Sparklyr/Hive:如何正确使用正则表达式(regexp_replace)? [英] Sparklyr/Hive: how to use regex (regexp_replace) correctly?

问题描述

考虑以下示例

dataframe_test<- data_frame(mydate = c('2011-03-01T00:00:04.226Z', '2011-03-01T00:00:04.226Z'))

# A tibble: 2 x 1
                    mydate
                     <chr>
1 2011-03-01T00:00:04.226Z
2 2011-03-01T00:00:04.226Z

sdf <- copy_to(sc, dataframe_test, overwrite = TRUE)

> sdf
# Source:   table<dataframe_test> [?? x 1]
# Database: spark_connection
                    mydate
                     <chr>
1 2011-03-01T00:00:04.226Z
2 2011-03-01T00:00:04.226Z

我想修改字符timestamp，使其具有更常规的格式.我尝试使用regexp_replace这样做，但失败了.

I would like to modify the character timestamp so that it has a more conventional format. I tried to do so using regexp_replace but it fails.

> sdf <- sdf %>% mutate(regex = regexp_replace(mydate, '(\\d{4})-(\\d{2})-(\\d{2})T(\\d{2}):(\\d{2}):(\\d{2}).(\\d{3})Z', '$1-$2-$3 $4:$5:$6.$7'))
> sdf
# Source:   lazy query [?? x 2]
# Database: spark_connection
                    mydate                    regex
                     <chr>                    <chr>
1 2011-03-01T00:00:04.226Z 2011-03-01T00:00:04.226Z
2 2011-03-01T00:00:04.226Z 2011-03-01T00:00:04.226Z

有什么想法吗?正确的语法是什么?

Any ideas? What is the correct syntax?

推荐答案

Spark SQL和Hive提供两种不同的功能:

Spark SQL and Hive provide two different functions:

• regexp_extract-包含字符串，模式和要提取的组的索引.
• regexp_replace-接受字符串，模式和替换字符串.

• regexp_extract - which takes string, pattern and the index of the group to be extracted.
• regexp_replace - which takes a string, pattern, and the replacement string.

前一个可用于提取具有索引语义. rel ="nofollow noreferrer">与 java.util.regex.Matcher

The former one can be used to extract a single group with the index semantics being the same as for java.util.regex.Matcher

对于regexp_replace模式，必须匹配整个字符串，如果不匹配，则返回输入字符串:

For regexp_replace pattern has to match a whole string and if there is no match, and the input string is returned:

sdf %>% mutate(
 regex = regexp_replace(mydate, '^([0-9]{4}).*', "$1"),
 regexp_bad = regexp_replace(mydate, '([0-9]{4})', "$1"))

## Source:   query [2 x 3]
## Database: spark connection master=local[8] app=sparklyr local=TRUE
## 
## # A tibble: 2 x 3
##                     mydate regex               regexp_bad
##                      <chr> <chr>                    <chr>
## 1 2011-03-01T00:00:04.226Z  2011 2011-03-01T00:00:04.226Z
## 2 2011-03-01T00:00:04.226Z  2011 2011-03-01T00:00:04.226Z

在使用regexp_extract时不是必需的:

sdf %>% mutate(regex = regexp_extract(mydate, '([0-9]{4})', 1))

## Source:   query [2 x 2]
## Database: spark connection master=local[8] app=sparklyr local=TRUE
## 
## # A tibble: 2 x 2
##                     mydate regex
##                      <chr> <chr>
## 1 2011-03-01T00:00:04.226Z  2011
## 2 2011-03-01T00:00:04.226Z  2011

此外，由于间接执行(R-> Java)，您必须转义两次:

Also, due to indirect execution (R -> Java), you have to escape twice:

sdf %>% mutate(
  regex = regexp_replace(
    mydate, 
    '^(\\\\d{4})-(\\\\d{2})-(\\\\d{2})T(\\\\d{2}):(\\\\d{2}):(\\\\d{2}).(\\\\d{3})Z$',
    '$1-$2-$3 $4:$5:$6.$7'))

通常会使用Spark日期时间函数:

Normally one would use Spark datetime functions:

spark_session(sc) %>%  
  invoke("sql",
    "SELECT *, DATE_FORMAT(CAST(mydate AS timestamp), 'yyyy-MM-dd HH:mm:ss.SSS') parsed from dataframe_test") %>% 
  sdf_register


## Source:   query [2 x 2]
## Database: spark connection master=local[8] app=sparklyr local=TRUE
## 
## # A tibble: 2 x 2
##                     mydate                  parsed
##                      <chr>                   <chr>
## 1 2011-03-01T00:00:04.226Z 2011-03-01 01:00:04.226
## 2 2011-03-01T00:00:04.226Z 2011-03-01 01:00:04.226

但可悲的是，sparklyr在这一领域似乎非常有限，并且将时间戳视为字符串.

but sadly sparklyr seems to be extremely limited in this area, and treats timestamps as strings.

另请参见使用hive命令更改DF中的字符串并使用sparklyr进行更改.

这篇关于Sparklyr/Hive:如何正确使用正则表达式(regexp_replace)?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！