在使用dplyr中的group_by（）时保留重复条目 library（dplyr）##激活data.table库 mydataWithWeeksAndWeights%计数（satisfactionLevel）它统计所有第46周条目的satisfactionLevel。问题在于，前三排的第46周是指2016年，其余的是指2017年。我想保留这些重复的条目。

我不能确定我的代码是否符合您的要求，因为您没有给出预期的输出，但我认为您需要做的是添加一个

年

列，并通过将其添加到

组中，以便区分2016年第46周和2017年第46周

编辑：如果您需要自动定义从您拥有的结束日期开始的年份，我将在@docendodiscimus的注释中添加位：

library(dplyr) ##activates the data.table library

mydataWithWeeksAndWeights <- data_frame(ended = c("14/11/2016",
                                                  "14/11/2016",
                                                  "14/11/2016",
                                                  "02/01/2017",
                                                  "02/01/2017",
                                                  "15/11/2017",
                                                  "15/11/2017",
                                                  "16/11/2017",
                                                  "16/11/2017"),
                                        week = c(46, 46, 46, 1, 1, 46, 46, 46, 46),
                                        satisfactionLevel = c("Very dissatisfied",
                                                              "Very satisfied",
                                                              "Satisfied",
                                                              "Dissatisfied",
                                                              "Very dissatisfied",
                                                              "Very satisfied",
                                                              "Very dissatisfied",
                                                              "Very Satisfied",
                                                              "Very satisfied"),
                                        weight = c(0, 1, 0.75, 0.25, 0, 1, 0, 1, 1))

库（dplyr）
mydataWithWeeksAndWeights下面是我要做的：将“ended”格式重新格式化为日期格式，并使用聚合函数：

library(dplyr)

mydataWithWeeksAndWeights <- data_frame(ended = c("14/11/2016",
                                                  "14/11/2016",
                                                  "14/11/2016",
                                                  "02/01/2017",
                                                  "02/01/2017",
                                                  "15/11/2017",
                                                  "15/11/2017",
                                                  "16/11/2017",
                                                  "16/11/2017"),
                                        week = c(46, 46, 46, 1, 1, 46, 46, 46, 46),
                                        satisfactionLevel = c("Very dissatisfied",
                                                              "Very satisfied",
                                                              "Satisfied",
                                                              "Dissatisfied",
                                                              "Very dissatisfied",
                                                              "Very satisfied",
                                                              "Very dissatisfied",
                                                              "Very Satisfied",
                                                              "Very satisfied"),
                                        weight = c(0, 1, 0.75, 0.25, 0, 1, 0, 1, 1))

mydataWithWeeksAndWeights$year <- format(as.Date(mydataWithWeeksAndWeights$ended,
                                                 "%d/%m/%Y"), "%Y")

pivotTable <- mydataWithWeeksAndWeights %>%
  group_by(week, year, weight) %>%
  count(satisfactionLevel)

#只是为了缩短df名称
df Try
mydataWithWeeksAndWeights%%>%groupby（week，weight）%%>%filter（n（）>1）
我想你应该区分年份。您可以使用
df%>%groupby（year\u week=format（截至日期（结束，“%d/%m/%Y”）、%Y-%W）、重量）%%>%count（满意度水平）
@docendodiscimus！！！Cheers@docendodiscimus你能提交你的评论作为答案吗？@greeconomist它已经以类似的方式在现有答案中实现了
# just to shorten df-name
df <- mydataWithWeeksAndWeights 

# reformat and add column with year
df[,"ended"] <- as.Date(df[[1]], format = "%d/%m/%Y")
df$year <- format(df[[1]], "%Y")

# actual aggregating
aggregate (df$weight, by = list(df$year, df$satisfactionLevel, df$week), FUN = sum)