总和中的R不同的列值(Sum by distinct column value in R)

我在R A非常的大数据帧，并想总结两列的其他列中的每个不同的值，例如说我们有过了一天在各种商店交易的数据帧的数据如下

shop <- data.frame('shop_id' = c(1, 1, 1, 2, 3, 3), 
  'shop_name' = c('Shop A', 'Shop A', 'Shop A', 'Shop B', 'Shop C', 'Shop C'), 
  'city' = c('London', 'London', 'London', 'Cardiff', 'Dublin', 'Dublin'), 
  'sale' = c(12, 5, 9, 15, 10, 18), 
  'profit' = c(3, 1, 3, 6, 5, 9))

这就是：

shop_id  shop_name    city      sale profit
   1     Shop A       London    12   3
   1     Shop A       London    5    1
   1     Shop A       London    9    3
   2     Shop B       Cardiff   15   6
   3     Shop C       Dublin    10   5
   3     Shop C       Dublin    18   9

而且我要总结的销售和利润各店给：

shop_id  shop_name    city      sale profit
   1     Shop A       London    26   7
   2     Shop B       Cardiff   15   6
   3     Shop C       Dublin    28   14

我目前使用下面的代码来做到这一点：

 shop_day <-ddply(shop, "shop_id", transform, sale=sum(sale), profit=sum(profit))
 shop_day <- subset(shop_day, !duplicated(shop_id))

该工作绝对没问题，但我说我的数据框大（140,000行，37列和近10万的唯一行，我想总结）和我的代码需要年龄运行，然后最后说，他们已经耗尽内存。

有谁知道的最有效的方式来做到这一点。

提前致谢！

Answer 1:

**强制性数据表的答案**

> library(data.table)
data.table 1.8.0  For help type: help("data.table")
> shop.dt <- data.table(shop)
> shop.dt[,list(sale=sum(sale), profit=sum(profit)), by='shop_id']
     shop_id sale profit
[1,]       1   26      7
[2,]       2   15      6
[3,]       3   28     14
> 

其中直到事情得到更大听起来不错，和良好的...

shop <- data.frame(shop_id = letters[1:10], profit=rnorm(1e7), sale=rnorm(1e7))
shop.dt <- data.table(shop)

> system.time(ddply(shop, .(shop_id), summarise, sale=sum(sale), profit=sum(profit)))
   user  system elapsed 
  4.156   1.324   5.514 
> system.time(shop.dt[,list(sale=sum(sale), profit=sum(profit)), by='shop_id'])
   user  system elapsed 
  0.728   0.108   0.840 
> 

如果你创建一个关键的data.table你获得额外的速度增加：

shop.dt <- data.table(shop, key='shop_id')

> system.time(shop.dt[,list(sale=sum(sale), profit=sum(profit)), by='shop_id'])
   user  system elapsed 
  0.252   0.084   0.336 
> 

Answer 2:

以下是如何使用基础R加快这样的操作：

idx <- split(1:nrow(shop), shop$shop_id)
a2 <- data.frame(shop_id=sapply(idx, function(i) shop$shop_id[i[1]]),
                 sale=sapply(idx, function(i) sum(shop$sale[i])), 
                 profit=sapply(idx, function(i) sum(shop$profit[i])) )

时间缩短到0.75秒VS 5.70秒，我的系统上ddply总结版本。

Answer 3:

我认为这样做最巧妙的方法是在dplyr

library(dplyr)
shop %>% 
  group_by(shop_id, shop_name, city) %>% 
  summarise_all(sum)

Answer 4:

为了以防万一，如果你有列一长串，使用summarize_if（）

总结所有列，如果数据类型为int

library(dplyr)
shop %>% 
  group_by(shop_id, shop_name, city) %>% 
  summarise_if(is.integer, sum)

文章来源: Sum by distinct column value in R