R-从现有列值创建和命名数据框

Question

I have a data frame that is structured like so, via dput :

structure(list(railroad = c("bnsf railway company", "bnsf railway company", 
"bnsf railway company", "bnsf railway company", "bnsf railway company", 
"bnsf railway company", "bnsf railway company", "bnsf railway company", 
"union pacific railroad", "union pacific railroad", "union pacific railroad", 
"union pacific railroad", "union pacific railroad", "union pacific railroad", 
"union pacific railroad", "union pacific railroad"), measure = 
c("cars.owned.by", 
"cars.owned.by", "cars.type", "cars.type", "cars.type", "train.speed", 
"train.speed", "terminal.dwell", "cars.owned.by", "cars.owned.by", 
"cars.type", "cars.type", "cars.type", "train.speed", "train.speed", 
"terminal.dwell"), category = c("system", "private", "box", "intermodal", 
"total", "intermodal", "all.trains", "entire.railroad", "system", 
"private", "box", "intermodal", "total", "intermodal", "all.trains", 
"entire.railroad"), irm = c(201510L, 201510L, 201510L, 201510L, 
201510L, 201510L, 201510L, 201510L, 201510L, 201510L, 201510L, 
201510L, 201510L, 201510L, 201510L, 201510L), mean = c(66623, 
149937.333, 11395, 16499, 236866, 33.3, 24.5, 25.267, 57618.333, 
195764.667, 22229.333, 14135.333, 293164.333, 31.933, 26.6, 27.6
)), row.names = c(1L, 3L, 6L, 9L, 14L, 15L, 20L, 32L, 127L, 129L, 
132L, 135L, 140L, 141L, 146L, 160L), class = "data.frame")

What I would like to do is the following:

1. Create separate data frames for each combination of measure and category , named by pasting measure and category separated by "." . So the first data frame would be called cars.owned.by.system and so on.

2. Rename the fifth column, mean of each data frame to the name of the data frame itself. So, for the first data frame it would be colnames(df)[5] <- cars.owned.by.system .

The desired output is 8 separate data frames, named as I mentioned above

I tried the following:

cars.owned.by.system <- df[df$category == "system",]
colnames(cars.owned.by.system)[5[ <- cars.owned.by.system

And it does the job, but I don't want to have to do this repetitively. I imagine there is a version of the canonical split-apply-combine approach that would work. Any advice or help would be much appreciated. Thanks.

Answer 1

Assuming df is your dataframe, I think this does it.

for(cat in unique(df$category)) {
  newdf<-paste("cars.owned.by.", cat, sep="")
  assign(newdf, df[df$category==cat,])
  eval(parse(text=paste("colnames(", newdf, ")[5] <- '", newdf, "'", sep="")))
}

Answer 2

What about a classical for loop:

# first create the pasted name to iterate the loop 
df$name <- paste(df$railroad,df$measure,sep='.')

# an empty list to have all your df
list_df <- list()

# the loop
for (i in df$name){
data <- df[which(df$name == i),]  # select the df of name
colnames(data)[4]<-i              # rename the mean
data<- data[,-5]                  # remove the useless name
list_df[[i]] <- data              # store in list
}

# here you can see all the df in a list
list_df

> list_df
$`bnsf railway company.cars.owned.by`
              railroad       measure category bnsf railway company.cars.owned.by                               name
1 bnsf railway company cars.owned.by   system                             201510 bnsf railway company.cars.owned.by
3 bnsf railway company cars.owned.by  private                             201510 bnsf railway company.cars.owned.by

$`bnsf railway company.cars.type`
               railroad   measure   category bnsf railway company.cars.type                           name
6  bnsf railway company cars.type        box                         201510 bnsf railway company.cars.type
9  bnsf railway company cars.type intermodal                         201510 bnsf railway company.cars.type
14 bnsf railway company cars.type      total                         201510 bnsf railway company.cars.type
... and so on  

# you can select each df, for example choosin its name
list_df$`bnsf railway company.cars.type`
                    railroad   measure   category bnsf railway company.cars.type                           name
6  bnsf railway company cars.type        box                         201510 bnsf railway company.cars.type
9  bnsf railway company cars.type intermodal                         201510 bnsf railway company.cars.type
14 bnsf railway company cars.type      total                         201510 bnsf railway company.cars.type

# and you're sure it's a df
class(list_df$`bnsf railway company.cars.type`)
[1] "data.frame"

Answer 3

Consider split to subset data frame by the two factors and then Map (wrapper to mapply ) to iterate elementwise through subsetted data frames and list's names.

Also, consider setNames() the left-hand version of colnames() to return the new named object in one call.

# CREATES NAMED LIST
df_list <- split(df, list(df$measure, df$category))

# RETURNS SAME LIST WITH RENAMED FIFTH COLUMN
df_list <- Map(function(sub, nm) setNames(sub, c("railroad", "measure", "category", "irm", nm)), 
               df_list, names(df_list))

# OUTPUT DFs 
df_list$cars.owned.by.all.trains

df_list$cars.type.all.trains

df_list$terminal.dwell.all.trains 
...

Answer 4

This will give you a named list of dataframes, which is almost certainly preferable to having them all separately in your global environment:

lst <- split(df, paste(df$measure, df$category, sep = ".")) %>% 
  purrr::imap(~`names<-`(.x, c(names(.x)[1:4], .y)))