R 如果<；0然后使用0

我在两件事上遇到了问题，但条件求和的问题更重要（我想不那么琐碎）。我想从每一行的总和中减去1，如果结果是，一种快速而肮脏的方法是将数据分组：

a %>% 
  mutate(
    lag_disp = lag(a$DISPENSED_DURATION, default=-1),
    change = DISPENSED_DURATION != lag_disp,
    group = cumsum(as.integer(DISPENSED_DURATION & change))
  ) %>% 
  group_by(group) %>% 
  mutate(DISP_INVT = ifelse(
    group == 1,
    pmax(1+cumsum(DISPENSED_DURATION-1), 0),
    pmax(cumsum(DISPENSED_DURATION-1), 0)
  )) %>% 
  ungroup() %>% 
  select(-c(lag_disp, change, group))

#      id1   id2 date       DISPENSED_DURATION DISP_INVT
#    <dbl> <dbl> <chr>                   <dbl>     <dbl>
#  1     1     1 2020-01-01                  4         4
#  2     1     2 2020-01-02                  0         3
#  3     1     3 2020-01-03                  0         2
#  4     1     4 2020-01-04                  0         1
#  5     1     5 2020-01-05                  0         0
#  6     1     6 2020-01-06                  0         0
#  7     1     7 2020-01-07                  0         0
#  8     1     8 2020-01-08                  4         3
#  9     1     9 2020-01-09                  0         2
# 10     1    10 2020-01-10                  0         1

a%>%
变异(
lag_disp=lag（一个$DISPENSED_持续时间，默认值=-1），
更改=分配的持续时间！=延迟分配，
组=总和（作为整数（分配的持续时间和变化））
) %>% 
分组依据（分组）%>%
变异（DISP_INVT=ifelse(
组==1，
pmax（1+cumsum（分配持续时间-1），0），
pmax（累计金额（分配持续时间-1），0）
)) %>% 
解组（）%>%
选择（-c（滞后显示、更改、组））
#id1 id2分配日期\u持续时间DISP\u INVT
#                              
#  1     1     1 2020-01-01                  4         4
#  2     1     2 2020-01-02                  0         3
#  3     1     3 2020-01-03                  0         2
#  4     1     4 2020-01-04                  0         1
#  5     1     5 2020-01-05                  0         0
#  6     1     6 2020-01-06                  0         0
#  7     1     7 2020-01-07                  0         0
#  8     1     8 2020-01-08                  4         3
#  9     1     9 2020-01-09                  0         2
# 10     1    10 2020-01-10                  0         1

---

我确信存在一种更简洁的方法，但这应该是可行的。

这里需要的是减少/迭代计算，而不是累积/向量，因为一行中的值取决于前一行的计算值

a[，DISP\u INVT:=Reduce（函数（prev，this）max（0，prev+this-1），
分配（持续时间，累计=真）]
A.
#id1 id2分配日期\u持续时间DISP\u INVT
#                              
#  1:     1     1 2020-01-01                  4         4
#  2:     1     2 2020-01-02                  0         3
#  3:     1     3 2020-01-03                  0         2
#  4:     1     4 2020-01-04                  0         1
#  5:     1     5 2020-01-05                  0         0
#  6:     1     6 2020-01-06                  0         0
#  7:     1     7 2020-01-07                  0         0
#  8:     1     8 2020-01-08                  4         3
#  9:     1     9 2020-01-09                  0         2
# 10:     1    10 2020-01-10                  0         1

谢谢你，威廉，这是可行的，不过我会选择@r2evans的答案，因为它使用data.table，但我已经对你的答案投了赞成票。非常感谢你。

    a[, DISP_INVT := cumsum(DISPENSED_DURATION-1), id1]

    id1 id2       date DISPENSED_DURATION DISP_INVT
 1:   1   1 2020-01-01                  4         3
 2:   1   2 2020-01-02                  0         2
 3:   1   3 2020-01-03                  0         1
 4:   1   4 2020-01-04                  0         0
 5:   1   5 2020-01-05                  0        -1
 6:   1   6 2020-01-06                  0        -2
 7:   1   7 2020-01-07                  0        -3
 8:   1   8 2020-01-08                  4         0
 9:   1   9 2020-01-09                  0        -1
10:   1   10 2020-01-10                 0        -2

    a[, DISP_INVT := ifelse(cumsum(DISPENSED_DURATION-1)<0,0,cumsum(DISPENSED_DURATION-1)), id1]

    id1 id2       date DISPENSED_DURATION DISP_INVT
 1:   1   1 2020-01-01                  4         3
 2:   1   2 2020-01-02                  0         2
 3:   1   3 2020-01-03                  0         1
 4:   1   4 2020-01-04                  0         0
 5:   1   5 2020-01-05                  0         0
 6:   1   6 2020-01-06                  0         0
 7:   1   7 2020-01-07                  0         0
 8:   1   8 2020-01-08                  4         0
 9:   1   9 2020-01-09                  0         0
10:   1   10 2020-01-10                 0         0

    a[, DISP_INVT := cumsum(ifelse(DISPENSED_DURATION-1<0,0,DISPENSED_DURATION-1)), id1]

    id1 id2       date DISPENSED_DURATION DISP_INVT
 1:   1   1 2020-01-01                  4         3
 2:   1   2 2020-01-02                  0         3
 3:   1   3 2020-01-03                  0         3
 4:   1   4 2020-01-04                  0         3
 5:   1   5 2020-01-05                  0         3
 6:   1   6 2020-01-06                  0         3
 7:   1   7 2020-01-07                  0         3
 8:   1   8 2020-01-08                  4         6
 9:   1   9 2020-01-09                  0         6
10:   1   10 2020-01-10                 0         6

a %>% 
  mutate(
    lag_disp = lag(a$DISPENSED_DURATION, default=-1),
    change = DISPENSED_DURATION != lag_disp,
    group = cumsum(as.integer(DISPENSED_DURATION & change))
  ) %>% 
  group_by(group) %>% 
  mutate(DISP_INVT = ifelse(
    group == 1,
    pmax(1+cumsum(DISPENSED_DURATION-1), 0),
    pmax(cumsum(DISPENSED_DURATION-1), 0)
  )) %>% 
  ungroup() %>% 
  select(-c(lag_disp, change, group))

#      id1   id2 date       DISPENSED_DURATION DISP_INVT
#    <dbl> <dbl> <chr>                   <dbl>     <dbl>
#  1     1     1 2020-01-01                  4         4
#  2     1     2 2020-01-02                  0         3
#  3     1     3 2020-01-03                  0         2
#  4     1     4 2020-01-04                  0         1
#  5     1     5 2020-01-05                  0         0
#  6     1     6 2020-01-06                  0         0
#  7     1     7 2020-01-07                  0         0
#  8     1     8 2020-01-08                  4         3
#  9     1     9 2020-01-09                  0         2
# 10     1    10 2020-01-10                  0         1