用数字分隔字符串

Question

我正在尝试按型号和引擎分隔我的列 VEHICLE_TYPE。代码可以是普通的 SQL 或 R 代码。

我的数据是这样的：

       MODEL           VEHICLE_TYPE
77        Bora               Bora 1.6
79      Ducato Ducato 15 120 Multijet
80      Ducato Ducato 15 120 Multijet
87       Astra         Astra 1.7 CDTI
88         406            406 2.0 HDi
89         406            406 2.0 HDi
90 Focus C-MAX   Focus C-MAX 1.6 TDCi
91 Focus C-MAX   Focus C-MAX 1.6 TDCi
92 Focus C-MAX   Focus C-MAX 1.6 TDCi
93 Focus C-MAX   Focus C-MAX 1.6 TDCi
94 Focus C-MAX   Focus C-MAX 1.6 TDCi
97    S-Klasse              S 320 CDI
98    S-Klasse              S 320 CDI
99    S-Klasse              S 320 CDI

我想收到这样的东西：

MODEL         VEHICLE TYPE
Bora          1.6
Ducato 15     120 Multijet
...           ...
Focus C-Max   1.6 TDCi

问题是，VEHICLE_TYPE 可以有不同的长度和不同数量的空格，我可以用它们来分隔。

我用 gsub 和 regex 试过了，没用，但是 strsplit工作了。与我真正想要的相去甚远，我运行没有想法，现在需要一些帮助。

> strsplit(as.character(test$VEHICLE_TYPE)," ")

[[1]]
[1] "Bora" "1.6"

[[2]]
[1] "Ducato"   "15"       "120"      "Multijet"

[[3]]
[1] "Ducato"   "15"       "120"      "Multijet"

[[4]]
[1] "Astra" "1.7"   "CDTI" 

[[5]]
[1] "406" "2.0" "HDi"

[[6]]
[1] "406" "2.0" "HDi"

[[7]]
[1] "Focus" "C-MAX" "1.6"   "TDCi" 

[[8]]
[1] "Focus" "C-MAX" "1.6"   "TDCi" 

[[9]]
[1] "Focus" "C-MAX" "1.6"   "TDCi" 

[[10]]
[1] "Focus" "C-MAX" "1.6"   "TDCi" 

[[11]]
[1] "Focus" "C-MAX" "1.6"   "TDCi" 

[[12]]
[1] "S"   "320" "CDI"

Answer 1

我猜有人会知道用比这更简单的正则表达式来做到这一点的方法，但由于我是正则表达式笨蛋，所以这是我的尝试。按 space 拆分，然后折叠第一个 "numeric" 值之后的所有内容。

library( magrittr )
df[['VEHICLE_TYPE']] %<>%
    strsplit( " " ) %>%
    sapply( function(x) paste(
        x[ grep( "[[:digit:]]", x )[1] : length(x) ],
        collapse = " " )
    )

结果

> df
# # A tibble: 14 × 2
# MODEL    VEHICLE_TYPE
# <chr>           <chr>
# 1         Bora             1.6
# 2       Ducato 15 120 Multijet
# 3       Ducato 15 120 Multijet
# 4        Astra        1.7 CDTI
# 5          406     406 2.0 HDi
# 6          406     406 2.0 HDi
# 7  Focus C-MAX        1.6 TDCi
# 8  Focus C-MAX        1.6 TDCi
# 9  Focus C-MAX        1.6 TDCi
# 10 Focus C-MAX        1.6 TDCi
# 11 Focus C-MAX        1.6 TDCi
# 12    S-Klasse         320 CDI
# 13    S-Klasse         320 CDI
# 14    S-Klasse         320 CDI

或者，如果您更喜欢按 last 数值而不是 first 进行拆分：

df[['VEHICLE_TYPE']] %<>%
    strsplit( " " ) %>%
    sapply( function(x) paste(
        x[ tail( grep( "[[:digit:]]", x ), 1 ) : length(x) ],
        collapse = " " )
    )

> df
# # A tibble: 14 × 2
# MODEL VEHICLE_TYPE
# <chr>        <chr>
# 1         Bora          1.6
# 2       Ducato 120 Multijet
# 3       Ducato 120 Multijet
# 4        Astra     1.7 CDTI
# 5          406      2.0 HDi
# 6          406      2.0 HDi
# 7  Focus C-MAX     1.6 TDCi
# 8  Focus C-MAX     1.6 TDCi
# 9  Focus C-MAX     1.6 TDCi
# 10 Focus C-MAX     1.6 TDCi
# 11 Focus C-MAX     1.6 TDCi
# 12    S-Klasse      320 CDI
# 13    S-Klasse      320 CDI
# 14    S-Klasse      320 CDI

编辑：如果您有一些行没有任何数值，您可能需要一些额外的修补：

df[['VEHICLE_TYPE']] %<>%
    strsplit( " " ) %>%
    sapply( function(x) paste(
        if( length( grep( "[[:digit:]]", x ) ) > 1L ) {
            x[ tail( grep( "[[:digit:]]", x ), 1 ) : length(x) ]
        } else { x },
        collapse = " " )
    )

Answer 2

正则表达式示例

with s(id,model,type) as (
select 77,'Bora','Bora 1.6' from dual union all
select 79,'Ducato','Ducato 15 120 Multijet' from dual union all
select 80 ,'Ducato','Ducato 15 120 Multijet' from dual union all
select 87 ,'Astra','Astra 1.7 CDTI' from dual union all
select 88 ,'406','406 2.0 HDi' from dual union all
select 89 ,'406','406 2.0 HDi' from dual union all
select 90 ,'Focus C-MAX','Focus C-MAX 1.6 TDCi' from dual union all
select 91 ,'Focus C-MAX','Focus C-MAX 1.6 TDCi' from dual union all
select 92 ,'Focus C-MAX','Focus C-MAX 1.6 TDCi' from dual union all
select 93 ,'Focus C-MAX','Focus C-MAX 1.6 TDCi' from dual union all
select 94 ,'Focus C-MAX','Focus C-MAX 1.6 TDCi' from dual union all
select 97  ,'S-Klasse','S 320 CDI' from dual union all
select 98  ,'S-Klasse','S 320 CDI' from dual union all
select 99  ,'S-Klasse','S 320 CDI' from dual 
)
select regexp_substr(type,'\d+(\.\d+)?\s*\w*$') /*cut part with model*/
from s

Answer 3

这是使用 gsub

的替代解决方案

df$VEHICLE_TYPE <- gsub(".+ ([0-9.]+(?: [^ ]+)?)$", "\1", df$VEHICLE_TYPE)

> df

#           MODEL VEHICLE_TYPE
#  1         Bora          1.6
#  2       Ducato 120 Multijet
#  3       Ducato 120 Multijet
#  4        Astra     1.7 CDTI
#  5          406      2.0 HDi
#  6          406      2.0 HDi
#  7  Focus C-MAX     1.6 TDCi
#  8  Focus C-MAX     1.6 TDCi
#  9  Focus C-MAX     1.6 TDCi
#  10 Focus C-MAX     1.6 TDCi
#  11 Focus C-MAX     1.6 TDCi
#  12    S-Klasse      320 CDI
#  13    S-Klasse      320 CDI
#  14    S-Klasse      320 CDI

我假设车辆类型总是在末尾并遵循以下模式：(1) 一组数字字符（0 到 9 和点），例如1.6 或 (2) 组数字字符和组任何其他字符的组合，由 space 分隔（例如 120 Multijet、2.0 HDi）

更新 : 处理 308 1.6i Flex 和 Cherokee 2.8 CRD 4x4

df$VEHICLE_TYPE <- gsub(".+ ([0-9.]+[a-z]?(?: [^ ]+)?(?: [^ ]+)?)$", "\1", df$VEHICLE_TYPE)

# OR, simply grep "number" and everything after
# df$VEHICLE_TYPE <- gsub(".+ ([0-9.]+[a-z]? .+)$", "\1", df$VEHICLE_TYPE)


> df

#          MODEL VEHICLE_TYPE
# 1         Bora          1.6
# 2       Ducato 120 Multijet
# 3       Ducato 120 Multijet
# 4        Astra     1.7 CDTI
# 5          406      2.0 HDi
# 6          406      2.0 HDi
# 7  Focus C-MAX     1.6 TDCi
# 8  Focus C-MAX     1.6 TDCi
# 9  Focus C-MAX     1.6 TDCi
# 10 Focus C-MAX     1.6 TDCi
# 11 Focus C-MAX     1.6 TDCi
# 12    S-Klasse      320 CDI
# 13    S-Klasse      320 CDI
# 14    S-Klasse      320 CDI
# 15         308    1.6i Flex
# 16    Cherokee  2.8 CRD 4x4

Answer 4

在Oracle中，可以使用正则表达式^(.*?)\s+(\d.*)$:

中的第一个和第二个匹配组

SELECT REGEXP_SUBSTR( vehicle_type, '^(.*?)\s+(\d.*)$', 1, 1, NULL, 1 )
         AS model,
       REGEXP_SUBSTR( vehicle_type, '^(.*?)\s+(\d.*)$', 1, 1, NULL, 2 )
         AS vehicle_type
FROM   your_table;

用数字分隔字符串

Separate a string by a number

regex

sql

oracle

r

strsplit