在 bash 中循环元组？
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Loop over tuples in bash?
提问by Frank
Is it possible to loop over tuples in bash?
是否可以在 bash 中循环元组？
As an example, it would be great if the following worked:
例如，如果以下内容有效，那就太好了：
for (i,j) in ((c,3), (e,5)); do echo "$i and $j"; done
Is there a workaround that somehow lets me loop over tuples?
是否有一种解决方法可以让我遍历元组？
回答by Eduardo Ivanec
$ for i in c,3 e,5; do IFS=","; set -- $i; echo and ; done
c and 3
e and 5
About this use of set(from man builtins):
关于set(from man builtins) 的这种用法：
Any arguments remaining after option processing are treated as values for the positional parameters and are assigned, in order, to $1, $2, ... $n
选项处理后剩余的任何参数都被视为位置参数的值，并按顺序分配给 $1, $2, ... $n
The IFS=","sets the field separator so every $igets segmented into $1and $2correctly.
该IFS=","设置字段分隔符所以每天$i被分成$1和$2正确。
Via this blog.
通过这个博客。
Edit: more correct version, as suggested by @SLACEDIAMOND:
编辑：更正确的版本，正如@SLACEDIAMOND 所建议的：
$ OLDIFS=$IFS; IFS=','; for i in c,3 e,5; do set -- $i; echo and ; done; IFS=$OLDIFS
c and 3
e and 5
回答by Grant Humphries
I believe this solution is a little cleaner than the others that have been submitted, h/t to thisbash style guide for illustrating how read can be used to split strings at a delimiter and assign them to individual variables.
我相信这个解决方案比其他已提交的解决方案更简洁，请参阅此bash 样式指南，以说明如何使用 read 在分隔符处拆分字符串并将它们分配给各个变量。
for i in c,3 e,5; do
IFS=',' read item1 item2 <<< "${i}"
echo "${item1}" and "${item2}"
done
回答by MZHm
Based on the answer given by @eduardo-ivanec without setting/resetting the IFS, one could simply do:
根据@eduardo-ivanec 给出的答案，无需设置/重置IFS，可以简单地执行以下操作：
for i in "c 3" "e 5"
do
set -- $i
echo and
done
The output:
输出：
c and 3
e and 5
回答by VasiliNovikov
Use associative array (also known as dictionary/hashMap):
使用关联数组（也称为字典/hashMap）：
declare -A pairs=(
[c]=3
[e]=5
)
for key in "${!pairs[@]}"; do
value="${pairs[$key]}"
echo "key is $key and value is $value"
done
Works for bash4.0+.
适用于 bash4.0+。
If you need triples instead of pairs, you can use the more general approach:
如果您需要三元组而不是对组，您可以使用更通用的方法：
animals=(dog cat mouse)
declare -A sound=(
[dog]=barks
[cat]=purrs
[mouse]=cheeps
)
declare -A size=(
[dog]=big
[cat]=medium
[mouse]=small
)
for animal in "${animals[@]}"; do
echo "$animal ${sound[$animal]} and it is ${size[$animal]}"
done
回答by user unknown
c=('a' 'c')
n=(3 4 )
for i in $(seq 0 $((${#c[*]}-1)))
do
echo ${c[i]} ${n[i]}
done
Might sometimes be more handy.
有时可能更方便。
To explain the uglypart, as noted in the comments:
为了解释该ugly部分，如评论中所述：
seq 0 2produces the sequence of numbers 0 1 2. $(cmd) is command substitution, so for this example the output of seq 0 2, which is the number sequence. But what is the upper bound, the $((${#c[*]}-1))?
seq 0 2产生数字序列 0 1 2。 $(cmd) 是命令替换，所以对于这个例子， 的输出seq 0 2，它是数字序列。但是上限是$((${#c[*]}-1))多少？
$((somthing)) is arithmetic expansion, so $((3+4)) is 7 etc. Our Expression is ${#c[*]}-1, so something - 1. Pretty simple, if we know what ${#c[*]}is.
$((somthing)) 是算术展开式，所以 $((3+4)) 是 7 等等。我们的表达式是${#c[*]}-1，所以某事 - 1。很简单，如果我们知道是什么${#c[*]}。
c is an array, c[*] is just the whole array, ${#c[*]} is the size of the array which is 2 in our case. Now we roll everything back: for i in $(seq 0 $((${#c[*]}-1)))is for i in $(seq 0 $((2-1)))is for i in $(seq 0 1)is for i in 0 1. Because the last element in the array has an index which is the length of the Array - 1.
c 是一个数组， c[*] 只是整个数组， ${#c[*]} 是数组的大小，在我们的例子中是 2。现在我们回滚一切：for i in $(seq 0 $((${#c[*]}-1)))is for i in $(seq 0 $((2-1)))is for i in $(seq 0 1)is for i in 0 1。因为数组中的最后一个元素有一个索引，它是数组的长度 - 1。
回答by kev
$ echo 'c,3;e,5;' | while IFS=',' read -d';' i j; do echo "$i and $j"; done
c and 3
e and 5
回答by Ole Tange
Using GNU Parallel:
使用 GNU 并行：
parallel echo {1} and {2} ::: c e :::+ 3 5
Or:
或者：
parallel -N2 echo {1} and {2} ::: c 3 e 5
Or:
或者：
parallel --colsep , echo {1} and {2} ::: c,3 e,5
回答by anubhava
Using printfin a process substitution:
使用printf在进程替换：
while read -r k v; do
echo "Key $k has value: $v"
done < <(printf '%s\n' 'key1 val1' 'key2 val2' 'key3 val3')
Key key1 has value: val1
Key key2 has value: val2
Key key3 has value: val3
Above requires bash. If bashis not being used then use simple pipeline:
以上要求bash。如果bash没有被使用，那么使用简单的管道：
printf '%s\n' 'key1 val1' 'key2 val2' 'key3 val3' |
while read -r k v; do echo "Key $k has value: $v"; done
回答by prodriguez903
do echo $key $value
done < file_discriptor
for example:
例如：
$ while read key value; do echo $key $value ;done <<EOF
> c 3
> e 5
> EOF
c 3
e 5
$ echo -e 'c 3\ne 5' > file
$ while read key value; do echo $key $value ;done <file
c 3
e 5
$ echo -e 'c,3\ne,5' > file
$ while IFS=, read key value; do echo $key $value ;done <file
c 3
e 5
回答by Diego Torres Milano
A bit more involved, but may be useful:
涉及更多，但可能有用：
a='((c,3), (e,5))'
IFS='()'; for t in $a; do [ -n "$t" ] && { IFS=','; set -- $t; [ -n "" ] && echo i= j=; }; done
