Question

运行以下代码后：

using: Data.table library: 
library(data.table)   # CRAN version 1.10.4 used
setDT(bb)[, D := bal / shift(bal, 6L)][seq(1L, nrow(bb), 5L), D := NA][]

OR

Using: Dplyr library: 
DF %>%
  group_by(g1=seq_along(bal) %% 5) %>%
  mutate(denominator=lag(bal)) %>%
  ungroup() %>%
  group_by(g2=(seq_along(bal) - 1) %/% 5) %>%
  mutate(denominator=lag(denominator),
     D=bal / denominator) %>%
  ungroup()

我得到以下输出：

  date      Bucket          balance  D
1/31/2013   bkt 0           10       NA
1/31/2013   bkt 1(10-20)    11       NA
1/31/2013   bkt 2(20-30)    12       NA
1/31/2013   bkt 3(30-40)    13       NA
1/31/2013   bkt 4(40+)      14       NA
2/28/2013   bkt 0           20       NA
2/28/2013   bkt 1(10-20)    30      3.00
2/28/2013   bkt 2(20-30)    40      3.63
2/28/2013   bkt 3(30-40)    50      4.16
2/28/2013   bkt 4(40+)      60      4.61
3/30/2013   bkt 0   10       NA
3/30/2013   bkt 1(10-20)    11      0.55
3/30/2013   bkt 2(20-30)    12      0.40
3/30/2013   bkt 3(30-40)    13      0.32
3/30/2013   bkt 4(40+)      15      0.30

现在的问题是：在Bucket专栏中每隔bkt 3和bkt 4，我需要在coloumn D打印100.

预期产量：在每个bkt 3和bkt 4，它应该在coloumn D中打印100.

  date      Bucket          balance  D
1/31/2013   bkt 0           10       NA
1/31/2013   bkt 1(10-20)    11       NA
1/31/2013   bkt 2(20-30)    12       NA
1/31/2013   bkt 3(30-40)    13       NA
1/31/2013   bkt 4(40+)      14       NA
2/28/2013   bkt 0           20       NA
2/28/2013   bkt 1(10-20)    30      3.00
2/28/2013   bkt 2(20-30)    40      3.63
2/28/2013   bkt 3(30-40)    50      100
2/28/2013   bkt 4(40+)      60      100
3/30/2013   bkt 0           10       NA
3/30/2013   bkt 1(10-20)    11      0.55
3/30/2013   bkt 2(20-30)    12      0.40
3/30/2013   bkt 3(30-40)    13      100
3/30/2013   bkt 4(40+)      15      100

Also tried: 
setDT(bb1)[Bucket ==  "bkt 3(30-40)" , D := 100 ]

输出 - 没有错误，也没有要打印的值。

请提前帮助，谢谢。

使用R的列上的条件

0 个答案: