分割特定の列の値が複数の列に

Question

-Dataset

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) 

R>データ

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


Output<-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 
from data group by type,aid,month,year,Outcome,ser_no") 

R>出力

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

• 私はaddメソッド列にしたいですから私の出力にはがあります。

  メソッドの列が4つしか値1.Cheque 2.DD 3.NetBank 4.Blankは現金

を表すことができ、私は（最後の4つの列を確認してください）以下の出力方法列の値を追加します。 sqldfなしでそれを行うことはできますか？ グループ内のメソッド値の出現を見つけようとしています。

例：GROUP BY句ごとに、行1は1 Chequeと1 DDの値を持ち、したがって、この数は1と表示されます。 NetbankとCash値がカウントあり、したがって存在しない行3は、このようにそこ2 Netbank値がカウント有するGROUP BY節当たり0。 ある2として表示されないNetbank、CashとCheque値がこのようにそこにないようにすると、カウントされた0

ありますデータテーブルと

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 

Answer 1

：ここ

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 

DT[,.N, by = .(type,method)]は異なる方法をカウントし、大規模な形式に変換しdcasdt。 あなたは、あなたの出力にマージすることができ

Output <- setDT(Output) 
Output[plouf, on = "type"] 

    type aid month year Outcome ser_no members entries UNITS amt LowestAmt HighestAmount Mean user_cnt Suggestion cash 
1: Type1 A1 JAN 2016 Positive A00001  2  2 300 2500  1000   1500 1250  20 0.10000000 0 
2: Type2 A2 FEB 2015 Negative A00002  2  2 700 5000  2000   3000 2500  15 0.13333333 0 
3: Type3 A3 MAR 2017 Medium A00003  2  2 1100 6500  2500   4000 3250  32 0.06250000 0 
4: Type4 A4 APR 2018 Neutral A00004  1  1 700 6000  6000   6000 6000  44 0.02272727 1 
    cheque dd netbank 
1:  1 1  0 
2:  1 1  0 
3:  0 0  2 
4:  0 0  0 

Answer 2

私はTOLOWERがsqldf.Soの下で働いて両方のオプションが含まれていないとして、「チェック」に関するケースの問題を解決することができませんでした。

全体の凝集が data.tableを使用して1つのステートメントで行うことができます

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") 

Answer 3

：場合

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)] 

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 

以上のちょうど4つの異なる値がmethodである私は、他のアプローチを示唆していますdcast()のようにして参加してください。