随机ARG为了for语句一个bash [英] Randomizing arg order for a bash for statement

问题描述

我有使用类似

对于我在* .TXT
做
   OPS .....
做

for i in *.txt do ops..... done

有成千上万的文件，他们总是在因* .TXT扩张字母顺序进行处理。

There are thousands of files and they are always processed in alphanumerical order because of '*.txt' expansion.

有没有一种简单的方法来随机的顺序，仍然确保我处理所有的文件只有一次？

Is there a simple way to random the order and still insure that I process all of the files only once?

推荐答案

假设文件名没有空格，刚刚替补的列表::的Util ::洗牌。

Assuming the filenames do not have spaces, just substitute the output of List::Util::shuffle.

for i in `perl -MList::Util=shuffle -e'$,=$";print shuffle<*.txt>'`; do
    ....
done

如果文件名也有空间，但不具有嵌入式换行符或反斜线，一次读取一行。

If filenames do have spaces but don't have embedded newlines or backslashes, read a line at a time.

perl -MList::Util=shuffle -le'$,=$\;print shuffle<*.txt>' | while read i; do
    ....
done

要完全安全的猛砸，使用NUL结尾的字符串。

To be completely safe in Bash, use NUL-terminated strings.

perl -MList::Util=shuffle -0 -le'$,=$\;print shuffle<*.txt>' |
while read -r -d '' i; do
    ....
done

---

不是很有效的，但它是可能的，如果希望这样做在纯击。 排序-R 做这样的事情，在内部。

Not very efficient, but it is possible to do this in pure Bash if desired. sort -R does something like this, internally.

declare -a a                     # create an integer-indexed associative array
for i in *.txt; do
    j=$RANDOM                    # find an unused slot
    while [[ -n ${a[$j]} ]]; do
        j=$RANDOM
    done
    a[$j]=$i                     # fill that slot
done
for i in "${a[@]}"; do           # iterate in index order (which is random)
    ....
done

还是使用传统的费雪耶茨洗牌。

Or use a traditional Fisher-Yates shuffle.

a=(*.txt)
for ((i=${#a[*]}; i>1; i--)); do
    j=$[RANDOM%i]
    tmp=${a[$j]}
    a[$j]=${a[$[i-1]]}
    a[$[i-1]]=$tmp
done
for i in "${a[@]}"; do
    ....
done

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