jq unique_by-选择剩余的元素 [英] jq unique_by - Choose remaining element

问题描述

我目前正在尝试通过REST API从PagerDuty获取事件列表，

I'm currently trying to get a list of incidents from PagerDuty via the REST api, which returns them in JSON array. I want to remove any duplicate events by using unique_by() on the incident_key. However, I want the first occurrence of the incident_key, and unique_by() is removing all but the last. Right now, if I have incident_number 849, 850, and 851, all with the same incident_key, unique_by() will return 851.

简单的例子:

[
{ "reference_key":"200", "id":"1" },
{ "reference_key":"200", "id":"2" },
{ "reference_key":"200", "id":"3" },
{ "reference_key":"201", "id":"4" },
{ "reference_key":"201", "id":"5" },
{ "reference_key":"201", "id":"6" }
]

我想做的是根据ID使用unique_by()来获得reference_key的首次出现.所以在这种情况下，我希望输出为

What I'm trying to do is use unique_by() to get the first occurrence of reference_key, based on the id. So in this case, I'd want the output to be

[
{ "reference_key":"200", "id":"1" },
{ "reference_key":"201", "id":"4" }
]

问题是我对此无能为力，而对于我目前正在尝试使用的数据，它返回的是最后一次而不是第一次的返回.

The problem is that I have no control over this, and with the data I'm currently trying to do this with, it is returning the last occurrence instead of the first, like so.

[
{ "reference_key":"200", "id":"3" },
{ "reference_key":"201", "id":"6" }
]

我尝试过使用反向，然后调用unique_by()，但是得到的结果是相同的.有什么办法可以对此进行控制吗?

I have tried using reverse and then calling unique_by(), but I am getting the same results. Is there any way to have some control over this?

推荐答案

也许您的jq版本不够新.使用jq 1.5:

Maybe your version of jq is not sufficiently recent. Using jq 1.5:

unique_by( .reference_key ) 

收益

[{"reference_key":"200","id":"1"},{"reference_key":"201","id":"4"}]

(截至2016年1月18日(7835a72)，内置的sort过滤器是稳定的；在此之前，稳定性取决于平台.)

(As of January 18, 2016 (7835a72), the builtin sort filter is stable; prior to that, stability was platform-dependent.)

如果您没有访问足够新版本的jq的权限，请考虑以下内容，这些版本已通过jq 1.3、1.4和1.5进行了测试:

If you don't have access to a sufficiently recent version of jq, then consider the following, which has been tested with jq 1.3, 1.4 and 1.5:

def bucketize(f):
  reduce .[] as $x ({}; .[$x|f] += [$x] );

bucketize(.reference_key) | .[][0]

或更经济:

reduce .[] as $x ({};
  $x.reference_key as $key 
  | if .[$key] then . else .[$key] = $x end)
| .[]

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