R 将特定列值拆分为多列_R - Fatal编程技术网

R 将特定列值拆分为多列

R 将特定列值拆分为多列,r,R,-数据集 ID<-c(1,2,3,4,5,6,7) method<-c("cheque","DD","DD","Cheque","NetBank","NetBank","Cash") type<-c("Type1","Type1","Type2","Type2","Type3","Type3","Type4") aid<-c("A1","A1","A2","A2","A3","A3","A4") month<-c("JAN","JAN","FEB","F

-数据集

ID<-c(1,2,3,4,5,6,7)
method<-c("cheque","DD","DD","Cheque","NetBank","NetBank","Cash")
type<-c("Type1","Type1","Type2","Type2","Type3","Type3","Type4")    
aid<-c("A1","A1","A2","A2","A3","A3","A4")  
month<-c("JAN","JAN","FEB","FEB","MAR","MAR","APR")
year<-c(2016,2016,2015,2015,2017,2017,2018)
Outcome<-c("Positive","Positive","Negative","Negative","Medium","Medium","Neutral")
ser_no<-c("A00001","A00001","A00002","A00002","A00003","A00003","A00004")
Units<-c(100,200,300,400,500,600,700)
amt<-c(1000,1500,2000,3000,4000,2500,6000)
user_cnt<-c(20,20,15,15,32,32,44)

data<-data.frame(ID=ID,type=type,aid=aid,month=month,year=year,Outcome=Outcome,ser_no=ser_no,Units=Units,amt=amt,user_cnt=user_cnt,method=method)

我想在输出中添加来自

数据的方法列
方法列只能有四个值1.支票2.DD 3.NetBank
4.空白表示现金


我想在下面的输出中添加一个方法列值（检查最后四列）。不使用sqldf

是否可以执行此操作。我正在尝试查找组中方法值的出现情况
示例：根据GROUP BY子句，第一行有
1张支票
和
1张DD
值，因此计数显示为
1
<代码>净银行和
现金
值不存在，因此计数为0。根据GROUP BY子句，第三行有
2个Netbank
值，因此计数显示为
2
，并且没有
Netbank
，
Cash
和
check
值，因此计数为0

type aid month year Outcome ser_no members entries UNITS amt LowestAmt HighestAmount Mean user_cnt Suggestion Cheque DD Netbank Cash Type1 A1 JAN 2016 Positive A00001 2 2 300 2500 1000 1500 1250 20 0.10000000 1 1 0 0 Type2 A2 FEB 2015 Negative A00002 2 2 700 5000 2000 3000 2500 15 0.13333333 1 1 0 0 Type3 A3 MAR 2017 Medium A00003 2 2 1100 6500 2500 4000 3250 32 0.06250000 0 0 2 0 Type4 A4 APR 2018 Neutral A00004 1 1 700 6000 6000 6000 6000 44 0.02272727 0 0 0 1
使用数据表：

library(data.table) DT <- setDT(data) DT[,method := tolower(method)] # to avoid different count with upper and lower case plouf<-dcast(DT[,.N, by = .(type,method)],type~ method) plouf[is.na(plouf)]<-0 type cash cheque dd netbank 1: Type1 0 1 1 0 2: Type2 0 1 1 0 3: Type3 0 0 0 2 4: Type4 1 0 0 0
使用数据表：

library(data.table) DT <- setDT(data) DT[,method := tolower(method)] # to avoid different count with upper and lower case plouf<-dcast(DT[,.N, by = .(type,method)],type~ method) plouf[is.na(plouf)]<-0 type cash cheque dd netbank 1: Type1 0 1 1 0 2: Type2 0 1 1 0 3: Type3 0 0 0 2 4: Type4 1 0 0 0

我无法解决“支票”上的案例问题，因为tolower不在sqldf下工作。因此包括两个选项

sqldf("select type ,aid ,month ,year ,Outcome ,ser_no ,count(distinct ID) as members ,count(type) as entries ,sum(UNITS) as UNITS ,sum(amt) as amt ,min(amt) as LowestAmt ,max(amt) as HighestAmount ,AVG(amt) as Mean ,user_cnt ,cast (count(distinct ID) as real)/user_cnt as Suggestion ,count(case when lower(method)='cheque' then method end ) as cheque ,count(case when method ='DD' then method end ) as DD ,count(case when method ='NetBank' then method end ) as NetBank ,count(case when method ='Cash' then method end ) as Cash from data group by type,aid,month,year,Outcome,ser_no")

我无法解决“支票”上的案例问题，因为tolower不在sqldf下工作。因此包括两个选项

sqldf("select type ,aid ,month ,year ,Outcome ,ser_no ,count(distinct ID) as members ,count(type) as entries ,sum(UNITS) as UNITS ,sum(amt) as amt ,min(amt) as LowestAmt ,max(amt) as HighestAmount ,AVG(amt) as Mean ,user_cnt ,cast (count(distinct ID) as real)/user_cnt as Suggestion ,count(case when lower(method)='cheque' then method end ) as cheque ,count(case when method ='DD' then method end ) as DD ,count(case when method ='NetBank' then method end ) as NetBank ,count(case when method ='Cash' then method end ) as Cash from data group by type,aid,month,year,Outcome,ser_no")

整个聚合可以使用
数据在一条语句中完成。表： library(data.table) setDT(data)[ , .(members = uniqueN(ID), entries = .N, UNITS = sum(Units), amt = sum(amt), LowestAmt = min(amt), HighestAmount = max(amt), Mean = mean(amt), user_cnt = first(user_cnt), Suggestion = uniqueN(ID) / first(user_cnt), Cheque = sum(tolower(method) == "cheque"), DD = sum(tolower(method) == "dd"), NetBank = sum(tolower(method) == "netbank"), Cash = sum(tolower(method) %in% c("cash", ""))), by = .(type, aid, month, year, Outcome, ser_no)] 如果method 中有4个以上的不同值，我建议使用其他方法，如dcast（）和join 整个聚合可以使用数据在一条语句中完成。表： library(data.table) setDT(data)[ , .(members = uniqueN(ID), entries = .N, UNITS = sum(Units), amt = sum(amt), LowestAmt = min(amt), HighestAmount = max(amt), Mean = mean(amt), user_cnt = first(user_cnt), Suggestion = uniqueN(ID) / first(user_cnt), Cheque = sum(tolower(method) == "cheque"), DD = sum(tolower(method) == "dd"), NetBank = sum(tolower(method) == "netbank"), Cash = sum(tolower(method) %in% c("cash", ""))), by = .(type, aid, month, year, Outcome, ser_no)] 如果method 中有4个以上的不同值，我建议使用其他方法，如dcast（）和join 有没有办法在一个查询本身中获得预期的输出，而不是附加它？我在六个列上分组，分别是类型、援助、月、年、结果、序号，而不仅仅是类型 dcast（data，type+month+year+Outcome+aid+seru no~method，fun.aggregate=length）您可以通过创建一个分组变量来使用相同的代码：DT[，grp:=粘贴（type，aid，month，year，Outcome，seru no，sep=“”）] 然后dcast（DT[，.N，by=（grp，method）]，grp~method）有没有办法在一个查询本身中获得预期的输出，而不是附加它？我在六个列上分组，分别是类型、援助、月、年、结果、序号，而不仅仅是类型 dcast（data，type+month+year+Outcome+aid+seru no~method，fun.aggregate=length）您可以通过创建一个分组变量来使用相同的代码：DT[，grp:=粘贴（type，aid，month，year，Outcome，seru no，sep=“”）] 然后dcast（DT[，.N，by=（grp，method）]，grp~method）在140万行的数据集上，它比PIG答案慢得多。清管器回答需要17秒，数据表需要1分14秒@在140万行的数据集上，UweIt比PIG答案慢得多。清管器回答需要17秒，数据表需要1分14秒@Uwe type aid month year Outcome ser_no members entries UNITS amt LowestAmt HighestAmount Mean user_cnt Suggestion Cheque DD NetBank Cash 1: Type1 A1 JAN 2016 Positive A00001 2 2 300 2500 1000 1500 1250 20 0.10000000 1 1 0 0 2: Type2 A2 FEB 2015 Negative A00002 2 2 700 5000 2000 3000 2500 15 0.13333333 1 1 0 0 3: Type3 A3 MAR 2017 Medium A00003 2 2 1100 6500 2500 4000 3250 32 0.06250000 0 0 2 0 4: Type4 A4 APR 2018 Neutral A00004 1 1 700 6000 6000 6000 6000 44 0.02272727 0 0 0 1