Question

我想将城市智慧数据导出到csv文件，因为我遵循以下静态方法

city=c("NYC","NYC","NYC","LONDON","LONDON","LONDON","PARIS","PARIS")
country=c("USA","USA","USA","UK","UK","UK","FRANCE","FRANCE")
year=c(2000,2000,2000,2010,2010,2010,2017,2017)
df=data.frame(city,country,year)

#1
city<-df[which(df$city == "NYC"),]
file_name = paste0("NYC",".csv")
write.csv(city,file_name,row.names=FALSE)

#2
city<-df[which(df$city == "LONDON"),]
file_name = paste0("LONDON",".csv")
write.csv(city,file_name,row.names=FALSE)

＃3>同样＃3 请帮助它转换为动态代码。提前致谢

Answer 1

我会做类似的事情：

for (cit in unique(df$city)) {
  write.csv(
    x = subset(df, city == cit),
    file = paste0(cit, ".csv"),
    row.names = FALSE
  )
}

<强> PS。通常小写文件名是首选，因此我会将file = paste0(cit, ".csv")转换为file = tolower(paste0(cit, ".csv"))

Answer 2

@ snoram的答案应该运作良好，所以请考虑这只是做同样事情的另一种方式。我是data.table的粉丝，我喜欢它的一个方面是调用函数和by调用：

library( data.table )
setDT( df )

df[ , write.csv( .SD,
                 paste0( city[1], ".csv" ),
                 row.names = FALSE ),
    by = city ]

.SD此处代表df的每个子集;每个值city一个。

city[1]从city获取.SD的第一个值（该列将一直填充一个值）。

Answer 3

你也可以使用split来破坏列表中的df，然后循环：

for (i in split(df,city)){
  write.csv(i,paste0(i$city[1],".csv"))
}

我们如何从单个df将数据导出到多个csv文件中

3 个答案: