示例数据：
product_id <- c("1000","1000","1000","1000","1000","1000", "1002","1002","1002","1002","1002","1002")
qty_ordered <- c(1,2,1,1,1,1,1,2,1,2,1,1)
price <- c(2.49,2.49,2.49,1.743,2.49,2.49, 2.093,2.093,2.11,2.11,2.11, 2.97)
date <- c("2/23/15","2/23/15", '3/16/15','3/16/15','5/16/15', "6/18/15", "2/19/15","3/19/15","3/19/15","3/19/15","3/19/15","4/19/15")
sampleData <- data.frame(product_id, qty_ordered, price, date)
我想在每次价格发生变化时确定。另外，我想总结（）这两个价格变化日期之间的总qty_ordered。例如，
对于product_id == "1000"，价格变动发生在2015年3月16日，从2.49美元到1.743美元。总的qty_ordered是1 + 2 + 1 = 4;
这两个最早的价格变动日期之间的差异是2015年2月23日至2015年3月16日，即21天。
所以新数据框应该是：
product_id sum_qty_ordered price date_diff
1000 4 2.490 21
1000 1 1.743 61
1000 2 2.490 33
以下是我的尝试：
**注意：对于这种情况，简单的“dplyr::group_by”将无效，因为它会忽略日期效果。
1）我从Determine when columns of a data.frame change value and return indices of the change找到了这段代码： 这是为了在每次价格变化时识别，这标识了每种产品价格变化的第一个日期。
IndexedChanged <- c(1,which(rowSums(sapply(sampleData[,3],diff))!=0)+1)
sampleData[IndexedChanged,]
但是，如果我使用该代码，我不确定如何计算每个条目的sum(qty_ordered)和日期差异。
2）我尝试编写一个WHILE循环来临时存储每批product_id，价格，日期范围（例如，一个product_id数据框的子集，一个价格和所有条目从价格变化的最早日期到变更之前的最后价格日期，
然后，汇总该子集以获得总和（sum_qty_ordered）和日期差异。
但是，我觉得我总是对WHILE和FOR感到困惑，所以我的代码中有一些问题。这是我的代码：
为以后的数据存储创建一个空数据框
NewData_Ready <- data.frame(
product_id = character(),
price = double(),
early_date = as.Date(character()),
last_date=as.Date(character()),
total_qty_demanded = double(),
stringsAsFactors=FALSE)
创建临时表以存储批次价格订单条目
temp_dataset <- data.frame(
product_id = character(),
qty_ordered = double(),
price = double(),
date=as.Date(character()),
stringsAsFactors=FALSE)
循环： 这很麻烦......可能没有意义，所以我确实对此有所帮助。
for ( i in unique(sampleData$product_id)){
#for each unique product_id in the dataset, we are gonna loop through it based on product_id
#for first product_id which is "1000"
temp_table <- sampleData[sampleData$product_id == "i", ] #subset dataset by ONE single product_id
#this dataset only has product of "1000" entries
#starting a new for loop to loop through the entire entries for this product
for ( p in 1:length(temp_table$product_id)){
current_price <- temp_table$price[p] #assign current_price to the first price value
#assign $2.49 to current price.
min_date <- temp_table$date[p] #assign the first date when the first price change
#assign 2015-2-23 to min_date which is the earliest date when price is $2.49
while (current_price == temp_table$price[p+1]){
#while the next price is the same as the first price
#that is, if the second price is $2.49 is the same as the first price of $2.49, which is TRUE
#then execute the following statement
temp_dataset <- rbind(temp_dataset, temp_table[p,])
#if the WHILE loop is TRUE, means every 2 entries have the same price
#then combine each entry when price is the same in temp_table with the temp_dataset
#if the WHILE loop is FALSE, means one entry's price is different from the next one
#then stop the statement at the above, but do the following
current_price <- temp_table$price[p+1]
#this will reassign the current_price to the next price, and restart the WHILE loop
by_idPrice <- dplyr::group_by(temp_dataset, product_id, price)
NewRow <- dplyr::summarise(
early_date = min(date),
last_date = max(date),
total_qty_demanded = sum(qty_ordered))
NewData_Ready <- rbind(NewData_Ready, NewRow)
}
}
}
我在相关问题上搜索了很多，但我还没有发现任何与此问题有关的内容。如果您有任何建议，请告诉我。 另外，请提供一些有关我的问题解决方案的建议。我非常感谢您的时间和帮助！
Here is my R version:
platform x86_64-apple-darwin13.4.0
arch x86_64
os darwin13.4.0
system x86_64, darwin13.4.0
status
major 3
minor 3.1
year 2016
month 06
day 21
svn rev 70800
language R
version.string R version 3.3.1 (2016-06-21)
nickname Bug in Your Hair
答案 0 :(得分：3)
使用data.table：
library(data.table)
setDT(sampleData)
一些预处理：
sampleData[, firstdate := as.Date(date, "%m/%d/%y")]
根据您计算日期差异的方式，我们最好为每一行创建一系列日期：
sampleData[, lastdate := shift(firstdate,type = "lead"), by = product_id]
sampleData[is.na(lastdate), lastdate := firstdate]
# Arun's one step: sampleData[, lastdate := shift(firstdate, type="lead", fill=firstdate[.N]), by = product_id]
然后为每次价格变化创建一个新ID：
sampleData[, price_id := cumsum(c(0,diff(price) != 0)), by = product_id]
然后按产品和价格运行计算您的分组功能：
sampleData[,
.(
price = unique(price),
sum_qty = sum(qty_ordered),
date_diff = max(lastdate) − min(firstdate)
),
by = .(
product_id,
price_id
)
]
product_id price_id price sum_qty date_diff
1: 1000 0 2.490 4 21 days
2: 1000 1 1.743 1 61 days
3: 1000 2 2.490 2 33 days
4: 1002 0 2.093 3 28 days
5: 1002 1 2.110 4 31 days
6: 1002 2 2.970 1 0 days
我认为1000的最后一次价格变动只有33天，前一次是61（不是60）。如果您在第一天包含22,62和34，则该行应显示为date_diff = max(lastdate) − min(firstdate) + 1