Question

我有一个一百万个名字的列表，我想在具有15万行的列的每个单元格中查找它们。我正在使用Grep逐一查找名称，如果在任何单元格中找到，请将单元格留空。我正在运行此循环1百万次，但是会花费很多时间。我如何加快循环速度？

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),]<- ""

    }

运行1000个名称的循环大约需要4分钟，因此运行100万个名称的循环大约需要4000分钟

Answer 1

我玩了一些，并用微基准测试获得了以下结果：

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

这些函数在下面定义，它们只是不同过程的包装。正如@RuiBarradas指出的那样，grep调用将执行3次。减少这种情况，就我而言，将执行时间减少了50％。

您的方法

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),] <- ""
  }

}

在for循环之前初始化列表
没错，这并没有极大地提高速度，可能是因为grep花费了很多时间。

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]
}

仅使用一次呼叫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]
}

强加于人的方法

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)]
}

数据

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")

Grep在for循环中以一百万次迭代检测名称，太慢

1 个答案: