B=("${A[@]:1:2}")
C=("${A[@]:1}")       # slice to the end of the array
echo "${B[@]}"        # bar a  b c
echo "${B[1]}"        # a  b c
echo "${C[@]}"        # bar a  b c 42
echo "${C[@]: -2:2}"  # a  b c 42 # The space before the - is necesssary

Note that the fact that "a b c" is one array element (and that it contains an extra space) is preserved.

请注意，“ab c”是一个数组元素（并且它包含一个额外的空格）这一事实被保留。

回答by Nicholas Sushkin

There is also a convenient shortcut to get all elements of the array starting with specified index. For example "${A[@]:1}" would be the "tail" of the array, that is the array without its first element.

还有一个方便的快捷方式来获取从指定索引开始的数组的所有元素。例如，“${A[@]:1}”将是数组的“尾部”，即没有第一个元素的数组。

version=4.7.1
A=( ${version//\./ } )
echo "${A[@]}"    # 4 7 1
B=( "${A[@]:1}" )
echo "${B[@]}"    # 7 1

回答by jandob

Array slicing like in Python (From the rebashlibrary):

像 Python 中的数组切片（来自rebash库）：

array_slice() {
    local __doc__='
    Returns a slice of an array (similar to Python).

    From the Python documentation:
    One way to remember how slices work is to think of the indices as pointing
    between elements, with the left edge of the first character numbered 0.
    Then the right edge of the last element of an array of length n has
    index n, for example:
    ```
    +---+---+---+---+---+---+
    | 0 | 1 | 2 | 3 | 4 | 5 |
    +---+---+---+---+---+---+
    0   1   2   3   4   5   6
    -6  -5  -4  -3  -2  -1
    ```

    >>> local a=(0 1 2 3 4 5)
    >>> echo $(array.slice 1:-2 "${a[@]}")
    1 2 3
    >>> local a=(0 1 2 3 4 5)
    >>> echo $(array.slice 0:1 "${a[@]}")
    0
    >>> local a=(0 1 2 3 4 5)
    >>> [ -z "$(array.slice 1:1 "${a[@]}")" ] && echo empty
    empty
    >>> local a=(0 1 2 3 4 5)
    >>> [ -z "$(array.slice 2:1 "${a[@]}")" ] && echo empty
    empty
    >>> local a=(0 1 2 3 4 5)
    >>> [ -z "$(array.slice -2:-3 "${a[@]}")" ] && echo empty
    empty
    >>> [ -z "$(array.slice -2:-2 "${a[@]}")" ] && echo empty
    empty

    Slice indices have useful defaults; an omitted first index defaults to
    zero, an omitted second index defaults to the size of the string being
    sliced.
    >>> local a=(0 1 2 3 4 5)
    >>> # from the beginning to position 2 (excluded)
    >>> echo $(array.slice 0:2 "${a[@]}")
    >>> echo $(array.slice :2 "${a[@]}")
    0 1
    0 1

    >>> local a=(0 1 2 3 4 5)
    >>> # from position 3 (included) to the end
    >>> echo $(array.slice 3:"${#a[@]}" "${a[@]}")
    >>> echo $(array.slice 3: "${a[@]}")
    3 4 5
    3 4 5

    >>> local a=(0 1 2 3 4 5)
    >>> # from the second-last (included) to the end
    >>> echo $(array.slice -2:"${#a[@]}" "${a[@]}")
    >>> echo $(array.slice -2: "${a[@]}")
    4 5
    4 5

    >>> local a=(0 1 2 3 4 5)
    >>> echo $(array.slice -4:-2 "${a[@]}")
    2 3

    If no range is given, it works like normal array indices.
    >>> local a=(0 1 2 3 4 5)
    >>> echo $(array.slice -1 "${a[@]}")
    5
    >>> local a=(0 1 2 3 4 5)
    >>> echo $(array.slice -2 "${a[@]}")
    4
    >>> local a=(0 1 2 3 4 5)
    >>> echo $(array.slice 0 "${a[@]}")
    0
    >>> local a=(0 1 2 3 4 5)
    >>> echo $(array.slice 1 "${a[@]}")
    1
    >>> local a=(0 1 2 3 4 5)
    >>> array.slice 6 "${a[@]}"; echo $?
    1
    >>> local a=(0 1 2 3 4 5)
    >>> array.slice -7 "${a[@]}"; echo $?
    1
    '
    local start end array_length length
    if [[  == *:* ]]; then
        IFS=":"; read -r start end <<<""
        shift
        array_length="$#"
        # defaults
        [ -z "$end" ] && end=$array_length
        [ -z "$start" ] && start=0
        (( start < 0 )) && let "start=(( array_length + start ))"
        (( end < 0 )) && let "end=(( array_length + end ))"
    else
        start=""
        shift
        array_length="$#"
        (( start < 0 )) && let "start=(( array_length + start ))"
        let "end=(( start + 1 ))"
    fi
    let "length=(( end - start ))"
    (( start < 0 )) && return 1
    # check bounds
    (( length < 0 )) && return 1
    (( start < 0 )) && return 1
    (( start >= array_length )) && return 1
    # parameters start with , so add 1 to $start
    let "start=(( start + 1 ))"
    echo "${@: $start:$length}"
}
alias array.slice="array_slice"

回答by Arindam Roychowdhury

Lets say I am reading an array from user, then I want to see element 3 to 7 both inclusive .

假设我正在从用户读取一个数组，然后我想看到元素 3 到 7 都包括在内。

cnt=0
while read var;
    do
    myarr[cnt]=$var
    cnt=$((cnt+1)) 
    done


echo ${myarr[@]:3:5}