如何在Stata或R中从观测值中删除特定单词？

Question

我在互联网上进行了一些研究，到目前为止，我只找到了使用substr来删除观察值的第一个/最后一个字母的可能性，但没有像CTRL+F查找和替换那样的命令。这里的技巧是我不一定知道我想要删除的单词在哪里！

我的数据集看起来像这样：

Hosp_code   Hosp    Hosplat Hosplon
RRK RRK - UNIVERSITY HOSPITALS BIRMINGHAM NHS FOUNDATION TRUST  52.453271   -1.9362835
RLU RLU - BIRMINGHAM WOMEN'S NHS FOUNDATION TRUST   52.453184   -1.9422432
5MX 5MX - HEART OF BIRMINGHAM TEACHING PCT  52.471575   -1.9367724
NO0 NO0 - HEALTHHARMONIE LIMITED    52.470965   -1.9243192
NLU NLU - SK:N (LASERCARE CLINICS LTD)  52.470838   -1.9220819
NXX NXX - SCRIVENS LTD  52.47148    -1.91341
AGL AGL - ADDITIONAL COMMUNITY MEDICAL SERVICES LTD 52.477343   -1.917197
5M1 5M1 - SOUTH BIRMINGHAM PCT  52.445922   -1.8928915
NQR NQR - PRIMECARE PRIMARY CARE    52.484113   -1.9173169
RXT RXT - BIRMINGHAM AND SOLIHULL MENTAL HEALTH NHS FOUNDATION TRUST    52.484113   -1.9173169
RRJ RRJ - THE ROYAL ORTHOPAEDIC HOSPITAL NHS FOUNDATION TRUST   52.421133   -1.9608273
RXK RXK - SANDWELL AND WEST BIRMINGHAM HOSPITALS NHS TRUST  52.48982    -1.9294268
RQ3 RQ3 - BIRMINGHAM CHILDREN'S HOSPITAL NHS FOUNDATION TRUST   52.485173   -1.8944604
RYW RYW - BIRMINGHAM COMMUNITY HEALTHCARE NHS TRUST 52.487323   -1.8858108
5PG 5PG - BIRMINGHAM EAST AND NORTH PCT 52.491369   -1.886036
NIT NIT - SOUTH DOC SERVICES LIMITED HQ 52.401796   -1.9620201
RR1 RR1 - HEART OF ENGLAND NHS FOUNDATION TRUST 52.477876   -1.8275305
NIS NIS - COVENTRY AND WARWICKSHIRE DIAGNOSTIC SERVICES LIMITED 52.462504   -1.8159336
NDT NDT - WEST MIDLANDS DIAGNOSTIC SERVICES LTD 52.462504   -1.8159336
5PF 5PF - SANDWELL PCT  52.523328   -2.0026388
TAJ TAJ - BLACK COUNTRY PARTNERSHIP NHS FOUNDATION TRUST    52.519255   -2.0188053
NEP NEP - TICCS ULTRASOUND LIMITED  52.510017   -1.8113152
NL7 NL7 - ASSURA VERTIS URGENT CARE CENTRES (BIRMINGHAM)    52.542091   -1.8778985
NNT NNT - ASSURA KINGSTANDING   52.542091   -1.8778985
5QW 5QW - SOLIHULL PCT  52.391695   -1.8081752
NR9 NR9 - JOHN TAYLOR HOSPICE COMMUNITY INTEREST COMPANY    52.527341   -1.8234016
RYK RYK - DUDLEY AND WALSALL MENTAL HEALTH PARTNERSHIP NHS TRUST    52.508312   -2.0844533

我想要删除以下内容： - 前三个字母（例如RRK -） - 任何提到的 "LTD" "LTD "LIMITED" "HQ" "LLP" "TRUST" "FOUNDATION TRUST" 有什么建议吗？

- pdx

---

请返回已翻译的文本。 - m-dz

只是出于好奇，你从哪里获取了NHS组织的经纬度？ - r.bot

我手动搜索了它们，哈哈哈。 - pdx

什么是susbstr？Stata有substr()和subinstr()。无论哪种方式，请说明这是Stata还是R语法，并纠正任何拼写错误。我知道Ctrl-F通常在Windows程序中只是“查找”，但请注意，许多读者可能会使用其他操作系统。 - Nick Cox

在Stata中，您可以尝试一长串的replace ... subinstr()命令/函数，或者较短的replace ... regexr()命令/函数。后者需要一些正则表达式的知识。 - lmo

Answer 1

这里是使用Stata的类似策略。在将来，请考虑使用dataex生成您的数据示例。

* Example generated by -dataex-. To install: ssc install dataex
clear
input str68 Hosp
"RRK - UNIVERSITY HOSPITALS BIRMINGHAM NHS FOUNDATION TRUST"      
"RLU - BIRMINGHAM WOMEN'S NHS FOUNDATION TRUST"                   
"5MX - HEART OF BIRMINGHAM TEACHING PCT"                          
"NO0 - HEALTHHARMONIE LIMITED"                                    
"NLU - SK:N (LASERCARE CLINICS LTD)"                              
"NXX - SCRIVENS LTD"                                              
"AGL - ADDITIONAL COMMUNITY MEDICAL SERVICES LTD"                 
"5M1 - SOUTH BIRMINGHAM PCT"                                      
"NQR - PRIMECARE PRIMARY CARE"                                    
"RXT - BIRMINGHAM AND SOLIHULL MENTAL HEALTH NHS FOUNDATION TRUST"
"RRJ - THE ROYAL ORTHOPAEDIC HOSPITAL NHS FOUNDATION TRUST"       
"RXK - SANDWELL AND WEST BIRMINGHAM HOSPITALS NHS TRUST"          
"RQ3 - BIRMINGHAM CHILDREN'S HOSPITAL NHS FOUNDATION TRUST"       
"RYW - BIRMINGHAM COMMUNITY HEALTHCARE NHS TRUST"                 
"5PG - BIRMINGHAM EAST AND NORTH PCT"                             
"NIT - SOUTH DOC SERVICES LIMITED HQ"                             
"RR1 - HEART OF ENGLAND NHS FOUNDATION TRUST"                     
"NIS - COVENTRY AND WARWICKSHIRE DIAGNOSTIC SERVICES LIMITED"     
"NDT - WEST MIDLANDS DIAGNOSTIC SERVICES LTD"                     
"5PF - SANDWELL PCT"                                              
"TAJ - BLACK COUNTRY PARTNERSHIP NHS FOUNDATION TRUST"            
"NEP - TICCS ULTRASOUND LIMITED"                                  
"NL7 - ASSURA VERTIS URGENT CARE CENTRES (BIRMINGHAM)"            
"NNT - ASSURA KINGSTANDING"                                       
"5QW - SOLIHULL PCT"                                              
"NR9 - JOHN TAYLOR HOSPICE COMMUNITY INTEREST COMPANY"            
"RYK - DUDLEY AND WALSALL MENTAL HEALTH PARTNERSHIP NHS TRUST"    
end

gen work = substr(Hosp, 7, .)
replace work = ustrregexra(work, " *(LTD|LIMITED|HQ|LLP|TRUST|FOUNDATION TRUST)", "")
leftalign // from SSC, to install, type: ssc install leftalign
list

- Robert Picard

---

ustrregexra()仅适用于Stata 14。如果您使用的是早期版本的Stata，请改用regexr()。 - Aline

Answer 2

这里有一个使用R语言的答案。有两种策略，第一种使用基本R函数，第二种使用stringr包。两种方法都使用了三个步骤：i）删除前7个字符；ii）删除我们不想要的模式；iii）修剪任何剩余的空格。

# a subset of your data.frame making your question reproducible
df <- structure(list(Hosp = c("NXX - SCRIVENS LTD", "AGL - ADDITIONAL COMMUNITY MEDICAL SERVICES LTD", 
                              "5M1 - SOUTH BIRMINGHAM PCT", "NQR - PRIMECARE PRIMARY CARE", 
                              "RXT - BIRMINGHAM AND SOLIHULL MENTAL HEALTH NHS FOUNDATION TRUST", 
                              "RRJ - THE ROYAL ORTHOPAEDIC HOSPITAL NHS FOUNDATION TRUST", 
                              "RXK - SANDWELL AND WEST BIRMINGHAM HOSPITALS NHS TRUST")),
                .Names = "Hosp", row.names = c(NA, -7L), class = "data.frame")

> df$Hosp
[1] "NXX - SCRIVENS LTD"                                              
[2] "AGL - ADDITIONAL COMMUNITY MEDICAL SERVICES LTD"                 
[3] "5M1 - SOUTH BIRMINGHAM PCT"                                      
[4] "NQR - PRIMECARE PRIMARY CARE"                                    
[5] "RXT - BIRMINGHAM AND SOLIHULL MENTAL HEALTH NHS FOUNDATION TRUST"
[6] "RRJ - THE ROYAL ORTHOPAEDIC HOSPITAL NHS FOUNDATION TRUST"       
[7] "RXK - SANDWELL AND WEST BIRMINGHAM HOSPITALS NHS TRUST" 

#base R functions -----------
gsub(" $", "", gsub("LTD|LIMITED|HQ|LLP|TRUST|FOUNDATION TRUST", "", substr(df$Hosp, 7, nchar(df$Hosp))))

# a function to do it
nice_hospname <- function(x){
  gsub(" $", "", gsub("LTD|LIMITED|HQ|LLP|TRUST|FOUNDATION TRUST", "", substr(x, 7, nchar(x))))
}
# you can use it with:
nice_hospname(df$Hosp)

# with stringr package --------
library(stringr)
df$Hosp %>% str_sub(7) %>% str_replace("LTD|LIMITED|HQ|LLP|TRUST|FOUNDATION TRUST", "") %>% str_trim()

[1] "SCRIVENS"                                   "ADDITIONAL COMMUNITY MEDICAL SERVICES"     
[3] "SOUTH BIRMINGHAM PCT"                       "PRIMECARE PRIMARY CARE"                    
[5] "BIRMINGHAM AND SOLIHULL MENTAL HEALTH NHS"  "THE ROYAL ORTHOPAEDIC HOSPITAL NHS"        
[7] "SANDWELL AND WEST BIRMINGHAM HOSPITALS NHS"

# a function to do it
nice_hospname2 <- function(x){
  x %>% str_sub(7) %>% str_replace("LTD|LIMITED|HQ|LLP|TRUST|FOUNDATION TRUST", "") %>% str_trim()
}
# you can use it with:
nice_hospname2(df$Hosp)


library(stringr)

希望这有所帮助。

- Vincent Bonhomme