Question

我有以下问题。在一个数据框架上，我每天都会对客户进行观察。在另一个我有他们的购买。我感兴趣的是他们在任何一天到目前为止购买了多少件商品。我用for循环解决了这个问题，但是想知道是否有更有效的方法？

让我们看一个例子：

# Same customer observed on 10 different occasions
customers<-data.frame(id=rep(1:10, 10), date=rep(11:20, each=10))
purchases<-data.frame(id=c(1,1,4,6,6,6), date=c(12, 14, 12, 9, 13, 17))

# I can achieve what I want if I add a cumulative sum column and run a for loop
purchases$count<-sapply(1:length(purchases$id), function(i) sum(purchases$id[i]==purchases$id[1:i]))

customers$count<-0
for(i in 1:nrow(purchases)){
     customers[(customers$id==purchases[i, "id"] & customers$date>=purchases[i, "date"]),"count"]<-purchases[i,"count"]
}

customers
    id date count
1    1   11     0
2    2   11     0
3    3   11     0
4    4   11     0
5    5   11     0
6    6   11     1
7    7   11     0
8    8   11     0
9    9   11     0
10  10   11     0
11   1   12     1
12   2   12     0
13   3   12     0
14   4   12     1
 .   .    .     .
 .   .    .     .
100  10   20    0

我想知道更快的方法是什么？

提前致谢。

Answer 1

这是一个基础R解决方案 - 但dplyr和data.table等软件包对此也很有用：

# make sure purchases table is ordered correctly to do cumulative count
cum_purchases <- cbind(purchases <- purchases[order(purchases$id, purchases$date),],
                       count = with(purchases, ave(id==id, id, FUN=cumsum)) )
cum_purchases
#   id date count
# 1  1   11             1
# 2  1   14             2
# 3  4   12             1
# 4  6    9             1
# 5  6   13             2
# 6  6   17             3
out <- merge(customers,cum_purchases, all.x=T) # "left join"
out
# note that this solution produces NA instead of 0 for no purchases
# you can change this with:
out$count[is.na(out$count)] <- 0
out[order(out$date,out$id),] # produces a correct version of the example output

R为您提供了许多计算方法。 （已编辑使用累积计数。）

R比较两列上的两个数据帧并增加第三列

1 个答案: