Cumsum重置和延遲

Question

我正在尋找解決方案來計算拖欠桶。我已經想出了重設cumsum的部分，但我堅持如何基於觸發器「延遲」cumsum;看到我的，我想做到哪裏我期望的結果是correct_bucket什麼例子：

df <- data.frame(id = c(1,1,1,1,2,2,3,3,3,3,4,4,4,4,5,5,5,5,5,5,5,5,5,5,6,6,6,6,7,7,7,7,7,8,8,8,8), 
      min_due = c(25,50,50,75,25,50,25,50,25,25,25,50,75,100,25,50,75,100,100,25,50,25,14.99,0,25,60,60,0,25,50,75,100,75,25,50,25,50), 
      payment = c(0,0,25,0,0,0,0,0,50,25,0,0,0,0,0,0,0,0,25,100,0,150,25,14.99,0,25,60,60,0,0,0,0,50,0,0,25,0), 
      past_due_amt = c(0,25,25,50,0,25,0,25,0,0,0,25,50,75,0,25,50,75,75,0,25,0,0,0,0,0,0,0,0,25,50,75,50,0,25,0,25), 
      correct_bucket = c(0,1,1,2,0,1,0,1,0,0,0,1,2,3,0,1,2,3,3,0,1,0,0,0,0,0,0,0,0,1,2,3,2,0,1,0,1)) 

correct_bucket的說明：這表明，通過ID，該min_due被滿足（或不能）由支付地爲大於或等於先前（滯後1）min_pay。例如：ID＃1的min_due爲25（在第1行），付款爲0（第2行），因此correct_bucket = 1.正如您所見，在每個示例中，正確存儲桶的值需要迭代取決於付款是否已付款以及付款金額。

想法？請詢問您需要的任何澄清問題，我近在咫尺，歡迎任何額外的幫助！

謝謝！

Answer 1

df$original_order = 1:nrow(df) #In case you need later. OPTIONAL 

#Obtain the incremental min_due for each id 
df$b2 = unlist(lapply(split(df, df$id), function(a) c(0, diff(a$min_due)))) 

#Function to get your values from incremental min_due 
ff = function(x){ 
x$b3 = 0 
    for (i in 2:NROW(x)){ 
     if (x$b2[i] > 0){ 
      x$b3[i] = x$b3[i-1] + 1 
     } 
     if (x$b2[i] == 0){ 
      x$b3[i] = x$b3[i-1] 
     } 
     if (x$b2[i] < 0){ 
      x$b3[i] = 0 
     } 
    } 
    return(x) 
} 

#Split df by id and use the above function on each sub group 
#'b3' is the value you want 
do.call(rbind, lapply(split(df, df$id), function(a) ff(a))) 

新FF

ff = function(x){ 
    x$b3 = 0 

    if(NROW(x) < 2){ 
     return(x) 
    } 

    for (i in 2:NROW(x)){ 
     if (x$b2[i] > 0){ 
      x$b3[i] = x$b3[i-1] + 1 
     } 
     if (x$b2[i] == 0){ 
      x$b3[i] = x$b3[i-1] 
     } 
     if (x$b2[i] < 0){ 
      x$b3[i] = 0 
     } 
    } 
    return(x) 
}