將 grep 結果折疊成一行，同時保留唯一值並刪除重複值？

我正在使用zsh但也對bash解決方案持開放態度。假設我有以下命令：

ls **/*/assessment(.) | xargs egrep "(^  \[\./.*\]$|^    input = .*$)"

這給了我一個類似於這樣的輸出：

<path-01-to>/assessment:  [./input_file_01]
<path-01-to>/assessment:    input = 'input_file_01.i'
<path-02-to>/assessment:  [./input_file_02]
<path-02-to>/assessment:    input = 'input_file_02.i'
<path-02-to>/assessment:  [./input_file_02b]
<path-02-to>/assessment:    input = 'input_file_02b.i'
<path-03-to>/assessment:  [./input_file_03]
<path-03-to>/assessment:    input = 'input_file_03.i'

我想折疊線，使每對 ( [./title] input) 對都在同一行。因此，預期的輸出將如下所示：

<path-01-to>/assessment:  [./input_file_01] input = 'input_file_01.i'
<path-02-to>/assessment:  [./input_file_02] input = 'input_file_02.i'
<path-02-to>/assessment:  [./input_file_02b] input = 'input_file_02b.i'
<path-03-to>/assessment:  [./input_file_03] input = 'input_file_03.i'

我看過使用tr -d "\n"，但它將所有內容組合成一行。我在想awk，或者不同的語言可能更適合這個。

編輯：

這是輸入文件的範例assessment：

路徑/01/to/assessment：

[Tests]
 [./input_file_01]
   type = RunApp
   input = 'input_file_01.i'
 [../]
[]

路徑/02/到/評估：

[Tests]
 [./input_file_02]
   type = RunApp
   input = 'input_file_02.i'
   cli_args = 'blah blah'
 [../]
 [./input_file_02b]
   type = CSVDiff
   input = 'input_file_02b.i'
 [../]
[]

這將在每個 UNIX 機器上的任何 shell 中使用任何 awk 工作：

$ awk '/^ *\[\.\//{title=$1} /^ *input =/{print FILENAME ":", title, "=", $NF}' */*/assessment
01/to/assessment: [./input_file_01] = 'input_file_01.i'
02/to/assessment: [./input_file_02] = 'input_file_02.i'
02/to/assessment: [./input_file_02b] = 'input_file_02b.i'

引用自：https://unix.stackexchange.com/questions/575177