排序 -n bash空间分隔项目数组的内容

Question

>我有这个结构：

.
├── b withe spaces.py
├── c.py
├── lib.py
├── show.sh
└── sub
    └── a.py

我想要这个结果：：

$ bash ./show.sh
    9 lines c.py              => info from outside script
   17 lines a.py              => info from outside script
  300 lines b withe spaces.py => info from outside script
 1589 lines lib.py            => info from outside script
 1915 lines total             => info from outside script

按wc -l排序很容易：：

$ wc -l *.py */*.py | sort -n
    9 c.py
   17 sub/a.py
  300 b withe spaces.py
 1589 lib.py
 1915 total

所以我的出发点是：

array=(
$(find ./ -name "*.py" -print0 | while read -d $'' file
  do
    llines=$(cat $file | wc -l)
    analyse='ouput analyse from some script'
    printf "%s =t%25s => %s" $llines $file $analyse
  done
  ))
for item in ${array[@]}; do echo $item; done| sort -n

Answer 1

你根本不需要这个数组：

find ./ -name "*.py" -print0 | while read -d $'' file ; do
    llines=$(wc -l < "$file")
    analyse='ouput analyse from some script'
    # ----- you missed quoting these ---->vvv
    printf "%s =t%s => %sn" "$llines" "$file" "$analyse"
done | sort -k1nr

Answer 2

你可以做更容易的事情：

find . -name "*sh" -print0 | wc  --files0-from=- -l | sort -n | sed 's/(^[0-9]+) (.*)/1 lines 2 => info from outside script/'

在这种情况下，您不会为每个文件分别执行wc，这将节省一点时间。此外，wc将为您计算总数。