Question

> tempDT <- data.table(colA = c("E","E","A","C","E","C","E","C","E"), colB = c(20,30,40,30,30,40,30,20,10), group = c(1,1,1,1,2,2,2,2,2), want = c(NA, 30, 40, 70,NA,40,70,20,30))
> tempDT
   colA colB group want
1:    E   20     1   NA
2:    E   30     1   30
3:    A   40     1   40
4:    C   30     1   70
5:    E   30     2   NA
6:    C   40     2   40
7:    E   30     2   70
8:    C   20     2   20
9:    E   10     2   30

我有列'colA''colB''组'：在每个'组'中，我想从下到上总结'colB'，直到'colA'为'E'。

Answer 1

根据预期的＆＃39;想要＆＃39;，我们创建一个游程ID列＆＃39; grp＆＃39;通过检查值是否为“E＆＃39;在＆＃39; colA＆＃39;中，然后创建＆＃39; want1＆＃39;作为＆＃39; colB＆＃39;的累积总和经过＆＃39; grp＆＃39;分组后和＆＃39; group＆＃39;，获取＆＃39; colA＆＃39;中duplicated元素的行索引（＆＃39; i1＆＃39;）并且也是＆＃39; E＆＃39;并分配＆＃39; colB＆＃39;值为＆＃39;想要1＆＃39;

tempDT[, grp:= rleid(colA=="E") * (colA != "E")
        ][grp!= 0, want1 := cumsum(colB), .(grp, group)]
i1 <- tempDT[, .I[colA=="E" & duplicated(colA)], group]$V1
tempDT[i1, want1 := colB][, grp := NULL][]
#    colA colB group want want1
#1:    E   20     1   NA    NA
#2:    E   30     1   30    30
#3:    A   40     1   40    40
#4:    C   30     1   70    70
#5:    E   30     2   NA    NA
#6:    C   30     2   30    30

Answer 2

希望这有帮助！

library(dplyr)

df %>%
  group_by(group) %>%
  mutate(row_num = n():1) %>%
  group_by(group) %>%
  mutate(sum_colB = sum(colB[row_num < row_num[which(colA=='E')]]),
         flag = ifelse(row_num >= row_num[which(colA=='E')], 0, 1),) %>%
  mutate(sum_colB = ifelse(flag==1 & row_num==1, sum_colB, ifelse(flag==0, NA, colB))) %>%
  select(-flag, -row_num) %>%
  data.frame()

输出为：

  colA colB group want sum_colB
1    E   20     1   NA       NA
2    E   30     1   30       NA
3    A   40     1   40       40
4    C   30     1   70       70
5    E   30     2   NA       NA
6    C   30     2   30       30

示例数据：

df <- structure(list(colA = structure(c(3L, 3L, 1L, 2L, 3L, 2L), .Label = c("A", 
"C", "E"), class = "factor"), colB = c(20, 30, 40, 30, 30, 30
), group = c(1, 1, 1, 1, 2, 2), want = c(NA, 30, 40, 70, NA, 
30)), .Names = c("colA", "colB", "group", "want"), row.names = c(NA, 
-6L), class = "data.frame")

Answer 3

有一种方法：行参考+总和

# input data
tempDT <- data.table(colA = c("E","E","A","C","E","C","E","C","E"), colB = c(20,30,40,30,30,40,30,20,10), group = c(1,1,1,1,2,2,2,2,2), want = c(NA, 30, 40, 70,NA,40,70,20,30))
tempDT

# find row reference previous row where colA is "E"
lastEpos <- function(i) tail(which(tempDT$colA[1:(i-1)] == "E"), 1)
tempDT[, rowRef := sapply(.I, lastEpos), by = "group"]

# sum up
sumEpos <- function(i) {
  valTEMP <- tempDT$rowRef[i]
  outputTEMP <- sum(tempDT$colB[(valTEMP+1):i])  # sum
  return(outputTEMP)
}
tempDT[, want1 := sapply(.I, sumEpos), by = "group"]

# deal with first row in every group
tempDT[, want1 := c(NA, want1[-1]), by = "group"]

# clean output
tempDT[, rowRef := NULL]
tempDT

R data.table逐行条件求和

3 个答案: