Question

我有一个适用于1行的data.table（data.frame）的函数，但不适用于完整的data.table。我想扩展函数以考虑输入data.table。

的所有行

论证的要点如下：

一个data.table（tryshort3），其中一个字段是一个字符串，需要用另一个data.table（mapping），MRE中的另一个字符串替换，如下所示：

#this is the original data.table
tryshort3 <- structure(list(country = c("AT", "AT", "MT", "DE", "CH", "XK"
), name = c("ASDF AG", "ASDF GMBH", "ASDF DF", "ASDF KG", "ASDF SA", 
"ASDF DAF"), address = c("ACDSTR. 3", "ACDSTR. 4", "ACDSTR. 5", 
"ACDSTR. 6", "ACDSTR. 7", "ACDSTR. 8")), .Names = c("country", 
"name", "address"), row.names = c(NA, -6L), class = c("data.table", 
"data.frame"))



#this is the "mapping
mapping <- structure(list(country = c("AT", "AT", "DE", "DE", "HU"), short.form = c("AG", 
"GMBH", "GMBH", "EV", "EV"), long.form = c("AKTIENGESELLSCHAFT", 
"GESELLSCHAFT MIT BESCHRANKTER HAFTUNG", "GESELLSCHAFT MIT BESCHRANKTER HAFTUNG", 
"EINGETRAGENE VEREIN", "EGYENI VALLALKOZO")), .Names = c("country", 
"short.form", "long.form"), row.names = c(NA, -5L), class = c("data.table", 
"data.frame"), sorted = "country")


#this is the function that I am using (please not that both data.tables are keyed, but that has currently no say in the output (just avoids throwing an error):

substituting_short_form <- function(input) {
  #supply one data.frame of 1 row, the other data.frame is external to the function
  #get country from input
  setkey(input,country)
  setkey(mapping,country)
  matched_country <- input$country
  #subset of mapping to only the country from the input
  matched_map <- mapping[country == matched_country]
  #get list of short.forms from matched 
  list_of_relevant_short_forms <- matched_map[,short.form]
  #which one matches will return true if there is any match, THIS IS A NUMBER THAT WILL HAVE TO BE MATCHED TO mapping again to retrieve the correct form
  #error catching for when there is no short form found, or no country found if there is no long form it does not matter!
  indextrue <- tryCatch(which(unlist(lapply(list_of_relevant_short_forms, function(y) grepl(y, input$name)))), error = function(e) return(input))
  #substitute
  pattern_to_substitute <- paste0("(\\s|^)", matched_map[indextrue,short.form], "(\\s|$)")
  pattern_to_replace <- paste0("\\1", matched_map[indextrue,long.form], "\\2")
  input$name[1] <- gsub(pattern = pattern_to_substitute, replacement = pattern_to_replace,input$name ,    perl = TRUE)
  return(input)
}

简而言之，这个函数正在做的是将tryshort3作为输入（目前仅与tryshort3[1,]一起使用）并替换字段＆＃34; name＆＃34;在...中找到的值mapping表，如下所示：

> tryshort3[1,]
   country    name   address
1:      AT ASDF AG ACDSTR. 3
> substituting_short_form(tryshort3[1,])
   country                    name   address
1:      AT ASDF AKTIENGESELLSCHAFT ACDSTR. 3

我想要的是，我提供完整的data.table作为输入，并获得相同的输出（相同长度的data.table），这是我的预期输出：

   country                    name   address
1:      AT ASDF AKTIENGESELLSCHAFT ACDSTR. 3
2:      AT ASDF GESELLSCHAFT MIT BESCHRANKTER HAFTUNG ACDSTR. 4
3:      CH ASDF SA ACDSTR. 7
4:      DE ASDF KG ACDSTR. 6
5:      MT ASDF DF ACDSTR. 5
6:      XK ASDF DAF ACDSTR. 8

我希望的解决方案可以来自函数apply(tryshort3, 1, function(x) substituting_short_form(x) )，可能使用data.tables的索引功能，或者可能使用来自gapply的{{1}}来自内部？

Answer 1

也许你可以尝试几个步骤：

# create the shortform variable in tryshort3
tryshort3[, short.form := sub(".+\\s([^s]+)$", "\\1", name)]

# add the info from mapping
tryshort3long <- merge(tryshort3, mapping, all.x=TRUE, by=c("country", "short.form"))

# replace the short form by long form in the name and suppress the variables you don't need 
# (thanks to @DavidArenburg for the simplification of the "replace" part!)
tryshort3long[!is.na(long.form), 
              name := paste(sub(" .*", "", name), long.form)
              ][, c("long.form", "short.form") := NULL]

tryshort3long
   # country                                       name   address
# 1:      AT                    ASDF AKTIENGESELLSCHAFT ACDSTR. 3
# 2:      AT ASDF GESELLSCHAFT MIT BESCHRANKTER HAFTUNG ACDSTR. 4
# 3:      CH                                    ASDF SA ACDSTR. 7
# 4:      DE                                    ASDF KG ACDSTR. 6
# 5:      MT                                    ASDF DF ACDSTR. 5
# 6:      XK                                   ASDF DAF ACDSTR. 8

NB：抱歉，我只是把它作为你的示例data.table而不是作为一个函数

Answer 2

apply的问题在于它会将其参数强制转换为矩阵。尝试一个简单的循环：

lst <- list()
for(i in 1:nrow(tryshort3)) lst[[i]] <- substituting_short_form(tryshort3[i,])
rbindlist(lst)
#    country                                       name   address
# 1:      AT                    ASDF AKTIENGESELLSCHAFT ACDSTR. 3
# 2:      AT ASDF GESELLSCHAFT MIT BESCHRANKTER HAFTUNG ACDSTR. 4
# 3:      MT                                    ASDF DF ACDSTR. 5
# 4:      DE                                    ASDF KG ACDSTR. 6
# 5:      CH                                    ASDF SA ACDSTR. 7
# 6:      XK                                   ASDF DAF ACDSTR. 8

扩展一个函数，该函数将data.table作为参数使用整个表（而不是子集）

2 个答案: