將行分組到r data.table中的當前行

Question

我有一個如下所示的數據集：

library(data.table)

set.seed(10)

n_rows <- 50

data <- data.table(id = 1:n_rows,
                   timestamp = Sys.Date() + as.difftime(1:n_rows, units = "days"),
                   subject = sample(letters[1:4], n_rows, replace = T),
                   response = sample(3, n_rows, replace = T)
                   )

head(data, 10)

    id  timestamp subject response
 1:  1 2016-05-17       c        2
 2:  2 2016-05-18       b        3
 3:  3 2016-05-19       b        1
 4:  4 2016-05-20       c        2
 5:  5 2016-05-21       a        1
 6:  6 2016-05-22       a        2
 7:  7 2016-05-23       b        2
 8:  8 2016-05-24       b        2
 9:  9 2016-05-25       c        2
10: 10 2016-05-26       b        2

我需要通過按日期對每個響應的出現進行求和的操作來進行一些分組。

下面的group by生成nth_test列。

new_vars <- data[, .(id, timestamp, nth_test = 1:.N, response), by=.(subject)]

    subject id  timestamp nth_test response
 1:       c  1 2016-05-17        1        2
 2:       c  4 2016-05-20        2        2
 3:       c  9 2016-05-25        3        2
 4:       c 11 2016-05-27        4        1
 5:       c 12 2016-05-28        5        1
 6:       c 14 2016-05-30        6        2
 7:       c 22 2016-06-07        7        2
 8:       c 26 2016-06-11        8        2
 9:       c 31 2016-06-16        9        3
10:       c 36 2016-06-21       10        1

但我不知道如何生成列resp_1，resp_2和resp_3，如下所示。

    subject id  timestamp nth_test response resp_1 resp_2 resp_3
 1:       c  1 2016-05-17        1        2      0      1      0
 2:       c  4 2016-05-20        2        2      0      2      0
 3:       c  9 2016-05-25        3        2      0      3      0
 4:       c 11 2016-05-27        4        1      1      3      0
 5:       c 12 2016-05-28        5        1      2      3      0
 6:       c 14 2016-05-30        6        2      2      4      0
 7:       c 22 2016-06-07        7        2      2      5      0
 8:       c 26 2016-06-11        8        2      2      6      0
 9:       c 31 2016-06-16        9        3      2      6      1
10:       c 36 2016-06-21       10        1      3      6      1

干杯

Answer 1

我們可以嘗試

Un1 <- unique(sort(data$response))
data[, c("nth_test", paste("resp", Un1, sep="_")) := c(list(1:.N),
         lapply(Un1, function(x) cumsum(x==response))) , .(subject)]
data[order(subject, timestamp)][subject=="c"]
#    id  timestamp subject response nth_test resp_1 resp_2 resp_3
# 1:  1 2016-05-17       c        2        1      0      1      0
# 2:  4 2016-05-20       c        2        2      0      2      0
# 3:  9 2016-05-25       c        2        3      0      3      0
# 4: 11 2016-05-27       c        1        4      1      3      0
# 5: 12 2016-05-28       c        1        5      2      3      0
# 6: 14 2016-05-30       c        2        6      2      4      0
# 7: 22 2016-06-07       c        2        7      2      5      0
# 8: 26 2016-06-11       c        2        8      2      6      0
# 9: 31 2016-06-16       c        3        9      2      6      1
#10: 36 2016-06-21       c        1       10      3      6      1
#11: 39 2016-06-24       c        1       11      4      6      1
#12: 40 2016-06-25       c        1       12      5      6      1
#13: 44 2016-06-29       c        2       13      5      7      1

Answer 2

我想看看如果在data.table是長格式的情況下完成cummax / cumsum會是什么樣子（在某些配置中可能更有效）：

> data[order(subject, timestamp)
+      ][, rCnt := 1:.N, .(subject, response)
+      ][, responseStr := sprintf('%s_%s', 'resp', response)
+      ][, dcast(.SD, id + timestamp + subject + response ~ responseStr, value.var='rCnt', fill=0)
+      ][, melt(.SD, id.vars=c('id', 'timestamp', 'subject', 'response'))
+      ][order(subject, timestamp)
+      ][, value := cummax(value), .(subject, variable)
+      ][, nth_test := 1:.N, .(subject, variable)
+      ][, dcast(.SD, id + timestamp + subject + response + nth_test ~ variable, value.var='value')
+      ][order(subject, timestamp)
+      ][subject == 'c'
+      ]
    id  timestamp subject response nth_test resp_1 resp_2 resp_3
 1:  1 2016-05-17       c        2        1      0      1      0
 2:  4 2016-05-20       c        2        2      0      2      0
 3:  9 2016-05-25       c        2        3      0      3      0
 4: 11 2016-05-27       c        1        4      1      3      0
 5: 12 2016-05-28       c        1        5      2      3      0
 6: 14 2016-05-30       c        2        6      2      4      0
 7: 22 2016-06-07       c        2        7      2      5      0
 8: 26 2016-06-11       c        2        8      2      6      0
 9: 31 2016-06-16       c        3        9      2      6      1
10: 36 2016-06-21       c        1       10      3      6      1
11: 39 2016-06-24       c        1       11      4      6      1
12: 40 2016-06-25       c        1       12      5      6      1
13: 44 2016-06-29       c        2       13      5      7      1
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