1

我正在清理数据库，其中一个字段是“国家”，但是数据库中的国家名称与我需要的输出不匹配。

我虽然使用 str_replace 函数，但我有超过 50 个国家需要修复，所以这不是最有效的方法。我已经准备了一个 CSV 文件，其中包含原始国家输入和我需要参考的输出。

这是我到目前为止所拥有的：

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="")
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2 回答 2

1

注意：其中的级别和标签factor必须是唯一的，否则不应包含重复项。

# 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

碱基R

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

数据表

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，例如 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) ))
于 2017-12-28T06:36:55.483 回答
1

过去曾使用类似的方法使用 .csv 文件进行批量替换来解决此类问题。

.csv 文件格式示例：

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")

一旦你有了“Replacements.csv”，你就可以用它来一次替换所有的名字stringi::replace_all_regex()。（对于它的价值，几乎整个包stringr本质上是对stringi.stringistringi

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="")

我尝试data.frame在可能的情况下使用上面的基本 R 语法以避免任何混淆，但如果我是为自己这样做，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")
于 2017-12-28T22:56:31.300 回答