How to group rows in into unique id [duplicate] - r

This question already has answers here:
How to get the maximum value by group
(5 answers)
Summarizing multiple columns with dplyr? [duplicate]
(5 answers)
Closed 8 months ago.
I have 1000+ rows shows like below
ID V1 V2 V3 V4 V5 V6 V7
KCT00094653 0 0 0 0 0 1 0
KCT00094653 0 0 0 1 0 0 0
KCT00094653 0 0 0 0 0 0 1
KCT00094653 0 0 1 0 0 0 0
KCT00140855 0 0 0 0 1 0 0
KCT00140855 0 0 0 0 0 0 1
KCT00140855 0 0 0 0 0 1 0
KCT00140855 0 0 0 1 0 0 0
KCT00162123 0 1 0 0 0 0 0
KCT00162123 1 0 0 0 0 0 0
KCT00323882 0 0 0 0 0 1 0
KCT00323882 0 0 0 1 0 0 0
KCT00323882 0 0 0 0 0 0 1
KCT00323882 0 0 0 0 1 0 0
KCT00323882 0 1 0 0 0 0 0
I trying to get all the 1 in a row for single Id
and the expected output is like below
ID V1 V2 V3 V4 V5 V6 V7
KCT00094653 0 0 1 1 0 1 1
KCT00140855 0 0 0 1 1 1 1
KCT00162123 1 1 0 0 0 0 0
KCT00323882 0 1 0 1 1 1 1

Related

R: Simulating ERGM model in R then generate adjacency matrix of that model

I use library(ergm) and library(igraph) and generate a ERGM network. But I want the adjacency matrix of that network. I am unable to find any function which can produce that.
library(ergm)
library(igraph)
g.use <- network(16,density=0.1,directed=FALSE)
#
# Starting from this network let's draw 3 realizations
# of a edges and 2-star network
#
g.sim <- simulate(~edges+kstar(2), nsim=3, coef=c(-1.8,0.03),
basis=g.use, control=control.simulate(
MCMC.burnin=1000,
MCMC.interval=100))
#g.sim[[3]]
summary(g.sim)
Is it possible to find the adjacency matrix from g.sim? and how?

EGRM package uses the network package and not the igraph package. You should maintain everythig in network and not load igraph as the two have some conflicting functions with same names.
In your case, you simulate 3 graphs thus you should have 3 adjacency matrices. The code is as below:
library(ergm)
g.use <- network(16,density=0.1,directed=FALSE)
g.sim <- simulate(~edges+kstar(2), nsim=3, coef=c(-1.8,0.03),
basis=g.use, control=control.simulate(
MCMC.burnin=1000,
MCMC.interval=100))
The code you want:
lapply(g.sim, as.matrix)
[[1]]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0
3 0 0 0 1 1 0 1 0 0 0 0 0 1 0 0 1
4 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0
5 1 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0
6 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0
7 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1
8 0 1 0 0 0 0 0 0 0 1 1 1 1 0 1 0
9 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1
10 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0
11 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0
12 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
13 0 0 1 0 1 0 0 1 0 1 1 0 0 0 0 1
14 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0
15 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
16 0 0 1 0 0 0 1 0 1 0 0 0 1 0 0 0
[[2]]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0
2 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
3 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0
4 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
5 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
6 0 0 0 0 0 0 0 1 1 0 1 1 1 0 0 1
7 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0
8 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0
9 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0
10 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0
11 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0
12 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1
13 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0
14 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
15 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0
16 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0
[[3]]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1
2 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0
3 0 0 0 0 1 0 0 0 0 1 1 0 0 0 1 0
4 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
5 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
6 0 1 0 0 0 0 1 0 1 0 0 0 1 0 1 0
7 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0
8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
10 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1
11 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1
12 0 1 0 0 0 0 0 0 0 0 0 0 1 1 1 0
13 1 1 0 1 0 1 0 0 0 0 0 1 0 0 0 0
14 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0
15 0 0 1 0 0 1 1 0 0 1 0 1 0 0 0 1
16 1 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0

Multiplying multiple columns with each other into a new dataframe in R

I want to multiply many of my binary variables into new columns, so called interactive variables. My dataset is structured like this:
YearCountry <- data.frame( Time = c("2000","2001", "2002", "2003",
"2000","2001", "2002", "2003",
"2000","2001", "2002", "2003"),
AL = c(1,1,1,1,0,0,0,0,0,0,0,0),
FR = c(0,0,0,0,1,1,1,1,0,0,0,0),
UK = c(0,0,0,0,0,0,0,0,1,1,1,1),
Y2000d = c(1,0,0,0,1,0,0,0,1,0,0,0),
Y2001d = c(0,1,0,0,0,1,0,0,0,1,0,0),
Y2002d = c(0,0,1,0,0,0,1,0,0,0,1,0),
Y2003d = c(0,0,0,1,0,0,0,1,0,0,0,1))
YearCountry
Time AL FR UK Y2000d Y2001d Y2002d Y2003d
1 2000 1 0 0 1 0 0 0
2 2001 1 0 0 0 1 0 0
3 2002 1 0 0 0 0 1 0
4 2003 1 0 0 0 0 0 1
5 2000 0 1 0 1 0 0 0
6 2001 0 1 0 0 1 0 0
7 2002 0 1 0 0 0 1 0
8 2003 0 1 0 0 0 0 1
9 2000 0 0 1 1 0 0 0
10 2001 0 0 1 0 1 0 0
11 2002 0 0 1 0 0 1 0
12 2003 0 0 1 0 0 0 1
I need to multiply the binary variable for each of the countries (AL,FR,UK) with each of the binary variables for a given year so that I get #country x #year new variables. In this case I have three countries and four years which gives 12 new variables. My full data contains 105 countries/regions and stretches over twenty years. I therefore need a general formula. I want data that looks like this
Interact <- data.frame(Time = c("2000","2001", "2002", "2003",
"2000","2001", "2002", "2003",
"2000","2001", "2002", "2003"),
Y2000xAL = c(1,0,0,0,0,0,0,0,0,0,0,0),
Y2001xAL = c(0,1,0,0,0,0,0,0,0,0,0,0),
Y2002xAL = c(0,0,1,0,0,0,0,0,0,0,0,0),
Y2003xAL = c(0,0,0,1,0,0,0,0,0,0,0,0),
Y2000xFR = c(0,0,0,0,1,0,0,0,0,0,0,0),
Y2001xFR = c(0,0,0,0,0,1,0,0,0,0,0,0),
Y2002xFR = c(0,0,0,0,0,0,1,0,0,0,0,0),
Y2003xFR = c(0,0,0,0,0,0,0,1,0,0,0,0),
Y2000xUk = c(0,0,0,0,0,0,0,0,1,0,0,0),
Y2001xUK = c(0,0,0,0,0,0,0,0,0,1,0,0),
Y2002xUK = c(0,0,0,0,0,0,0,0,0,0,1,0),
Y2003xUK = c(0,0,0,0,0,0,0,0,0,0,0,1))
Interact
Time Y2000xAL Y2001xAL Y2002xAL Y2003xAL Y2000xFR Y2001xFR Y2002xFR Y2003xFR Y2000xUk Y2001xUK Y2002xUK Y2003xUK
1 2000 1 0 0 0 0 0 0 0 0 0 0 0
2 2001 0 1 0 0 0 0 0 0 0 0 0 0
3 2002 0 0 1 0 0 0 0 0 0 0 0 0
4 2003 0 0 0 1 0 0 0 0 0 0 0 0
5 2000 0 0 0 0 1 0 0 0 0 0 0 0
6 2001 0 0 0 0 0 1 0 0 0 0 0 0
7 2002 0 0 0 0 0 0 1 0 0 0 0 0
8 2003 0 0 0 0 0 0 0 1 0 0 0 0
9 2000 0 0 0 0 0 0 0 0 1 0 0 0
10 2001 0 0 0 0 0 0 0 0 0 1 0 0
11 2002 0 0 0 0 0 0 0 0 0 0 1 0
12 2003 0 0 0 0 0 0 0 0 0 0 0 1

Here's an approach with dplyr::across. We can make the final result into a plain data.frame with purrr:invoke as demonstrated in this answer.
library(dplyr)
library(purrr)
YearCountry %>%
mutate(across(AL:UK, ~ . * select(cur_data(), Y2000d:Y2003d))) %>%
select(-(Y2000d:Y2003d)) %>%
invoke(.f = data.frame) %>%
rename_with(~str_replace(.,"\\.",""))
Time ALY2000d ALY2001d ALY2002d ALY2003d FRY2000d FRY2001d FRY2002d FRY2003d UKY2000d UKY2001d UKY2002d UKY2003d
1 2000 1 0 0 0 0 0 0 0 0 0 0 0
2 2001 0 1 0 0 0 0 0 0 0 0 0 0
3 2002 0 0 1 0 0 0 0 0 0 0 0 0
4 2003 0 0 0 1 0 0 0 0 0 0 0 0
5 2000 0 0 0 0 1 0 0 0 0 0 0 0
6 2001 0 0 0 0 0 1 0 0 0 0 0 0
7 2002 0 0 0 0 0 0 1 0 0 0 0 0
8 2003 0 0 0 0 0 0 0 1 0 0 0 0
9 2000 0 0 0 0 0 0 0 0 1 0 0 0
10 2001 0 0 0 0 0 0 0 0 0 1 0 0
11 2002 0 0 0 0 0 0 0 0 0 0 1 0
12 2003 0 0 0 0 0 0 0 0 0 0 0 1

1) model.matrix We split the names by the number of characters in them (the countries have 2 characters in their names and the years have 6) and paste pluses in each. (Alternately use Plus(grep("^..$", nms, value = TRUE)) to get the country names and use that in place of spl["2"] and similarly Plus(grep("^Y....d$", nms, value = TRUE)) in place of spl["6"].)
c(`2` = "AL+FR+UK", `6` = "Y2000d+Y2001d+Y2002d+Y2003d")
and from that the formula:
~(AL + FR + UK):(Y2000d + Y2001d + Y2002d + Y2003d) + 0
and then compute its model matrix.
The formula could also be expanded to one accepted by lm by modifying the sprintf format so we might not even need to create the model matrix. For example, if we had a response vector R then we could write: s <- sprintf("R ~ (%s)*(%s)", spl["2"], spl["4"]); fo <- formula(s); lm(fo, YearCountry) to include all variables and the interactions of countries and year as well as an intercept.
Plus <- function(x) paste(x, collapse = "+")
nms <- names(YearCountry)[-1]
spl <- sapply(split(nms, nchar(nms)), Plus)
s <- sprintf("~ (%s):(%s)+0", spl["2"], spl["6"])
fo <- formula(s)
model.matrix(fo, YearCountry)
giving this matrix:
AL:Y2000d AL:Y2001d AL:Y2002d AL:Y2003d FR:Y2000d FR:Y2001d FR:Y2002d FR:Y2003d UK:Y2000d UK:Y2001d UK:Y2002d UK:Y2003d
1 1 0 0 0 0 0 0 0 0 0 0 0
2 0 1 0 0 0 0 0 0 0 0 0 0
3 0 0 1 0 0 0 0 0 0 0 0 0
4 0 0 0 1 0 0 0 0 0 0 0 0
5 0 0 0 0 1 0 0 0 0 0 0 0
6 0 0 0 0 0 1 0 0 0 0 0 0
7 0 0 0 0 0 0 1 0 0 0 0 0
8 0 0 0 0 0 0 0 1 0 0 0 0
9 0 0 0 0 0 0 0 0 1 0 0 0
10 0 0 0 0 0 0 0 0 0 1 0 0
11 0 0 0 0 0 0 0 0 0 0 1 0
12 0 0 0 0 0 0 0 0 0 0 0 1
attr(,"assign")
[1] 1 2 3 4 5 6 7 8 9 10 11 12
Alternately we can write it compactly like this:
Plus <- function(x) paste(x, collapse = "+")
nms <- names(YearCountry)
s <- sprintf("~ (%s):(%s)+0", Plus(nms[2:4]), Plus(nms[5:8]))
fo <- formula(s)
model.matrix(fo, YearCountry)
2) eList Another approach is to use list comprehensions. With the eList package we can do this:
library(eList)
DF(for(i in YearCountry[2:4]) for(j in YearCountry[5:8]) i*j)
giving this data frame. Use as.matrix(...) on it if you want a matrix.
AL.Y2000d AL.Y2001d AL.Y2002d AL.Y2003d FR.Y2000d FR.Y2001d FR.Y2002d FR.Y2003d UK.Y2000d UK.Y2001d UK.Y2002d UK.Y2003d
1 1 0 0 0 0 0 0 0 0 0 0 0
2 0 1 0 0 0 0 0 0 0 0 0 0
3 0 0 1 0 0 0 0 0 0 0 0 0
4 0 0 0 1 0 0 0 0 0 0 0 0
5 0 0 0 0 1 0 0 0 0 0 0 0
6 0 0 0 0 0 1 0 0 0 0 0 0
7 0 0 0 0 0 0 1 0 0 0 0 0
8 0 0 0 0 0 0 0 1 0 0 0 0
9 0 0 0 0 0 0 0 0 1 0 0 0
10 0 0 0 0 0 0 0 0 0 1 0 0
11 0 0 0 0 0 0 0 0 0 0 1 0
12 0 0 0 0 0 0 0 0 0 0 0 1
3) listcompr listcompr is another list comprehension package. Note that the development version of this package is needed in order to use bycol=. Replace gen.named.matrix with gen.named.data.frame if you want a data frame.
# devtools::github_github("patrickroocks/listcompr")
library(listcompr)
nms <- names(YearCountry)
gen.named.matrix("{nms[i]}.{nms[j]}", YearCountry[[i]] * YearCountry[[j]],
i = 2:4, j = 5:8, bycol = TRUE)

Filling a table with additional columns if they don't exist

I've the following difficult problem. Here short example of my data. Assume that I've two data sets (my real example has something about 20). The data frames result as a list computed by a self written function with lapply. So, I put the data frames in my example in a list, too. Then I "rbind" them to compute a frequency table.
df1 <- data.frame(rev(seq(12:0)), paste0("a=",sample(0:12, 13, replace=T)))
colnames(df1) <- c("k", "a")
df2 <- data.frame(rev(seq(12:0)), paste0("a=",sample(0:12, 13, replace=T)))
colnames(df2) <- c("k", "a")
list_df <- list(df1,df2)
df_combine<- plyr::ldply(list_df, rbind)
freq_foo <- table(df_combine$k,df_combine$a)
I get a frequency table of the following form.
a=0 a=11 a=12 a=2 a=5 a=6 a=7 a=8 a=3 a=9
1 1 0 0 0 0 0 0 1 0 0
2 1 0 0 0 0 0 0 0 0 1
3 1 0 0 0 0 1 0 0 0 0
4 0 0 0 1 0 1 0 0 0 0
5 0 0 0 1 1 0 0 0 0 0
6 0 0 0 0 0 0 1 0 0 1
7 0 1 1 0 0 0 0 0 0 0
8 1 0 0 0 0 1 0 0 0 0
9 0 0 0 0 0 0 2 0 0 0
10 0 0 1 0 1 0 0 0 0 0
11 1 1 0 0 0 0 0 0 0 0
12 0 0 0 0 0 0 1 0 1 0
13 1 0 1 0 0 0 0 0 0 0
I want to extend and manipulate my table in the following way:
First the table should go over a range of a=0 to a=15. So if there is a missing column, it should be added. And 2nd) I want to order the columns from 0 to 15.
For the first problem I tried
if(freq_foo$paste0("a=",0:15) == F){freq_foo$paste("a=",0:15) <- 0}
but this should work only for data frames and not for tables. Also. i've no idea how to order the columns with an ascending order. The data type isnt important to me because I just want to use the output for further calculations. So, it can also be a data frame instead of a table.

#convert freq_foo table to dataframe
df <- as.data.frame.matrix(freq_foo)
#add all zeros column for missing column name in 0:15 series
df[, paste0("a=", c(0:15)[!(c(0:15) %in% as.numeric(gsub(".*=(\\d+)", "\\1", names(df))))])] <- 0
#order columns from 0 to 15
df <- df[, order(as.numeric(gsub(".*=(\\d+)", "\\1", names(df))))]
Output is:
a=0 a=1 a=2 a=3 a=4 a=5 a=6 a=7 a=8 a=9 a=10 a=11 a=12 a=13 a=14 a=15
1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0
2 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0
3 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0
4 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0
5 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0
6 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
7 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0
8 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0
9 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0
10 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
11 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
12 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0
13 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0
(Edit: Updated code after getting a requirement clarification from OP)

r row names selection using columns

Suppose i have this matrix
0 1 2 3 4 5 6 98 183 385 419 420 422 423 469 470 35698 35709 35729 37415
0 0 1 1 1 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1
1 1 0 1 0 0 1 1 0 0 1 0 0 0 0 1 0 0 0 0 0
2 1 1 0 1 1 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0
3 1 0 1 0 1 1 0 1 1 0 1 1 1 1 0 0 1 0 0 1
4 0 0 1 1 0 1 1 1 0 0 1 1 1 0 0 1 0 1 1 0
5 0 1 0 1 1 0 1 1 0 0 0 1 0 0 0 1 0 0 1 0
6 1 1 0 0 1 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0
98 0 0 0 1 1 1 1 0 0 0 0 1 0 0 0 1 0 0 1 0
183 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1
385 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
419 0 0 1 1 1 0 0 0 0 0 0 1 1 0 0 0 1 1 0 0
420 0 0 0 1 1 1 0 1 0 0 1 0 0 0 0 0 1 1 0 0
422 0 0 1 1 1 0 0 0 1 0 1 0 0 1 1 0 0 0 0 1
423 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1
469 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1
470 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0
35698 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0
35709 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0
35729 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0
37415 1 0 0 1 0 0 0 0 1 0 0 0 1 1 1 0 0 0 0 0
I am getting a value from another program let us say
x=3.
I want to choose the name of rows where x == 1 i.e. where the value of 3 is 1.
Output will be : 0,2,4,5,98,183,419,420,422,423,35698,37415.
And I don't want to pass "3" directly into the command. I want to pass the variable x so that if this number varies I could get the output accordingly.
Can anyone help me, please? thanks in advance

x=matrix(c(1,1,2,5,6,6,5,7,7,8,3,3,1,9,20,20,4,7,9,5),4,5,dimnames = list(c(letters[1:4]),c(LETTERS[1:5])))
you'r requirement is row names then
rownames(x)[x[,"D"]==20]
here '20' is you'r input value and D is you'r searching column.

using lappy and elseif command

Using R I have a table, lets say 'locations'
head(locations, n=10)
apillar fender fwheel fdoor compart rdoor rwheel boot
1 0 0 0 0 0 0 0 1
2 0 0 0 1 0 0 0 0
3 0 0 0 0 1 0 0 0
4 0 1 0 0 0 0 0 0
5 1 0 1 0 0 0 0 0
6 1 0 0 1 0 0 0 0
7 0 0 0 0 0 0 0 0
8 0 0 0 0 1 0 0 0
9 0 0 0 1 0 0 0 0
10 0 0 0 0 0 1 0 0
now i want to create a new variable "cat" which groups the impacts into category locations.
I have been using if, elseif and else command, but I cannot get it to work.
The command is:
cat <- lapply(locations, function(x) if (apillar|fender|fwheel == 1)print("front") else if (fdoor|compart|rdoor == 1)print("middle") else if(rwheel|boot ==1)print("rear") else print("NA")
such that cat should read rear, middle, middle, middle, front etc

When vectors of TRUE or FALSE statements are involved, I usually prefer not to work with if to avoid loops. I find conditional referencing to be more elegant in this case. See below.
locations <- read.table(header=TRUE, text=
"apillar fender fwheel fdoor compart rdoor rwheel boot
1 0 0 0 0 0 0 0 1
2 0 0 0 1 0 0 0 0
3 0 0 0 0 1 0 0 0
4 0 1 0 0 0 0 0 0
5 1 0 1 0 0 0 0 0
6 1 0 0 1 0 0 0 0
7 0 0 0 0 0 0 0 0
8 0 0 0 0 1 0 0 0
9 0 0 0 1 0 0 0 0
10 0 0 0 0 0 1 0 0")
locations$cat <- NA
within(locations,{
cat[apillar|fender|fwheel] <- "front"
cat[fdoor|compart|rdoor] <- "middle"
cat[rwheel|boot] <- "rear"
})
Result:
apillar fender fwheel fdoor compart rdoor rwheel boot cat
1 0 0 0 0 0 0 0 1 rear
2 0 0 0 1 0 0 0 0 middle
3 0 0 0 0 1 0 0 0 middle
4 0 1 0 0 0 0 0 0 front
5 1 0 1 0 0 0 0 0 front
6 1 0 0 1 0 0 0 0 middle
7 0 0 0 0 0 0 0 0 <NA>
8 0 0 0 0 1 0 0 0 middle
9 0 0 0 1 0 0 0 0 middle
10 0 0 0 0 0 1 0 0 middle
Cheers!

Corrected your own code:
locations$cat= with(locations, ifelse(apillar|fender|fwheel, "front", ifelse(fdoor|compart|rdoor,"middle",ifelse(rwheel|boot, "rear", "NA"))) )
> locations
apillar fender fwheel fdoor compart rdoor rwheel boot cat
1 0 0 0 0 0 0 0 1 rear
2 0 0 0 1 0 0 0 0 middle
3 0 0 0 0 1 0 0 0 middle
4 0 1 0 0 0 0 0 0 front
5 1 0 1 0 0 0 0 0 front
6 1 0 0 1 0 0 0 0 front
7 0 0 0 0 0 0 0 0 NA
8 0 0 0 0 1 0 0 0 middle
9 0 0 0 1 0 0 0 0 middle
10 0 0 0 0 0 1 0 0 middle
>

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