扩展以data.table为参数的函数以使用完整的表(而不是子集)
扩展以data.table为参数的函数以使用完整的表(而不是子集)
提问于 2016-02-18 13:27:43
对于1行的data.table (data.frame)，我有一个可以工作的函数，但对完整的data.table不起作用。我想对函数进行扩展，以考虑到输入data.table的所有行。
这一论点的要点如下：
字段为字符串的data.table (tryshort3)需要替换为来自另一个data.table (mapping)的另一个字符串，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$name , perl = TRUE)
return(input)
}简而言之，此函数所做的是将tryshort3 asn作为输入(目前只使用tryshort3[1,])，并在字段中替换mapping表中找到的值，如下所示：
> tryshort3[1,]
country name address
1: AT ASDF AG ACDSTR. 3
> substituting_short_form(tryshort3[1,])
country name address
1: AT ASDF AKTIENGESELLSCHAFT ACDSTR. 3我想要的是，作为输入提供完整的data.table，并得到相同的输出(长度相同的data.table )，下面是我的预期输出：
country name address
1: AT ASDF AKTIENGESELLSCHAFT ACDSTR. 3
2: AT ASDF GESELLSCHAFT MIT BESCHRANKTER HAFTUNG ACDSTR. 4
3: CH ASDF SA ACDSTR. 7
4: DE ASDF KG ACDSTR. 6
5: MT ASDF DF ACDSTR. 5
6: XK ASDF DAF ACDSTR. 8我想要的解决方案是函数apply(tryshort3, 1, function(x) substituting_short_form(x) )中的一些内容，也许使用两个data.tables的索引功能，或者从内部使用来自nlme的gapply？
Stack Overflow用户
回答已采纳
发布于 2016-02-18 14:14:00
也许你可以尝试几个步骤：
# create the shortform variable in tryshort3
tryshort3[, short.form := sub(".+\\s([^s]+)$", "\\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注:对不起，我只是把它放在你的例子data.table上，而不是作为一个函数
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原文链接：
https://stackoverflow.com/questions/35482781
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