通过R中的id将一组观察转移到下一组答案

【问题标题】：shift observation of one group into next group by id in R通过R中的id将一组观察转移到下一组
【发布时间】：2021-01-21 12:12:10
【问题描述】：

假设我有一个这样的数据框：

contracts

        Dates Last.Price Last.Price.1  id        carry
1  1998-11-30      94.50        98.50  QS -0.040609137
2  1998-11-30      31.32        32.13  HO -0.025210084
3  1998-12-31      95.50        98.00  QS -0.025510204
4  1998-12-31      34.00        34.28  HO -0.008168028
5  1999-01-29     100.00       100.50  QS -0.004975124
6  1999-01-29      33.16        33.42  HO -0.007779773
7  1999-02-26     100.25       100.25  QS  0.000000000
8  1999-02-26      32.29        32.37  HO -0.002471424
9  1999-02-26      10.88        11.00  CO -0.010909091
10 1999-03-31     131.50       130.75  QS  0.005736138
11 1999-03-31      44.68        44.00  HO  0.015454545
12 1999-03-31      15.24        15.16  CO  0.005277045

我想计算每个月每个 id 的权重。我有一个功能可以做到这一点。我使用 dplyr 来实现这一点：

library(dplyr)
library(lubridate)

contracts <- contracts %>%
  mutate(Dates = ymd(Dates)) %>%
  group_by(Dates) %>%
  mutate(weights = weight(carry))

给出：

contracts 
        Dates Last.Price Last.Price.1  id        carry     weights
1  1998-11-30      94.50        98.50  QS -0.040609137 0.616979910
2  1998-11-30      31.32        32.13  HO -0.025210084 0.383020090
3  1998-12-31      95.50        98.00  QS -0.025510204 0.757468623
4  1998-12-31      34.00        34.28  HO -0.008168028 0.242531377
5  1999-01-29     100.00       100.50  QS -0.004975124 0.390056023
6  1999-01-29      33.16        33.42  HO -0.007779773 0.609943977
7  1999-02-26     100.25       100.25  QS  0.000000000          NA
8  1999-02-26      32.29        32.37  HO -0.002471424 0.184703218
9  1999-02-26      10.88        11.00  CO -0.010909091 0.815296782
10 1999-03-31     131.50       130.75  QS  0.057361377 0.057361377
11 1999-03-31      44.68        44.00  HO  0.015454545 0.015454545
12 1999-03-31      15.24        15.16  CO  0.005277045 0.005277045

现在我想要权重的滞后，以便在 11 月计算的权重在 12 月应用。所以我基本上想按组移动权重列，该组是日期。所以 11 月的值最终成为 12 月的值，依此类推。

现在我还希望按 id 匹配移位，这样如果包含新 id，则第一次出现 id 的组将在滞后列中具有 NA。

所需的输出如下：

desired
        Dates Last.Price Last.Price.1  id        carry     weights          w
1  1998-11-30      94.50        98.50  QS -0.040609137 0.616979910         NA
2  1998-11-30      31.32        32.13  HO -0.025210084 0.383020090         NA
3  1998-12-31      95.50        98.00  QS -0.025510204 0.757468623 0.61697991
4  1998-12-31      34.00        34.28  HO -0.008168028 0.242531377 0.38302009
5  1999-01-29     100.00       100.50  QS -0.004975124 0.390056023 0.75746862
6  1999-01-29      33.16        33.42  HO -0.007779773 0.609943977 0.24253138
7  1999-02-26     100.25       100.25  QS  0.000000000          NA 0.39005602
8  1999-02-26      32.29        32.37  HO -0.002471424 0.184703218 0.60994398
9  1999-02-26      10.88        11.00  CO -0.010909091 0.815296782         NA
10 1999-03-31     131.50       130.75  QS  0.057361377 0.057361377         NA
11 1999-03-31      44.68        44.00  HO  0.015454545 0.015454545 0.18470322
12 1999-03-31      15.24        15.16  CO  0.005277045 0.005277045 0.81529678

注意 1999 年 2 月。CO 有一个 NA，因为它首先出现在 2 月。

现在看看 1999 年 3 月，CO 具有来自 Februray 的值，QS 具有 NA 只是因为 2 月的值是 NA（由于除以 0）。

这个可以吗？

数据：


contracts  <- read.table(text = "Dates, Last.Price, Last.Price.1, id,carry
1998-11-30, 94.500, 98.500, QS, -0.0406091371
1998-11-30, 31.320, 32.130, HO, -0.0252100840
1998-12-31, 95.500, 98.000, QS, -0.0255102041
1998-12-31, 34.000, 34.280, HO, -0.0081680280
1999-01-29, 100.000, 100.500, QS, -0.0049751244
1999-01-29, 33.160, 33.420, HO, -0.0077797726
1999-02-26, 100.250, 100.250, QS,  0.0000000000
1999-02-26, 32.290, 32.370, HO, -0.0024714242
1999-02-26, 10.880, 11.000, CO, -0.0109090909
1999-03-31, 131.500, 130.750, QS,  0.0057361377
1999-03-31, 44.680, 44.000, HO,  0.0154545455
1999-03-31, 15.240, 15.160, CO,  0.0052770449", sep = ",", header = T)


desired <- read.table(text = "Dates,Last.Price,Last.Price.1,id,carry,weights,w
1998-11-30,94.5,98.5, QS,-0.0406091371,0.616979909839741,NA
1998-11-30,31.32,32.13, HO,-0.025210084,0.383020090160259,NA
1998-12-31,95.5,98, QS,-0.0255102041,0.757468623182272,0.616979909839741
1998-12-31,34,34.28, HO,-0.008168028,0.242531376817728,0.383020090160259
1999-01-29,100,100.5, QS,-0.0049751244,0.390056023188584,0.757468623182272
1999-01-29,33.16,33.42, HO,-0.0077797726,0.609943976811416,0.242531376817728
1999-02-26,100.25,100.25, QS,0,NA,0.390056023188584
1999-02-26,32.29,32.37, HO,-0.0024714242,0.184703218189261,0.609943976811416
1999-02-26,10.88,11, CO,-0.0109090909,0.815296781810739,NA
1999-03-31,131.5,130.75, QS,0.057361377,0.057361377,NA
1999-03-31,44.68,44, HO,0.0154545455,0.0154545455,0.184703218189261
1999-03-31,15.24,15.16, CO,0.0052770449,0.0052770449,0.815296782", sep = ",", header = TRUE)

权重函数：

weight <- function(vec) {
  neg <- which(vec<0)
  w <- vec
  w[neg] <- vec[vec<0] / sum(vec[vec<0])
  w[-neg] <- vec[vec>=0] / sum(vec[vec>=0])
  w
}

【问题讨论】：

听起来你需要 dplyr 的 lag 和 lead 函数。

标签： r date dplyr data.table data-manipulation

【解决方案1】：

contracts %>%
  group_by(Dates) %>%
  mutate(weights = weight(carry)) %>%
  arrange(Dates) %>%
  group_by(id) %>%
  mutate(w = dplyr::lag(weights)) %>%
  ungroup()
# # A tibble: 12 x 7
#    Dates      Last.Price Last.Price.1 id       carry   weights       w
#    <chr>           <dbl>        <dbl> <chr>    <dbl>     <dbl>   <dbl>
#  1 1998-11-30       94.5         98.5 " QS" -0.0406    0.617    NA    
#  2 1998-11-30       31.3         32.1 " HO" -0.0252    0.383    NA    
#  3 1998-12-31       95.5         98   " QS" -0.0255    0.757     0.617
#  4 1998-12-31       34           34.3 " HO" -0.00817   0.243     0.383
#  5 1999-01-29      100          100.  " QS" -0.00498   0.390     0.757
#  6 1999-01-29       33.2         33.4 " HO" -0.00778   0.610     0.243
#  7 1999-02-26      100.         100.  " QS"  0       NaN         0.390
#  8 1999-02-26       32.3         32.4 " HO" -0.00247   0.185     0.610
#  9 1999-02-26       10.9         11   " CO" -0.0109    0.815    NA    
# 10 1999-03-31      132.         131.  " QS"  0.00574   0.00574 NaN    
# 11 1999-03-31       44.7         44   " HO"  0.0155    0.0155    0.185
# 12 1999-03-31       15.2         15.2 " CO"  0.00528   0.00528   0.815

注意事项：

我使用了dplyr::lag 而不仅仅是lag，因为它可能与stats::lag 混淆，后者的行为与dplyr::lag 明显不同。虽然大多数情况下它会正常工作，但它会一直工作直到它不起作用......而且它通常不会警告你:-)
无论月份如何，这都会滞后 Dates。我假设您确定Dates 总是非常频繁。如果您认为有可能存在差距（其中滞后的逐行不正确），那么您需要将年/月分解为一个新字段并加入自身，而不是执行lag。

【讨论】：