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社区首页 >问答首页 >使用AWS雅典娜的时间戳差异

使用AWS雅典娜的时间戳差异
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Stack Overflow用户
提问于 2020-01-29 08:20:56
回答 1查看 508关注 0票数 2

我正在我的jupyter笔记本中使用AWS雅典娜运行一个SQL查询,如下所示。它涉及计算时间戳之间的差异,如下所示。

代码语言:javascript
复制
query_demog = """


select ad.subject_id, ad.hadm_id, i.icustay_id ,
       date_diff('second', timestamp '1970-01-01 00:00:00', ad.admittime) as admittime,
       date_diff('second', timestamp '1970-01-01 00:00:00', ad.dischtime) as dischtime,
       ROW_NUMBER() over (partition by ad.subject_id order by i.intime asc) as adm_order,
       case when i.first_careunit='NICU' then 5
            when i.first_careunit='SICU' then 2
            when i.first_careunit='CSRU' then 4
            when i.first_careunit='CCU' then 6
            when i.first_careunit='MICU' then 1
            when i.first_careunit='TSICU' then 3
       end as unit,
       date_diff('second', timestamp '1970-01-01 00:00:00', i.intime) as intime,
       date_diff('second', timestamp '1970-01-01 00:00:00', i.outtime) as outtime,
       i.los,
from mimiciii.admissions ad,
     mimiciii.icustays i,
     mimiciii.patients p
where ad.hadm_id=i.hadm_id and p.subject_id=i.subject_id 
order by subject_id asc, intime asc

"""

效果很好。现在,当我包含另一行具有类似时间戳差异时,我会得到一个错误。

代码语言:javascript
复制
query_demog = """


select ad.subject_id, ad.hadm_id, i.icustay_id ,date_diff('second', timestamp '1970-01-01 00:00:00', ad.admittime) as admittime, date_diff('second', timestamp '1970-01-01 00:00:00', ad.dischtime) as dischtime, ROW_NUMBER() over (partition by ad.subject_id order by i.intime asc) as adm_order, case when i.first_careunit='NICU' then 5 when i.first_careunit='SICU' then 2 when i.first_careunit='CSRU' then 4 when i.first_careunit='CCU' then 6 when i.first_careunit='MICU' then 1 when i.first_careunit='TSICU' then 3 end as unit,  date_diff('second', timestamp '1970-01-01 00:00:00', i.intime) as intime, date_diff('second', timestamp '1970-01-01 00:00:00', i.outtime) as outtime, i.los,

 EXTRACT(EPOCH FROM (i.intime-p.dob)::INTERVAL)/86400 as age





from mimiciii.admissions ad, mimiciii.icustays i, mimiciii.patients p

where ad.hadm_id=i.hadm_id and p.subject_id=i.subject_id 

order by subject_id asc, intime asc

"""

包含行EXTRACT(EPOCH FROM (i.intime-p.dob)::INTERVAL)/86400 as age将创建一个错误,如下所示。

调用StartQueryExecution操作时发生错误(InvalidRequestException):第3行:37:不匹配的输入“:”期望{'.',‘’),‘'[’‘,'AT’‘,'+',’‘,'*','%',’%‘}无法回滚

我不知道为什么包含EXTRACT(EPOCH FROM (i.intime-p.dob)::INTERVAL)/86400 as age会产生错误

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回答 1

Stack Overflow用户

回答已采纳

发布于 2020-01-29 08:28:21

内置的to_unixtime()应该可以工作:

代码语言:javascript
复制
to_unixtime(i.intime-p.dob)/86400 as age
票数 2
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/59963000

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