使用引用表替换多个值

Question

我正在清理数据库，其中一个字段是“country”但是我的数据库中的国家/地区名称与我需要的输出不匹配。

我虽然使用str_replace函数，但我有50多个国家需要修复，所以它不是最有效的方法。我已经准备了一个CSV文件，其中包含原始国家/地区输入和我需要的输出以供参考。

这是我到目前为止：

library(stringr)
library(dplyr)
library(tidyr)
library(readxl)
database1<- read_excel("database.xlsx") 
database1$country<str_replace(database1$country,"USA","United States")
database1$country<str_replace(database1$country,"UK","United Kingdom")
database1$country<str_replace(database1$country,"Bolivia","Bolivia,Plurinational State of")
write.csv(database1, "test.csv", row.names=FALSE, fileEncoding = 'UTF 8', na="")

Answer 1

注意：factor中的级别和标签必须是唯一的，或者不应包含重复项。

# database1 <- read_excel("database.xlsx")  ## read database excel book
old_names <- c("USA", "UGA", "CHL") ## country abbreviations
new_names <- c("United States", "Uganda", "Chile")  ## country full form

基地R.

database1 <- within( database1, country <- factor( country, levels = old_names, labels = new_names ))

Data.Table

library('data.table')
setDT(database1)
database1[, country := factor(country, levels = old_names, labels = new_names)]

database1
#          country
# 1: United States
# 2:        Uganda
# 3:         Chile
# 4: United States
# 5:        Uganda
# 6:         Chile
# 7: United States
# 8:        Uganda
# 9:         Chile

数据

database1 <- data.frame(country = c("USA", "UGA", "CHL", "USA", "UGA", "CHL", "USA", "UGA", "CHL"))
#    country
# 1     USA
# 2     UGA
# 3     CHL
# 4     USA
# 5     UGA
# 6     CHL
# 7     USA
# 8     UGA
# 9     CHL

编辑：您可以创建一个命名向量countries，而不是两个变量，如old_names和new_names。

countries <- c("USA", "UGA", "CHL")
names(countries) <- c("United States", "Uganda", "Chile")
within( database1, country <- factor( country, levels = countries, labels = names(countries) ))

Answer 2

过去曾使用类似的方法使用.csv文件进行批量替换，从而解决了类似的问题。

示例.csv文件格式：

library(data.table)

## Generate example replacements csv file to see the format used
Replacements <- data.table(Old = c("USA","UGA","CHL"),
                           New = c("United States", "Uganda", "Chile"))

fwrite(Replacements,"Replacements.csv")

获得“Replacements.csv”后，您可以使用stringi::replace_all_regex()一次性替换所有名称。 （对于它的价值，几乎整个stringr包基本上是对stringi调用的包装。由于stringi运行速度稍快且具有更多功能，我更喜欢坚持使用stringi。）See stringi vs stringr blog by HRBRMSTR

library(data.table)
library(readxl)
library(stringi)

## Read in list of replacements
Replacements <- fread("Replacements.csv")

## Read in file to be cleaned
database1<- read_excel("database.xlsx")

## Perform Replacements
database1$countries <- stringi::stri_replace_all_regex(database1$countries,
                                              "^"%s+%Replacements$Old%s+%"$",
                                              Replacements$New,
                                              vectorize_all = FALSE)

## Write CSV
write.csv(database1, "test.csv", row.names=FALSE, fileEncoding = 'UTF 8', na="")

我试图在可能的地方使用基本R data.frame语法，以避免任何混淆，但如果我为自己做这个，我会坚持完整的data.table语法如下：

library(data.table)
library(readxl)
library(stringi)

## Read in list of replacements
Replacements <- fread("Replacements.csv")

## Read in file to be cleaned
database1<- read_excel("database.xlsx")

## Perform Replacements
database1[, countries := stri_replace_all_regex(countries,"^"%s+%Replacements[,Old]%s+%"$",
                                              Replacements[,New],
                                              vectorize_all = FALSE)]
## Write CSV
fwrite(database1,"test.csv")