R将一列的所有值除以同一列但不同组中的值

我有一个数据表，如下所示-

library(data.table)
dt <- structure(list(date = structure(c(17956L, 17959L, 17960L, 
                                  17962L, 17963L, 17966L, 17967L, 17968L, 17969L, 17970L, 17973L, 
                                  17974L, 17975L, 17976L, 17977L, 17980L, 17981L, 17982L, 17983L, 
                                  17984L, 17956L, 17959L, 17960L, 17961L, 17962L, 17963L, 17966L, 
                                  17967L, 17968L, 17980L, 17981L, 17982L, 17983L, 17984L), class = c("IDate", "Date")), 
               group = c("A", "A", "A", "A", 
                          "A", "A", "A", "A", "A", "A", "A", "A", 
                          "A", "A", "A", "A", "A", "A", "A", "A", 
                          "B", "B", "B", "B", "B", "B", "B", "B", 
                          "B", "B", "B", "B", "B", "B"), 
               value = c(43.7425, 
                         43.9625, 43.8825, 43.125, 43.2275, 44.725, 45.2275, 45.4275, 
                         45.9325, 46.53, 47.005, 46.6325, 47.04, 48.7725, 47.7625, 47.185, 
                         46.6975, 47.1175, 47.18, 47.4875, 12.31, 12.51, 12.7, 12.4, 12.63, 
                         12.93, 13.18, 13.23, 13.35, 14.27, 14.5, 14.25, 13.88, 13.71)), 
          row.names = c(NA, -34L), class = c("data.table", "data.frame"))
> dt
          date group   value
 1: 2019-03-01     A 43.7425
 2: 2019-03-04     A 43.9625
 3: 2019-03-05     A 43.8825
 5: 2019-03-07     A 43.1250
 6: 2019-03-08     A 43.2275
 7: 2019-03-11     A 44.7250
 8: 2019-03-12     A 45.2275
 9: 2019-03-13     A 45.4275
10: 2019-03-14     A 45.9325
11: 2019-03-15     A 46.5300
12: 2019-03-18     A 47.0050
13: 2019-03-19     A 46.6325
14: 2019-03-20     A 47.0400
15: 2019-03-21     A 48.7725
16: 2019-03-22     A 47.7625
17: 2019-03-25     A 47.1850
18: 2019-03-26     A 46.6975
19: 2019-03-27     A 47.1175
20: 2019-03-28     A 47.1800
21: 2019-03-29     A 47.4875
22: 2019-03-01     B 12.3100
23: 2019-03-04     B 12.5100
24: 2019-03-05     B 12.7000
25: 2019-03-06     B 12.4000
26: 2019-03-07     B 12.6300
27: 2019-03-08     B 12.9300
28: 2019-03-11     B 13.1800
29: 2019-03-12     B 13.2300
30: 2019-03-13     B 13.3500
31: 2019-03-25     B 14.2700
32: 2019-03-26     B 14.5000
33: 2019-03-27     B 14.2500
34: 2019-03-28     B 13.8800
35: 2019-03-29     B 13.7100

请注意-

在<代码>比率列中，我展示了计算方法，以便于理解。我只需要

比率

列中的比率值

组由日期列编制索引

这两列中都缺少值。稍微做一点错误处理会有所帮助

---

我希望有一种简单的方法可以使用

data.table

库来解决这个问题。

我可以提供一种简洁的方法

library(tidyr)
library(dplyr)
df1 <- df %>%   
  pivot_wider(names_from = group, values_from = value) %>% 
  mutate(ratio = B/A)

library（tidyr）
图书馆（dplyr）
df1%
透视图（名称从=组，值从=值）%>%
突变（比率=B/A）

以下是使用data.table的一行解决方案。它会处理丢失的值

setDT(dt)[, ratio := value/value[group=="A"] , date]
          date group   value     ratio
 1: 2019-03-01     A 43.7425 1.0000000
 2: 2019-03-04     A 43.9625 1.0000000
 3: 2019-03-05     A 43.8825 1.0000000
 4: 2019-03-07     A 43.1250 1.0000000
 5: 2019-03-08     A 43.2275 1.0000000
 6: 2019-03-11     A 44.7250 1.0000000
 7: 2019-03-12     A 45.2275 1.0000000
 8: 2019-03-13     A 45.4275 1.0000000
 9: 2019-03-14     A 45.9325 1.0000000
10: 2019-03-15     A 46.5300 1.0000000
11: 2019-03-18     A 47.0050 1.0000000
12: 2019-03-19     A 46.6325 1.0000000
13: 2019-03-20     A 47.0400 1.0000000
14: 2019-03-21     A 48.7725 1.0000000
15: 2019-03-22     A 47.7625 1.0000000
16: 2019-03-25     A 47.1850 1.0000000
17: 2019-03-26     A 46.6975 1.0000000
18: 2019-03-27     A 47.1175 1.0000000
19: 2019-03-28     A 47.1800 1.0000000
20: 2019-03-29     A 47.4875 1.0000000
21: 2019-03-01     B 12.3100 0.2814197
22: 2019-03-04     B 12.5100 0.2845607
23: 2019-03-05     B 12.7000 0.2894092
24: 2019-03-06     B 12.4000        NA
25: 2019-03-07     B 12.6300 0.2928696
26: 2019-03-08     B 12.9300 0.2991151
27: 2019-03-11     B 13.1800 0.2946898
28: 2019-03-12     B 13.2300 0.2925211
29: 2019-03-13     B 13.3500 0.2938749
30: 2019-03-25     B 14.2700 0.3024266
31: 2019-03-26     B 14.5000 0.3105091
32: 2019-03-27     B 14.2500 0.3024354
33: 2019-03-28     B 13.8800 0.2941925
34: 2019-03-29     B 13.7100 0.2887076
          date group   value     ratio

我们也可以这样做

library(data.table)
dt[order(date, group), ratio := value/first(value), date]

---

我的建议是在

date

上执行

dt

与子集

dt[group==“A”]

的更新联接。这将自动处理丢失的值：

dt[dt[group == "A"], on = "date", ratio := x.value / i.value][]

请注意，这种方法在第24行返回

NA

，因为

A组中没有相应的日期
2019-03-06

基准
由于提出了几种解决方案，我想知道在执行速度和内存消耗方面有什么区别：

“分组”：按日期分组的分组
“加入”：按照此答案中的建议更新加入
“重塑”、“重塑2”：的
data.table
版本，但经过大量扩展，以两种风格返回与其他答案相同的结果
对于基准测试，将
测试台
包用于不同的问题大小。还模拟了缺失值对日期的影响，这些缺失值会导致某些日期不匹配

这是通过分别为每个组
A
和
B
创建单独的数据表来实现的，其行数比指定的问题大小
n
多10%。从这两个data.table
n
行中，对每一行进行采样，并将其组合到实际的基准数据中。table
dt0
dt0
有2行
n
行

还请注意，每个基准测试运行都从一个新的
dt0
副本开始，因为有些方法会修改输入数据


请注意对数时间刻度

对于较小规模的问题，小组方法最快。随着问题规模的增加，这种方法被其他方法取代，特别是连接方法。对于n=107，join的速度大约是group的七倍。令人惊讶的是，尽管数据被前后重塑，但重塑方法是第二快的

虽然该组的速度相对较慢，问题规模较大，但内存占用最少（
mem\u alloc
）

请注意，即使对于最大的问题大小，所有计时都在20秒以下。因此，只有在多次重复该操作时，任何速度差异都可能很重要。
我仍然不确定您想要的输出是什么。维度是多少？收音机看起来像是B组的值除以A组的相同日期。这是正确的。结果的形状有些不同。
data.table
等价物将是
dcast（dt，date~group）[，ratio:=B/A]
。不幸的是，答案没有显示输出，这隐藏了一个事实，即答案在第24行返回了1.0的
比率，尽管在
A
组中没有相应的2019-03-06年
日期。这与此处返回
NA
的其他答案形成对比，包括。
dt[dt[group == "A"], on = "date", ratio := x.value / i.value][]

          date group   value     ratio
 1: 2019-03-01     A 43.7425 1.0000000
 2: 2019-03-04     A 43.9625 1.0000000
 3: 2019-03-05     A 43.8825 1.0000000
 4: 2019-03-07     A 43.1250 1.0000000
 5: 2019-03-08     A 43.2275 1.0000000
 6: 2019-03-11     A 44.7250 1.0000000
 7: 2019-03-12     A 45.2275 1.0000000
 8: 2019-03-13     A 45.4275 1.0000000
 9: 2019-03-14     A 45.9325 1.0000000
10: 2019-03-15     A 46.5300 1.0000000
11: 2019-03-18     A 47.0050 1.0000000
12: 2019-03-19     A 46.6325 1.0000000
13: 2019-03-20     A 47.0400 1.0000000
14: 2019-03-21     A 48.7725 1.0000000
15: 2019-03-22     A 47.7625 1.0000000
16: 2019-03-25     A 47.1850 1.0000000
17: 2019-03-26     A 46.6975 1.0000000
18: 2019-03-27     A 47.1175 1.0000000
19: 2019-03-28     A 47.1800 1.0000000
20: 2019-03-29     A 47.4875 1.0000000
21: 2019-03-01     B 12.3100 0.2814197
22: 2019-03-04     B 12.5100 0.2845607
23: 2019-03-05     B 12.7000 0.2894092
24: 2019-03-06     B 12.4000        NA
25: 2019-03-07     B 12.6300 0.2928696
26: 2019-03-08     B 12.9300 0.2991151
27: 2019-03-11     B 13.1800 0.2946898
28: 2019-03-12     B 13.2300 0.2925211
29: 2019-03-13     B 13.3500 0.2938749
30: 2019-03-25     B 14.2700 0.3024266
31: 2019-03-26     B 14.5000 0.3105091
32: 2019-03-27     B 14.2500 0.3024354
33: 2019-03-28     B 13.8800 0.2941925
34: 2019-03-29     B 13.7100 0.2887076
          date group   value     ratio

library(bench)
library(ggplot2)
bm <- press(
  n = 10^(3:7)
  , {
    nx <- as.integer(n * 1.1)
    dates <- seq(as.IDate("1970-01-01"), by = 1L, length.out = nx)
    dtA <- data.table(date = dates,  group = "A", value = (1:nx) * pi)
    dtB <- data.table(date = dates,  group = "B", value = (1:nx) * 2*pi)
    set.seed(123)
    dt0 <- rbind(dtA[sample(nx, n)], dtB[sample(nx, n)])
    setorder(dt0, group, date)
    mark(
      join = {
        dt <- copy(dt0)
        dt[dt[group == "A"], on = "date", ratio := x.value / i.value]
      }, 
      group = {
        dt <- copy(dt0)
        dt[, ratio := value/value[group=="A"] , date]  
      },
      reshape = {
        dt <- copy(dt0)
        dcast(dt, date ~ group)[, c("ratioA", "ratioB") := .(A/A, B/A)][
          , melt(.SD, measure.vars = list(value = c("A", "B"), ratio = c("ratioA", "ratioB")), 
                 variable.name = "group")][
                   !(is.na(value) & is.na(ratio))][
                     , group := c("A", "B")[group]]
      },
      reshape2 = {
        dt <- copy(dt0)
        dcast(dt, date ~ group)[, c("ratioA", "ratioB") := .(rep(1.0, .N), B/A)][
          , melt(.SD, measure.vars = patterns(value = "^[AB]", ratio = "^ratio"), 
                 variable.name = "group")][
                   , group := c("A", "B")[group]][
                     !is.na(value)]
      },
      check = function(x,y) all.equal(x, y, check.attributes = FALSE),
      min_iterations = 3L
    )
  }
)

 bm[, 1:10]

# A tibble: 20 x 10
   expression        n      min   median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc total_time
   <bch:expr>    <dbl> <bch:tm> <bch:tm>     <dbl> <bch:byt>    <dbl> <int> <dbl>   <bch:tm>
 1 join           1000   3.19ms   3.88ms  243.      355.42KB    2.10    116     1   477.13ms
 2 group          1000   2.27ms   2.81ms  342.      137.24KB    2.05    167     1    488.5ms
 3 reshape        1000   4.74ms   6.26ms  155.      805.56KB    2.12     73     1   472.21ms
 4 reshape2       1000   4.68ms   6.18ms  159.      797.66KB    2.09     76     1    479.4ms
 5 join          10000   4.86ms   5.96ms  161.         1.9MB    0        81     0   503.38ms
 6 group         10000  18.68ms  19.77ms   49.7     930.32KB    4.52     22     2   442.34ms
 7 reshape       10000   9.13ms  11.59ms   83.9       4.29MB    0        42     0   500.68ms
 8 reshape2      10000  10.58ms  12.83ms   76.7       4.21MB    0        39     0   508.78ms
 9 join         100000  23.43ms  28.14ms   35.2      17.41MB    0        18     0   512.06ms
10 group        100000 187.33ms 192.88ms    5.18      9.13MB    2.59      2     1   385.76ms
11 reshape      100000   51.8ms  57.67ms   17.4      39.31MB    2.17      8     1   460.72ms
12 reshape2     100000  50.59ms  56.78ms   17.3      38.55MB    0         9     0   520.46ms
13 join        1000000 183.66ms 184.12ms    5.40    172.53MB    0         3     0    555.4ms
14 group       1000000    1.83s    1.98s    0.486    86.12MB    2.43      3    15      6.18s
15 reshape     1000000 473.52ms 492.22ms    2.05    389.47MB    0         3     0      1.46s
16 reshape2    1000000 498.48ms 505.92ms    1.97    381.84MB    0         3     0      1.52s
17 join       10000000    2.01s    2.44s    0.432     1.68GB    0.576     3     4      6.95s
18 group      10000000   18.41s    18.7s    0.0531   860.1MB    2.76      3   156     56.46s
19 reshape    10000000    6.07s    6.46s    0.142      3.8GB    0.237     3     5      21.1s
20 reshape2   10000000    6.01s    6.02s    0.161     3.73GB    0.322     3     6     18.65s

autoplot(bm)