交换键和值并累积值

Question

我有以下JSON代码段：

{
"a": [ 1, "a:111" ],
"b": [ 2, "a:111", "irrelevant" ],
"c": [ 1, "a:222" ],
"d": [ 1, "b:222" ],
"e": [ 2, "b:222", "irrelevant"]
}

我想将键与数组的第二个值交换，并累积具有相同值的键，丢弃第二个值之后的可能值：

{ "a:111": [ [ 1, "a" ], [ 2, "b" ] ],
"a:222": [ [ 1, "c" ] ],
"b:222": [ [ 1, "d" ], [ 2, "e" ] ]
}

我最初的解决方案如下：

echo  '{
"a": [ 1, "a:111" ],
"b": [ 2, "a:111", "irrelevant" ],
"c": [ 1, "a:222" ],
"d": [ 1, "b:222" ],
"e": [ 2, "b:222", "irrelevant"]
}' 
| jq 'to_entries
| map({(.value[1]|tostring) : [[.value[0], .key]]})
| reduce .[] as $o ({}; reduce ($o|keys)[] as $key (.; .[$key] += $o[$key]))'

这会产生所需的结果，但可能不是很健壮，难以阅读且过长。我想有一个使用with_entries更具可读性的解决方案，但现在我还没有

。

Answer 1

简短的jq方法：

jq 'reduce to_entries[] as $o ({};
.[$o.value[1]] += [[$o.value[0], $o.key]])' input.json

输出：

{
"a:111": [
[
1,
"a"
],
[
2,
"b"
]
],
"a:222": [
[
1,
"c"
]
],
"b:222": [
[
1,
"d"
],
[
2,
"e"
]
]
}