如何将多个列与grep合并并求和r

Question

I have following dataframe in r

Engine   General   Ladder.winch   engine.phe   subm.gear.box   aux.engine   pipeline.maintain    pipeline    pipe.line    engine.mpd
 1        12        22             2            4               2             4                    5            6             7

and so on with more than 10000 rows.

Now,I want to combine columns and add values to reduce the columns into broader categories. eg Engine,engine.phe,aux.engine,engine.mpd should be combined into Engine category and all the values to be added. likewise pipeline.maintain,pipeline,pipe.line to be combined into Pipeline And rest columns to be added under General Category.

Desired dataframe would be

 Engine      Pipeline       General
   12          15             38

How can I do it in r?

Answer 1

Many ways in which you can do it, this is a more straight forward approach

# Example data.frame
dtf <- structure(list(Engine = c(1, 0, 1), 
   General = c(12, 3, 15), Ladder.winch = c(22, 28, 26), 
    engine.phe = c(2, 1, 0), subm.gear.box = c(4, 4, 10), 
    aux.engine = c(2, 3, 1), pipeline.maintain = c(4, 5, 1), 
    pipeline = c(5, 5, 2), pipe.line = c(6, 8, 2), engine.mpd = c(7, 8, 19)),
    .Names = c("Engine", "General", "Ladder.winch", "engine.phe", 
      "subm.gear.box", "aux.engine", "pipeline.maintain", 
      "pipeline", "pipe.line", "engine.mpd"), 
    row.names = c(NA, -3L), class = "data.frame")

with(dtf, data.frame(Engine=Engine+engine.phe+aux.engine+engine.mpd,
                   Pipeline=pipeline.maintain+pipeline+pipe.line,
                    General=General+Ladder.winch+subm.gear.box))

#   Engine Pipeline General
# 1     12       15      38
# 2     12       18      35
# 3     21        5      51

# a more generalized and 'greppy' solution
cnames <- tolower(colnames(dtf))
data.frame(Engine=rowSums(dtf[, grep("eng", cnames)]),
         Pipeline=rowSums(dtf[, grep("pip", cnames)]),
          General=rowSums(dtf[, !grepl("eng|pip", cnames)]))

Answer 2

It is mostly better to store you data in long format. Therefore, my proposal would to approach your problem as below:

1 - get your data in long format

library(reshape2)
dfl <- melt(df)

2 - create 'engine' and 'pipeline'-vectors

e_vec <- c("Engine","engine.phe","aux.engine","engine.mpd")
p_vec <- c("pipeline.maintain","pipeline","pipe.line")

3 - create a category column

dfl$newcat <- c("general","engine","pipeline")[1 + dfl$variable %in% e_vec + 2*(dfl$variable %in% p_vec)]

The result:

> dfl
            variable value   newcat
1             Engine     1   engine
2            General    12  general
3       Ladder.winch    22  general
4         engine.phe     2   engine
5      subm.gear.box     4  general
6         aux.engine     2   engine
7  pipeline.maintain     4 pipeline
8           pipeline     5 pipeline
9          pipe.line     6 pipeline
10        engine.mpd     7   engine

Now you can use aggregate to get the final result:

> aggregate(value ~ newcat, dfl, sum)
    newcat value
1   engine    12
2  general    38
3 pipeline    15

Answer 3

Here is an option by extracting the concerned words from the names of the column, and using tapply to get the sum . The str_extract_all returns a list ('lst'). Replace those elements which are having zero length with 'GENERAL', Then, using a group by function ie tapply , unlist the dataset, and use the grouping variables ie replicated 'lst' and the row of 'df1' get the sum

library(stringr)
lst <- str_extract_all(toupper(sub("(pipe)\\.", "\\1", names(df1))),
          "ENGINE|PIPELINE|GENERAL")
lst[lengths(lst)==0] <- "GENERAL"
t(tapply(unlist(df1), list(unlist(lst)[col(df1)], row(df1)), FUN = sum))
#   ENGINE  GENERAL PIPELINE 
#1      12       38       15 

Answer 4

myfactors = ifelse(grepl("engine", names(df), ignore.case = TRUE), "Engine",
                   ifelse(grepl("pipe|pipeline", names(df), ignore.case = TRUE), "Pipeline",
                          "General"))
data.frame(lapply(split.default(df, myfactors), rowSums))
#  Engine General Pipeline
#1     12      38       15
#2     12      35       18
#3     21      51        5

df is the data from this answer