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))
当我调用以下函数pivotTable <- mydataWithWeeksAndWeights %>% group_by(week, weight) %>% count(satisfactionLevel)
时,它会计算所有第46周条目的satisflevel。问题是前三行的第46周是指2016年,剩下的是指2017年。我想保留这些重复的条目。
答案 0 :(得分:2)
我无法确定我的代码是否符合您的要求,因为您没有提供预期的输出,但我认为您需要做的是添加year
列并将其添加到{{ 1}}以便您区分2016年第46周和2017年第46周。
编辑:如果您需要从结束日期自动定义年份,我在@ docendodiscimus的评论中添加一点:
group_by
答案 1 :(得分:0)
以下是我要做的事情:将“结束”重新格式化为日期格式并使用聚合函数:
# 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)
希望这有帮助!