Question

我们有一个data.frame d2_cleaned

> dput(d2_cleaned[1:3,c(2,4,5)])
structure(list(Customer.ID = c(110531L, 110531L, 110531L), Time.Spent.Watching = c(16032, 
10919, 236), Video.ID.v26 = c(3661L, 4313L, 3661L)), .Names = c("Customer.ID", 
"Time.Spent.Watching", "Video.ID.v26"), row.names = c(333515L, 
333516L, 333522L), class = "data.frame")
>

我们有另一个df = distinct_customers_after_cleaning，第一列是唯一的User.ID，因此每个用户都会被代表一次。其余的列（忽略col2）都是独特的电影。 df看起来像这样：

> dput(distinct_customers_after_cleaning[1:5,1:5])
structure(list(Customer.ID = c(110531L, 318721L, 468491L, 568071L, 
1390371L), Hits.After.Cleaning = c(58L, 44L, 98L, 6L, 5L), `3661` = c(0, 
0, 0, 0, 0), `4313` = c(0, 0, 0, 0, 0), `3661.1` = c(0, 0, 0, 
0, 0)), .Names = c("Customer.ID", "Hits.After.Cleaning", "3661", 
"4313", "3661.1"), row.names = c(NA, 5L), class = "data.frame")
>

我想要的是填写df distinct_customers_after_cleaning的值。为此，我想获取d2_cleaned的每一行Time.Spent.Watching并将其汇总到distinct_customers_after_cleaning的正确位置。要找到合适的地方，我需要匹配用户和电影ID：if（d2_cleaned [i，＆＃39; Customer.ID＆＃39;] == distinct_customers_after_cleaning [j，＆＃39; Customer.ID＆＃39;]） if（d2_cleaned [i，＆＃39; Video.ID.v26＆＃39;] ==姓名（distinct_customers_after_cleaning [y]））这是我使用的for循环：

#fill in rows
for (j in 1 : 10000) { 
  print(j)
  for (i in 1 : nrow(d2_cleaned)) {

    if (d2_cleaned[i, 'Customer.ID'] == distinct_customers_after_cleaning[j, 'Customer.ID']) {

      for (y in 1:ncol(distinct_customers_after_cleaning)) {

        if (d2_cleaned[i, 'Video.ID.v26'] == names(distinct_customers_after_cleaning[y])) {

          distinct_customers_after_cleaning[j, y] <- distinct_customers_after_cleaning[j, y] + d2_cleaned[i,'Time.Spent.Watching']

        }
      }
    }
  }
}

虽然这段代码可以按照我的意愿运行，但速度非常慢（需要4天才能完成所有数据）。您能否推荐一个更好的解决方案，可能包括aggregate？

Answer 1

您可以使用dplyr和tidyr来实现目标..有一个例子......

library(dplyr)
library(tidyr)
df <- data.frame(CUSTOMERS = c('u1','u1','u2','u1','u2','u3'),
                 VIDEOS = c('v1','v1', 'v1', 'v2','v3','v2'),
                 TIME = c(7, 12, 9,  2, 6, 4))

df_summary <- df %>% group_by(CUSTOMERS, VIDEOS) %>% summarise(TIME_SUM = sum(TIME))

df_summary_spread <- df_summary %>% spread(key = 'VIDEOS', value = 'TIME_SUM')

<强> DF：

      CUSTOMERS VIDEOS TIME
1        u1     v1    7
2        u1     v1   12
3        u2     v1    9
4        u1     v2    2
5        u2     v3    6
6        u3     v2    4

<强> df_summary：

  CUSTOMERS VIDEOS TIME_SUM
  <fct>     <fct>     <dbl>
1 u1        v1          19.
2 u1        v2           2.
3 u2        v1           9.
4 u2        v3           6.
5 u3        v2           4.

<强> df_summary_spread：

  CUSTOMERS    v1    v2    v3
  <fct>     <dbl> <dbl> <dbl>
1 u1          19.    2.   NA 
2 u2           9.   NA     6.
3 u3          NA     4.   NA

从一个data.frame聚合到另一个

1 个答案: