缩进线(树)到类似路径的线 [英] Indented lines (tree) to path-like lines

问题描述

我有如下结构的输入文件:

I have input files with the structure like the next:

a1
  b1
    c1
    c2
    c3
  b2
    c1
      d1
      d2
  b3
  b4
a2
a3
  b1
  b2
    c1
    c2

每一级缩进2个空格.所需的输出是:

Each level is indented by 2 spaces. The needed output is:

a1/b1/c1
a1/b1/c2
a1/b1/c3
a1/b2/c1/d1
a1/b2/c1/d2
a1/b3
a1/b4
a2
a3/b1
a3/b2/c1
a3/b2/c2

它就像一个文件系统，如果下一行有更大的缩进，当前的就像一个目录"，当有相同的缩进时它就像一个文件".需要打印文件"的完整路径.

It is like a filesystem, if the next line have bigger indentation, the current one is like a "directory" and when have same indentation it is like a "file". Need print full paths of "files".

试图在没有任何高级语言的情况下解决这个问题，例如 python、perl - 仅使用基本的 bash 命令.

Trying to solve this without any high-level language, like python, perl - with only basic bash commands.

我当前的代码/想法基于递归函数调用并使用堆栈，但在逻辑"方面存在问题.代码当前输出下一个:

My current code/idea is based on recursive function call and working with a stack, but have problem with the "logic". The code currently outputs the next:

a1 b1 c1
a1 b1
a1

DD: line 8: [0-1]: bad array subscript

只有第一行是可以的 - 所以处理递归是错误的......

only the 1st line is OK - so handling the recursion is wrong...

input="ifile.tree"

#stack array
declare -a stack

#stack manipulation
pushstack() { stack+=("$1"); }
popstack() { unset stack[${#stack[@]}-1]; }
printstack() { echo "${stack[*]}"; }

#recursive function
checkline() {
    local uplev=$1

    #read line - if no more lines - print the stack and return
    read -r level text || (printstack; exit 1) || return

    #if the current line level is largest than previous level
    if [[ $uplev < $level ]]
    then
        pushstack "$text"
        checkline $level    #recurse
    fi

    printstack
    popstack
}

# MAIN PROGRAM

# change the input from indented spaces to
# level_number<space>text
(
    #subshell - change IFS
    IFS=,
    while read -r spaces content
    do
        echo $(( (${#spaces} / 2) + 1 )) "$content"
    done < <(sed 's/[^ ]/,&/' < "$input")

) | (   #pipe to another subshell
    checkline 0 #recurse by levels
)

Sry 长代码 - 有人可以帮忙吗?

Sry for the long code - can anybody help?

推荐答案

有趣的问题.

这个 awk(可能是单行)命令可以完成这项工作:

this awk (could be one-liner) command does the job:

awk -F'  ' 'NF<=p{for(i=1;i<=p;i++)printf "%s%s", a[i],(i==p?RS:"/")
            if(NF<p)for(i=NF;i<=p;i++) delete a[i]}
            {a[NF] =$NF;p=NF }
            END{for(i=1;i<=NF;i++)printf "%s%s", a[i],(i==NF?RS:"/")}' file

你可以看到上面有重复的代码，你可以把它们提取成一个函数.

you can see above, there are duplicated codes, you can extract them into a function if you like.

用您的数据进行测试:

kent$  cat f
a1
  b1
    c1
    c2
    c3
  b2
    c1
      d1
      d2
  b3
  b4
a2
a3
  b1
  b2
    c1
    c2

kent$  awk -F'  ' 'NF<=p{for(i=1;i<=p;i++)printf "%s%s", a[i],(i==p?RS:"/")
if(NF<p)for(i=NF;i<=p;i++) delete a[i]}
{a[NF] =$NF;p=NF }END{for(i=1;i<=NF;i++)printf "%s%s", a[i],(i==NF?RS:"/")} ' f
a1/b1/c1
a1/b1/c2
a1/b1/c3
a1/b2/c1/d1
a1/b2/c1/d2
a1/b3
a1/b4
a2
a3/b1
a3/b2/c1
a3/b2/c2    

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