Reputation: 3
I have a list of 1 million names and i want to look them up in each cell of a column having 150k rows. i am using Grep to lookup the names one by one and if found in any cell, make the cell blank. i am running this loop 1 million times, but it will take a lot of time. How can i speed up the loop?
install.packages("babynames")
install.packages("randomNames")
names = babynames::babynames ###creating a random dataset for this example
temp_new2= data.frame(names$name) ##temp_new2 is a single column name dataframe
random_names<-strsplit((randomNames(n=1000,
which.names="first",
name.sep=" ",
sample.with.replacement=TRUE,
return.complete.data=FALSE
)
),"\n")
count = 0
t=0
list_of_names = list()
for (i in random_names)
{
if (length(grep(paste0("\\b",i,"\\b"),temp_new2$cleaned_names,ignore.case = TRUE)) != 0)
{
p = length(grep(paste0("\\b",i,"\\b"),temp_new2$cleaned_names,ignore.case = TRUE))
print(i)
list_of_names = append(list_of_names,i)
}
else
{t=0
p=0
}
count = count + p
temp_new2[grep(paste0("\\b",i,"\\b"),temp_new2$cleaned_names,ignore.case = TRUE),]<- ""
}
it takes about 4 mins to run a loop of 1000 names, so it will take 4000 mins to run a loop of 1 million names
Upvotes: 0
Views: 69
Reputation: 7724
I played around a little bit and got the following results with microbenchmark:
microbenchmark::microbenchmark(your_fun(), fun_initialize_list(), fun_list_one_grep(), fun_lapply())
Unit: milliseconds
expr min lq mean median uq max neval
your_fun() 51.02420 52.61047 55.19147 54.20093 55.98069 77.55637 100
fun_initialize_list() 50.86644 52.81099 55.52799 54.23134 56.37564 102.21945 100
fun_list_one_grep() 25.68943 26.31398 28.51748 27.73832 28.46759 56.01566 100
fun_lapply() 25.22339 26.02261 27.83738 27.26183 27.90310 43.80443 100
The functions are defined below and are simply a wrapper around the different procedures. As @RuiBarradas already pointed out, the grep
call is execute 3 times.
Reducing this, reduces the execution time by 50% in my case.
Your approach
your_fun <- function() {
count <- 0
t <- 0
list_of_names <- list()
for (i in random_names) {
if (length(grep(paste0("\\b",i,"\\b"), temp_new2$cleaned_names,ignore.case = TRUE)) != 0) {
p <- length(grep(paste0("\\b",i,"\\b"), temp_new2$cleaned_names,ignore.case = TRUE))
list_of_names <- append(list_of_names,i)
} else {
t <- 0
p <- 0
}
count <- count + p
temp_new2[grep(paste0("\\b",i,"\\b"),temp_new2$cleaned_names,ignore.case = TRUE),] <- ""
}
}
Initializing the list before the for-loop
You are right, that did not improve the speed tremendously, probably because grep
takes so much time.
fun_initialize_list <- function() {
count <- 0
t <- 0
list_of_names <- logical(length(random_names))
k <- 0
for (i in random_names) {
k <- k + 1
if (length(grep(paste0("\\b",i,"\\b"), temp_new2$cleaned_names,ignore.case = TRUE)) != 0) {
p <- length(grep(paste0("\\b",i,"\\b"), temp_new2$cleaned_names,ignore.case = TRUE))
list_of_names[k] <- TRUE
} else {
t <- 0
p <- 0
list_of_names[k] <- FALSE
}
count <- count + p
temp_new2[grep(paste0("\\b",i,"\\b"),temp_new2$cleaned_names,ignore.case = TRUE),] <- ""
}
list_of_names <- random_names[list_of_names]
}
Using only one call for grep
fun_list_one_grep <- function() {
count <- 0
t <- 0
list_of_names <- logical(length(random_names))
k <- 0
for (i in random_names) {
k <- k + 1
name_match <- grep(paste0("\\b",i,"\\b"), temp_new2$cleaned_names, ignore.case = TRUE)
len_match <- length(name_match)
if (len_match != 0) {
p <- len_match
list_of_names[k] <- TRUE
} else {
t <- 0
p <- 0
list_of_names[k] <- FALSE
}
count <- count + p
temp_new2[name_match, ] <- ""
}
list_of_names <- random_names[list_of_names]
}
Approach with lapply
fun_lapply <- function() {
random_matches <- lapply(random_names, function(i) {
grep(paste0("\\b",i,"\\b"), temp_new2$cleaned_names, ignore.case = TRUE)
})
temp_new2[unlist(random_matches), ] <- ""
count <- length(unique(unlist(random_matches)))
list_of_names <- random_names[!sapply(random_matches, is.null)]
}
Data
names = babynames::babynames ###creating a random dataset for this example
temp_new2 = data.frame(cleaned_names = names$name[1:1000],
stringsAsFactors = FALSE) ##temp_new2 is a single column name dataframe
set.seed(23)
random_names <- strsplit((
randomNames::randomNames(
n = 100,
which.names = "first",
name.sep = " ",
sample.with.replacement = TRUE,
return.complete.data = FALSE
)), "\n")
Upvotes: 1