Json 交换键和值以及累积值
我有以下JSON片段:Json 交换键和值以及累积值,json,linux,key-value,reduce,jq,Json,Linux,Key Value,Reduce,Jq,我有以下JSON片段: { "a": [ 1, "a:111" ], "b": [ 2, "a:111", "irrelevant" ], "c": [ 1, "a:222" ], "d": [ 1, "b:222" ], "e": [ 2, "b:222", "irrelevant"] } 我想用数组的第二个值交换键,并用相同的值累积键,丢弃第二个值之后可能出现的值: { "a:111": [ [ 1, "a" ], [ 2, "b" ] ], "a:222": [
{
"a": [ 1, "a:111" ],
"b": [ 2, "a:111", "irrelevant" ],
"c": [ 1, "a:222" ],
"d": [ 1, "b:222" ],
"e": [ 2, "b:222", "irrelevant"]
}
我想用数组的第二个值交换键,并用相同的值累积键,丢弃第二个值之后可能出现的值:
{ "a:111": [ [ 1, "a" ], [ 2, "b" ] ],
"a:222": [ [ 1, "c" ] ],
"b:222": [ [ 1, "d" ], [ 2, "e" ] ]
}
我的初步解决方案如下:
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,但可能不是非常健壮、难以阅读且过长。我想有一种使用with_条目的更具可读性的解决方案,但目前我还没有找到它。Short
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"
]
]
}
谢谢现在看起来好多了:-)