我正在尝试遵循一些有关何时将数据分组到图表的规则。 我将如何从这个数据框开始：
# A tibble: 11 x 8
assay year qtr invalid valid total_assays hfr predicted_inv
<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 test_case 2016. 1. 2. 36. 38. 0.0350 1.33
2 test_case 2016. 2. 1. 34. 35. 0.0350 1.23
3 test_case 2016. 3. 0. 25. 25. 0.0350 0.875
4 test_case 2016. 4. 2. 23. 25. 0.0350 0.875
5 test_case 2017. 1. 1. 29. 30. 0.0350 1.05
6 test_case 2017. 2. 2. 24. 26. 0.0350 0.910
7 test_case 2017. 3. 0. 23. 23. 0.0350 0.805
8 test_case 2017. 4. 1. 20. 21. 0.0350 0.735
9 test_case 2018. 1. 2. 33. 35. 0.0350 1.23
10 test_case 2018. 2. 5. 28. 33. 0.0350 1.16
11 test_case 2018. 3. 4. 9. 13. 0.0350 0.455
对此：
assay year qtr invalid valid total_assays hfr predicted_inv co_inv co_val co_prd_inv trend
1 test_case 2016 1 2 36 38 0.035 1.330 2 36 1.330 No
2 test_case 2016 2 1 34 35 0.035 1.225 3 70 2.555 No
3 test_case 2016 3 0 25 25 0.035 0.875 3 95 3.430 No
4 test_case 2016 4 2 23 25 0.035 0.875 5 118 4.305 Yes
5 test_case 2017 1 1 29 30 0.035 1.050 1 29 1.050 No
6 test_case 2017 2 2 24 26 0.035 0.910 3 53 1.960 No
7 test_case 2017 3 0 23 23 0.035 0.805 3 76 2.765 No
8 test_case 2017 4 1 20 21 0.035 0.735 4 96 3.500 No
9 test_case 2018 1 2 33 35 0.035 1.225 6 129 4.725 Yes
10 test_case 2018 2 5 28 33 0.035 1.155 5 28 1.155 Yes
11 test_case 2018 3 4 9 13 0.035 0.455 4 9 0.455 No
规则很简单。对于每一行，如果invalid或Forecast_inv的累加总和为5或更大，则趋势为“是”，并且将重置所有三个参数（无效，有效，predicted_inv）的累加总和，然后从下一行重新开始。最后，将对分组（co_ *）进行趋势分析。
我已经尝试过使用dplyr解决方案，但是当我尝试同时创建多个相互依赖的变量时，却不断出错。
现在，我正在尝试仅将3个参数作为向量的自定义函数，但是我一直被迫构建循环...我希望使用一种易于阅读的dplyr解决方案。
这里是垃圾：
egdf1 <- structure(list(assay = c("test_case", "test_case", "test_case",
"test_case", "test_case", "test_case", "test_case", "test_case",
"test_case", "test_case", "test_case"), year = c(2016, 2016,
2016, 2016, 2017, 2017, 2017, 2017, 2018, 2018, 2018), qtr = c(1,
2, 3, 4, 1, 2, 3, 4, 1, 2, 3), invalid = c(2, 1, 0, 2, 1, 2,
0, 1, 2, 5, 4), valid = c(36, 34, 25, 23, 29, 24, 23, 20, 33,
28, 9), total_assays = c(38, 35, 25, 25, 30, 26, 23, 21, 35,
33, 13), hfr = c(0.035, 0.035, 0.035, 0.035, 0.035, 0.035, 0.035,
0.035, 0.035, 0.035, 0.035), predicted_inv = c(1.33, 1.225, 0.875,
0.875, 1.05, 0.91, 0.805, 0.735, 1.225, 1.155, 0.455)), .Names = c("assay",
"year", "qtr", "invalid", "valid", "total_assays", "hfr", "predicted_inv"
), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,
-11L))
egdf2 <- structure(list(assay = c("test_case", "test_case", "test_case",
"test_case", "test_case", "test_case", "test_case", "test_case",
"test_case", "test_case", "test_case"), year = c(2016L, 2016L,
2016L, 2016L, 2017L, 2017L, 2017L, 2017L, 2018L, 2018L, 2018L
), qtr = c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L), invalid = c(2L,
1L, 0L, 2L, 1L, 2L, 0L, 1L, 2L, 5L, 4L), valid = c(36L, 34L,
25L, 23L, 29L, 24L, 23L, 20L, 33L, 28L, 9L), total_assays = c(38L,
35L, 25L, 25L, 30L, 26L, 23L, 21L, 35L, 33L, 13L), hfr = c(0.035,
0.035, 0.035, 0.035, 0.035, 0.035, 0.035, 0.035, 0.035, 0.035,
0.035), predicted_inv = c(1.33, 1.225, 0.875, 0.875, 1.05, 0.91,
0.805, 0.735, 1.225, 1.155, 0.455), co_inv = c(2L, 3L, 3L, 5L,
1L, 3L, 3L, 4L, 6L, 5L, 4L), co_val = c(36L, 70L, 95L, 118L,
29L, 53L, 76L, 96L, 129L, 28L, 9L), co_prd_inv = c(1.33, 2.555,
3.43, 4.305, 1.05, 1.96, 2.765, 3.5, 4.725, 1.155, 0.455), trend = c("No",
"No", "No", "Yes", "No", "No", "No", "No", "Yes", "Yes", "No"
)), .Names = c("assay", "year", "qtr", "invalid", "valid", "total_assays",
"hfr", "predicted_inv", "co_inv", "co_val", "co_prd_inv", "trend"
), class = "data.frame", row.names = c(NA, -11L))
答案 0 :(得分：1)
使用MESS软件包中的函数cumsumbinning设置阈值的值，该阈值不得超过累积组总和（在您的示例中为5）。请记住，在第9行中，因为加2到4越过阈值5，则会创建另一个组，而在您想要的输出中，您需要在下一行进行重置。
library(MESS)
egdf1 %>%
group_by(group = cumsumbinning(invalid, 5)) %>%
mutate(
co_inv = cumsum(invalid),
co_val = cumsum(valid),
co_prd_inv = cumsum(predicted_inv),
trend = ifelse(group - lag(group, default = 0) > 1, "yes", "no")
)
输出
assay year qtr invalid valid total_assays hfr predicted_inv group co_inv co_val co_prd_inv trend
<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <int> <dbl> <dbl> <dbl> <chr>
1 test_case 2016 1 2 36 38 0.035 1.33 1 2 36 1.33 no
2 test_case 2016 2 1 34 35 0.035 1.23 1 3 70 2.56 no
3 test_case 2016 3 0 25 25 0.035 0.875 1 3 95 3.43 no
4 test_case 2016 4 2 23 25 0.035 0.875 1 5 118 4.30 no
5 test_case 2017 1 1 29 30 0.035 1.05 2 1 29 1.05 yes
6 test_case 2017 2 2 24 26 0.035 0.91 2 3 53 1.96 no
7 test_case 2017 3 0 23 23 0.035 0.805 2 3 76 2.76 no
8 test_case 2017 4 1 20 21 0.035 0.735 2 4 96 3.5 no
9 test_case 2018 1 2 33 35 0.035 1.23 3 2 33 1.23 yes
10 test_case 2018 2 5 28 33 0.035 1.16 4 5 28 1.16 yes
11 test_case 2018 3 4 9 13 0.035 0.455 5 4 9 0.455 yes
答案 1 :(得分：1)
使用的基本R解决方案：
Reduce