将数据帧写入R中的多个csv

Question

我有以下数据框，其中包含有关不同状态的信息。

long=c(-106.61291,-106.61291,-106.61291,-81.97224,-81.97224,-81.97224,-84.4277,-84.4277,-84.4277)
lat=c(35.04333,35.04333,35.04333,33.37378,33.37378,33.37378,33.64073,33.64073,33.64073)
city=c("Albuquerque","Albuquerque","Albuquerque","Augusta","Augusta","Augusta","Atlanta","Atlanta","Atlanta")
date=c("2017-08-22","2017-08-23","2017-09-24","2017-09-28","2017-10-24","2017-09-22","2017-11-12","2017-010-14","2017-09-03")
value=c(12,10.8,18.3,12.4,43,21,12,32.1,14)

df<-data.frame(long,lat,city,date,value)

问题：我想将每个城市信息写在单独的 csv 中。每个 csv 应该看起来 如下所示。 最终输出：

Albuquerque.csv
       long      lat        city       date value
1 -106.6129 35.04333 Albuquerque 2017-08-22  12.0
2 -106.6129 35.04333 Albuquerque 2017-08-23  10.8
3 -106.6129 35.04333 Albuquerque 2017-09-24  18.3

Augusta.csv
      long      lat    city       date value
1 -81.97224 33.37378 Augusta 2017-09-28  12.4
2 -81.97224 33.37378 Augusta 2017-10-24  43.0
3 -81.97224 33.37378 Augusta 2017-09-22  21.0

Atlanta.csv
      long      lat    city        date value
1 -84.4277 33.64073 Atlanta  2017-11-12  12.0
2 -84.4277 33.64073 Atlanta 2017-010-14  32.1
3 -84.4277 33.64073 Atlanta  2017-09-03  14.0

提前致谢！

Answer 1

# Split dataframe by city
split_df <- split(df, list(df$city))

# Write out separate CSV for each city
for (city in names(split_df)) {
    write.csv(split_df[[city]], paste0(city, ".csv"))
}

Answer 2

long=c(-106.61291,-106.61291,-106.61291,-81.97224,-81.97224,-81.97224,-84.4277,-84.4277,-84.4277)
lat=c(35.04333,35.04333,35.04333,33.37378,33.37378,33.37378,33.64073,33.64073,33.64073)  
city=c("Albuquerque","Albuquerque","Albuquerque","Augusta","Augusta","Augusta","Atlanta","Atlanta","Atlanta")
date=c("2017-08-22","2017-08-23","2017-09-24","2017-09-28","2017-10-24","2017-09-22","2017-11-12","2017-010-14","2017-09-03")
value=c(12,10.8,18.3,12.4,43,21,12,32.1,14)

df<-data.frame(long,lat,city,date,value)

dflist <- split(df , f = df$city)

sapply(names(dflist), 
 function (x) write.csv(dflist[[x]], file=paste(x, "csv", sep=".") )   )

Answer 3

有几种不同的方法可以做到这一点，但同时为所有城市做到这一点的一种非常快速的方法是利用基础 R 中的 apply 系列函数 - 特别是 lapply。

long=c(-106.61291,-106.61291,-106.61291,-81.97224,-81.97224,-81.97224,-84.4277,-84.4277,-84.4277)
lat=c(35.04333,35.04333,35.04333,33.37378,33.37378,33.37378,33.64073,33.64073,33.64073)
city=c("Albuquerque","Albuquerque","Albuquerque","Augusta","Augusta","Augusta","Atlanta","Atlanta","Atlanta")
date=c("2017-08-22","2017-08-23","2017-09-24","2017-09-28","2017-10-24","2017-09-22","2017-11-12","2017-010-14","2017-09-03")
value=c(12,10.8,18.3,12.4,43,21,12,32.1,14)

df<-data.frame(long,lat,city,date,value)

# create a convenience function to split your data and export to csv
split_into_csv <- function(x) {
tmp <- df[df$city == x,]
write.csv(tmp, file = paste0(x,".csv"))}

# Apply split_into_csv over elements of list with lapply
lapply(levels(df$city), split_into_csv)
# Check output in director
dir()
[1] "Albuquerque.csv" "Atlanta.csv"     "Augusta.csv"  

Answer 4

您可以使用 base R 或 dplyr 包来做到这一点。

dplyr 方式：

dplyr::filter(df, city == 'Albuquerque') %>% write.csv(file = 'Albuquerque.csv', row.names = FALSE)
dplyr::filter(df, city == 'Augusta') %>% write.csv(file = 'Augusta.csv', row.names = FALSE)
dplyr::filter(df, city == 'Atlanta') %>% write.csv(file = 'Atlanta.csv', row.names = FALSE)

基础R：

write.csv(df[df$city == 'Albuquerque', ], file = 'Albuquerque.csv', row.names = FALSE)
write.csv(df[df$city == 'Augusta', ], file = 'Augusta.csv', row.names = FALSE)
write.csv(df[df$city == 'Atlanta', ], file = 'Atlanta.csv', row.names = FALSE)

如果您开始获取更多城市，您可以使用 for 循环。

for (city in c('Albuquerque', 'Augusta', 'Atlanta')) {
  write.csv(df[df$city == city, ], file = paste0(city, '.csv'))
}