我正在尝试使用bigrquery和dbplyr获取week of the year that a date corresponds to(即与lubridate::week()
相同,即
library(lubridate)
library(dbplyr)
library(bigrquery)
week("2015-08-11")
# [1] 32
但是我正在使用bigrquery
和dbplyr
使用lubridate::week()
,我知道
transactions %>%
select(item, date) %>%
mutate(week = week(date)) %>%
collect()
Error: Function not found: week at [1:30] [invalidQuery]
所以我尝试了这种自制解决方案
transactions %>%
select(item, date) %>%
mutate(week = strftime(date, format = "%V")) %>%
collect()
Error: Syntax error: Expected ")" but got keyword AS at [1:54] [invalidQuery]
In addition: Warning message:
Named arguments ignored for SQL strftime
以及另一个(非常丑陋的)自制解决方案
transactions %>%
select(item, date) %>%
mutate(week = as.numeric((as.Date(date) - as.Date(paste0(substr(date, 1, 4), "-01-01"))), units="days") %/% 7) %>%
collect()
Error in as.numeric((as.Date(date) - as.Date(paste0(substr(date, 1, :
unused argument (units = "days")
但是我似乎找不到使用bigquery和dbplyr获取星期数的方法
答案 0 :(得分:2)
我似乎找不到使用bigquery获取星期数的方法
看起来像您在下面的BigQuery Standard SQL函数
EXTRACT(WEEK FROM date)
您可以使用WEEK或WEEK(
在https://cloud.google.com/bigquery/docs/reference/standard-sql/date_functions#extract上查看有关日期部分的更多信息