【发布时间】：2020-08-14 10:36:16
【问题描述】：
我有一些数据，我将 mdo 值除以前一组中 mdo 实例的计数。
我也在计算 sog avg。
但我想计算与结果 (mdo/count) 值相同的实例发生的 sog avg。
library(dplyr)
library(lubridate)
library(purrr)
df <- tibble(mydate = as.Date(c("2019-05-11 23:01:00", "2019-05-11 23:02:00", "2019-05-11 23:03:00", "2019-05-11 23:04:00",
"2019-05-12 23:05:00", "2019-05-12 23:06:00", "2019-05-12 23:07:00", "2019-05-12 23:08:00",
"2019-05-13 23:09:00", "2019-05-13 23:10:00", "2019-05-13 23:11:00", "2019-05-13 23:12:00",
"2019-05-14 23:13:00", "2019-05-14 23:14:00", "2019-05-14 23:15:00", "2019-05-14 23:16:00",
"2019-05-15 23:17:00", "2019-05-15 23:18:00", "2019-05-15 23:19:00", "2019-05-15 23:20:00",
"2019-05-15 23:21:00", "2019-05-15 23:22:00", "2019-05-15 23:23:00", "2019-05-15 23:24:00",
"2019-05-15 23:25:00")),
mdo = c(1500, 1500, 1500, 1500,
1500, 1500, NA, 0,
0, 0, 900, 900, NA, NA, 1100, 1100,
1100, 200, 200, 200,200,
1100, 1100, 1100, 0
),
sog = c(12, 12, 12, 11, 10,9,
2,8.8, 8.7, 7.8, 11, 11, 12, 11,
9.54, 9.8, 10.4,4, 4, 4.5, 3.6,
7, 8, 9, 0))
df1 <- df %>%
mutate(grp = data.table::rleid(mdo))
df1 <- df1 %>%
#Keep only non-NA value
filter(!is.na(mdo)) %>%
#count occurence of each grp
count(grp, name = 'count') %>%
#Shift the count to the previous group
mutate(count = lag(count)) %>%
#Join with the original data
right_join(df1, by = 'grp') %>%
arrange(grp)
group_mdo <- df1 %>%
select(grp, mdo) %>%
unique() %>%
mutate(prev_mdo = lag(mdo, na.rm=TRUE)) %>%
select(-mdo) %>%
tidyr::fill(prev_mdo, .direction = "down")
df1 <- df1 %>%
left_join(group_mdo, by = "grp") %>%
mutate(result = ifelse(prev_mdo != 0, mdo / count, 0)) %>%
mutate(sog_avg = ifelse(prev_mdo != 0, map_dbl(.x = grp - 1, ~ mean(sog[grp == .x], na.rm=TRUE), na.rm=TRUE), NA))
现在的结果是：
grp count mydate mdo sog prev_mdo result sog_avg
1 NA 2019-05-11 1500 12 NA NA NA
1 NA 2019-05-11 1500 12 NA NA NA
1 NA 2019-05-11 1500 12 NA NA NA
1 NA 2019-05-11 1500 11 NA NA NA
1 NA 2019-05-12 1500 10 NA NA NA
1 NA 2019-05-12 1500 9 NA NA NA
2 NA 2019-05-12 NA 2 1500 NA 11
3 6 2019-05-12 0 8.8 1500 0 2
3 6 2019-05-13 0 8.7 1500 0 2
3 6 2019-05-13 0 7.8 1500 0 2
4 3 2019-05-13 900 11 0 0 NA
4 3 2019-05-13 900 11 0 0 NA
5 NA 2019-05-14 NA 12 900 NA 11
5 NA 2019-05-14 NA 11 900 NA 11
6 2 2019-05-14 1100 9.54 900 550 11.5
6 2 2019-05-14 1100 9.8 900 550 11.5
6 2 2019-05-15 1100 10.4 900 550 11.5
7 3 2019-05-15 200 4 1100 66.7 9.91
7 3 2019-05-15 200 4 1100 66.7 9.91
7 3 2019-05-15 200 4.5 1100 66.7 9.91
7 3 2019-05-15 200 3.6 1100 66.7 9.91
8 4 2019-05-15 1100 7 200 275 4.03
8 4 2019-05-15 1100 8 200 275 4.03
8 4 2019-05-15 1100 9 200 275 4.03
9 3 2019-05-15 0 0 1100 0 8
我想要的结果：
grp count mydate mdo sog prev_mdo result sog_avg
1 NA 2019-05-11 1500 12 NA NA NA
1 NA 2019-05-11 1500 12 NA NA NA
1 NA 2019-05-11 1500 12 NA NA NA
1 NA 2019-05-11 1500 11 NA NA NA
1 NA 2019-05-12 1500 10 NA NA NA
1 NA 2019-05-12 1500 9 NA NA NA
2 NA 2019-05-12 NA 2 1500 NA NA
3 6 2019-05-12 0 8.8 1500 0 0
3 6 2019-05-13 0 8.7 1500 0 0
3 6 2019-05-13 0 7.8 1500 0 0
4 3 2019-05-13 900 11 0 0 0
4 3 2019-05-13 900 11 0 0 0
5 NA 2019-05-14 NA 12 900 NA NA
5 NA 2019-05-14 NA 11 900 NA NA
6 2 2019-05-14 1100 9.54 900 550 11
6 2 2019-05-14 1100 9.8 900 550 11
6 2 2019-05-15 1100 10.4 900 550 11
7 3 2019-05-15 200 4 1100 66.7 9.91
7 3 2019-05-15 200 4 1100 66.7 9.91
7 3 2019-05-15 200 4.5 1100 66.7 9.91
7 3 2019-05-15 200 3.6 1100 66.7 9.91
8 4 2019-05-15 1100 7 200 275 4.03
8 4 2019-05-15 1100 8 200 275 4.03
8 4 2019-05-15 1100 9 200 275 4.03
9 3 2019-05-15 0 0 1100 0 0
result 为 0 时，sog_avg 应为 0，result 为 na 时，sog avg 应为 na。
如果使用之前的组数计算结果，则应该使用之前的值计算 sog avg。
所以，例如：
mdo = 1100，结果为 550，因为前一个非空组中的计数为 2（mdo 值 900）。
1100 / 2 = 550 。此时 sog avg 应为 (11 + 11) / 2 = 11，因为在前一个非空组中计数为 2。
【问题讨论】：
-
您需要
df1 %>% mutate(sog_avg = ifelse(is.na(result) | result == 0, result, sog_avg))吗？ -
@RonakShah：嗨，不，我不想要这个。我想按照我所说的正确计算 sog_avg 然后好的，如果结果是 nan 或 zero ，是的
-
写
lag(mdo, na.rm=TRUE)的时候想做什么？dplyr::lag没有名为na.rm =的参数 -
@Edo：是的，你是对的，不过你可以忽略它。
-
为什么计数
NA是grp == 5？我认为这没有意义。应该是 2。