Question

我遇到了一个问题，我需要$Access_score sum $lives!=0和$Access_score如果sum $lives=0和＆＃39; Trx＆＃的加权平均值$IMS_ID & $IMS_PLAN_ID 39;作为按sc2=aggregate(nonsc1, by=list(nonsc1$PRSC_CID,nonsc1$IMSPayerPlanId), FUN=function(x) if x$Lives=0 {colMeans(x$Breo_Access_score)} else {0} ) dt <- data.frame(IMS_ID=c(222,222,222,222,333,333,333,333),IMS_PLAN_ID=c(234,234,235,235,234,234,235,235),PLAN_ID=c(1234,678,1234,678,1234,678,1234,678),IMS_STATE=c('CA','CA','CA','CA','TX','TX','TX','TX'),PLAN_STATE=c('CA','CA','CA','CA','TX','TX','TX','TX'),ACCESS_SCORE=c(2,4,2,4,2,4,2,4),Lives=c(0,0,1000,200,0,0,1000,200),Trx=c(10,10,20,20,30,30,40,40));

分组的每个值的唯一值

IMS_ID  IMS_PLAN_ID TRX     ACCESS_SCORE
222     234          10           3
222     235          20       2.3333
333     234          30          3
333     235          40       2.3333

我的输出应该是：

class Episode {

var title: String?
var description: String?
var location: String?
var discount: String?
var star: Int?

init() {
    self.title = ""
    self.description = ""
    self.location = ""
    self.discount = ""
    self.star = 0
}

Answer 1

根据您的新样本输入和预期输出，我们可以这样做：

library(data.table);
setDT(dt)[,{ s <- sum(Lives); cbind(.SD[1L,.(Trx)],ACCESS_SCORE=if (s==0) mean(ACCESS_SCORE) else sum(ACCESS_SCORE*Lives)/s); },.(IMS_ID,IMS_PLAN_ID)];
##    IMS_ID IMS_PLAN_ID Trx ACCESS_SCORE
## 1:    222         234  10     3.000000
## 2:    222         235  20     2.333333
## 3:    333         234  30     3.000000
## 4:    333         235  40     2.333333

这是使用by()的基本R解决方案：

df <- as.data.frame(dt);
keys <- c('IMS_ID','IMS_PLAN_ID');
do.call(rbind,by(df,df[keys],function(g) { s <- sum(g$Lives); cbind(g[1L,c(keys,'Trx')],ACCESS_SCORE=if (s==0) mean(g$ACCESS_SCORE) else sum(g$ACCESS_SCORE*g$Lives)/s); }));
##   IMS_ID IMS_PLAN_ID Trx ACCESS_SCORE
## 1    222         234  10     3.000000
## 5    333         234  30     3.000000
## 3    222         235  20     2.333333
## 7    333         235  40     2.333333

根据R中的条件按列聚合特定列分组

1 个答案: