Question

我想按组聚合日期。但是，每个观察可以属于几个组（例如，观察1属于组A和B）。我找不到用data.table实现这个目标的好方法。目前，我为每个可能的组创建了一个逻辑变量，如果观察属于该组，则该变量取值TRUE。我正在寻找一种比下面提供的更好的方法。我还想知道如何使用tidyverse实现这一目标。

library(data.table)
# Data
set.seed(1)
TF <- c(TRUE, FALSE)
time <- rep(1:4, each = 5)
df <- data.table(time = time, x = rnorm(20), groupA = sample(TF, size = 20, replace = TRUE),
                                             groupB = sample(TF, size = 20, replace = TRUE),
                                             groupC = sample(TF, size = 20, replace = TRUE))

# This should be nicer and less repetitive
df[groupA == TRUE, .(A = sum(x)), by = time][
  df[groupB == TRUE, .(B = sum(x)), by = time], on = "time"][
    df[groupC == TRUE, .(C = sum(x)), by = time], on = "time"]

# desired output
time          A          B         C
1:    1         NA  0.9432955 0.1331984
2:    2  1.2257538  0.2427420 0.1882493
3:    3 -0.1992284 -0.1992284 1.9016244
4:    4  0.5327774  0.9438362 0.9276459

Answer 1

以下是data.table的解决方案：

df[, lapply(.SD[, .(groupA, groupB, groupC)]*x, sum), time]
# > df[, lapply(.SD[, .(groupA, groupB, groupC)]*x, sum), time]
#    time     groupA     groupB    groupC
# 1:    1  0.0000000  0.9432955 0.1331984
# 2:    2  1.2257538  0.2427420 0.1882493
# 3:    3 -0.1992284 -0.1992284 1.9016244
# 4:    4  0.5327774  0.9438362 0.9276459

或（thx to @ chinsoon12 for the comment）更多编程：

df[, lapply(.SD*x, sum), by=.(time), .SDcols=paste0("group", c("A","B","C"))]

如果您想要长格式的结果，您可以这样做：

df[, colSums(.SD*x), by=.(time), .SDcols=paste0("group", c("A","B","C"))]
### with indicator for the group:
df[, .(colSums(.SD*x), c("A","B","C")), by=.(time), .SDcols=paste0("group", c("A","B","C"))]

Answer 2

我认为这里以长格式工作更容易。首先，我将观察结果收集到长格式，然后仅保留观察属于相应组的值。然后我删除逻辑列，并将组重命名为单个字母。然后我在不同的群组和时间进行汇总（总结在dplyr）。最后，我回到了广泛的格式。

library(dplyr)
library(tidyr)

set.seed(1)
TF <- c(TRUE, FALSE)
time <- rep(1:4, each = 5)


df <- data.frame(time = time, x = rnorm(20), groupA = sample(TF, size = 20, replace = TRUE),
                 groupB = sample(TF, size = 20, replace = TRUE),
                 groupC = sample(TF, size = 20, replace = TRUE))


df %>% 
  gather(group, belongs, groupA:groupC) %>% 
  filter(belongs) %>% 
  select(-belongs) %>% 
  mutate(group = gsub("group", "", group)) %>% 
  group_by(time, group) %>% 
  summarise(x = sum(x)) %>% 
  spread(group, x)

输出

# A tibble: 4 x 4
# Groups:   time [4]
   time       A      B     C
  <int>   <dbl>  <dbl> <dbl>
1     1  NA      0.943 0.133
2     2   1.23   0.243 0.188
3     3  -0.199 -0.199 1.90 
4     4   0.533  0.944 0.928

Answer 3

选项可以将tidyr和dplyr个包与data.table结合使用。尝试以长格式处理数据，然后将其更改为宽格式。

library(dplyr)
library(tidyr)

melt(df, id.vars = c("time", "x")) %>%
  filter(value) %>%
  group_by(time, variable) %>%
  summarise(sum = sum(x)) %>%
  spread(variable, sum)

# # A tibble: 4 x 4
# # Groups: time [4]
# time  groupA groupB groupC
# * <int>   <dbl>  <dbl>  <dbl>
# 1     1  NA      0.943  0.133
# 2     2   1.23   0.243  0.188
# 3     3 - 0.199 -0.199  1.90 
# 4     4   0.533  0.944  0.928

如果每个观察可以属于多个组，则进行聚合

3 个答案: