如何避免复杂的for循环？

Question

我目前正在使用一系列大型数据集,并且我正在努力改进我在R中编写脚本的方式.我倾向于主要使用for循环,我知道这些循环很麻烦且很慢,尤其是非常大的数据集.

我听过很多人推荐apply()系列来避免复杂的for循环,但是我很难用它们一次性应用多个函数.

这是一些简单的示例数据:

A <- data.frame('Area' = c(4, 6, 5),
                'flow' = c(1, 1, 1))
B <- data.frame('Area' = c(6, 8, 4),
                'flow' = c(1, 2, 1))
files <- list(A, B)
frames <- list('A', 'B')

我想要做的是通过'flow'变量对数据进行排序,然后为每个数据点所代表的总'flow'和'area'部分添加列,最后再添加两列每个变量的累积百分比.

目前我用这个循环:

sort_files <- list()
n <- 1
for(i in files){
  name <- frames[n]
  nom <- paste(name,'_sorted', sep = '')
  data <- i[order(-i$flow),]
  area <- sum(i$Area)
  total <- sum(i$flow)
  data$area_portion <- (data$Area/area)*100
  data$flow_portion <- (data$flow/total)*100
  data$cum_area <- cumsum(data$area_portion)
  data$cum_flow <- cumsum(data$flow_portion)
  assign(nom, data)
  df <- get(paste(name,'_sorted', sep = ''))
  sort_files[[nom]] <- df
  n <- n + 1
}

哪个有效,但看起来过于复杂和丑陋,而且我相信它会比更好的脚本运行得慢得多.

如何简化和排除上述代码？

这是预期的输出:

sort_files

$`A_sorted`
  Area flow area_portion flow_portion  cum_area  cum_flow
1    4    1     26.66667     33.33333  26.66667  33.33333
2    6    1     40.00000     33.33333  66.66667  66.66667
3    5    1     33.33333     33.33333 100.00000 100.00000

$B_sorted
  Area flow area_portion flow_portion  cum_area cum_flow
2    8    2     44.44444           50  44.44444       50
1    6    1     33.33333           25  77.77778       75
3    4    1     22.22222           25 100.00000      100

Answer 1

使用lapply了循环files,并dplyr mutate增加新的列

library(dplyr)

setNames(lapply(files, function(x) 
          x %>%
            arrange(desc(flow)) %>%
            mutate(area_portion = Area/sum(Area)*100, 
                   flow_portion = flow/sum(flow) * 100, 
                   cum_area = cumsum(area_portion),
                   cum_flow = cumsum(flow_portion))
),paste0(frames, "_sorted"))


#$A_sorted
#  Area flow area_portion flow_portion  cum_area  cum_flow
#1    4    1     26.66667     33.33333  26.66667  33.33333
#2    6    1     40.00000     33.33333  66.66667  66.66667
#3    5    1     33.33333     33.33333 100.00000 100.00000

#$B_sorted
#  Area flow area_portion flow_portion  cum_area cum_flow
#1    8    2     44.44444           50  44.44444       50
#2    6    1     33.33333           25  77.77778       75
#3    4    1     22.22222           25 100.00000      100

或完全去tidyverse的方式,我们可以改变lapply与map和setNames与set_names

library(tidyverse)

map(set_names(files, str_c(frames, "_sorted")), 
  . %>% arrange(desc(flow)) %>%
  mutate(area_portion = Area/sum(Area)*100, 
         flow_portion = flow/sum(flow) * 100, 
         cum_area = cumsum(area_portion),
         cum_flow = cumsum(flow_portion)))

tidyverse根据@Moody_Mudskipper的一些指示更新了方法.

Answer 2

你也可以先定义一个函数..

f <- function(data) {

  # sort data by flow
  data <- data[order(data['flow'], decreasing = TRUE), ]

  # apply your functions
  data["area_portion"] <- data['Area'] / sum(data['Area']) * 100
  data["flow_portion"] <- data['flow'] / sum(data['flow']) * 100
  data["cum_area"] <- cumsum(data['area_portion'])
  data["cum_flow"] <- cumsum(data['flow_portion'])
  data
  }

..并使用lapply,啊,适用f于您的列表

out <- lapply(files, f)
out
#[[1]]
#  Area flow area_portion flow_portion  cum_area  cum_flow
#1    4    1     26.66667     33.33333  26.66667  33.33333
#2    6    1     40.00000     33.33333  66.66667  66.66667
#3    5    1     33.33333     33.33333 100.00000 100.00000

#[[2]]
#  Area flow area_portion flow_portion  cum_area cum_flow
#2    8    2     44.44444           50  44.44444       50
#1    6    1     33.33333           25  77.77778       75
#3    4    1     22.22222           25 100.00000      100

如果要更改名称,out可以使用setNames

out <- setNames(lapply(files, f), paste0(c("A", "B"), "_sorted"))
# or
# out <- setNames(lapply(files, f), paste0(unlist(frames), "_sorted"))