查看矢量化对以找到 return 个匹配值
Look across vectorised pairs to return matching values
与我在这里问的一个问题相关：
示例初始数据：
person start_loc start_date end_date.1 end_date.2 end_date.3 end_loc.1 end_loc.2 end_loc.3
1 1 a 2021-02-10 2021-02-17 <NA> <NA> g
2 2 a 2021-01-30 2020-09-29 2020-12-12 2021-02-04 a a g
3 3 g 2020-12-04 <NA> <NA> <NA>
4 4 r 2020-12-09 2020-12-12 2020-12-14 2021-01-05 c c g
5 5 t 2021-03-22 2021-03-25 2021-03-29 <NA> b t
6 6 b 2021-04-04 2021-04-07 2021-04-09 <NA> b t
example <- structure(list(person = 1:6, start_loc = c("a", "a", "g", "r",
"t", "b"), start_date = structure(c(18668, 18657, 18600, 18605,
18708, 18721), class = "Date"), end_date.1 = structure(c(18675,
18534, NA, 18608, 18711, 18724), class = "Date"), end_date.2 = structure(c(NA,
18608, NA, 18610, 18715, 18726), class = "Date"), end_date.3 = structure(c(NA,
18662, NA, 18632, NA, NA), class = "Date"), end_loc.1 = c("g",
"a", "", "c", "b", "b"), end_loc.2 = c("", "a", "", "c", "t",
"t"), end_loc.3 = c("", "g", "", "g", "", "")), class = "data.frame", row.names = c(NA,
-6L))
我的数据是这样排列的，每个 person 和一个 start_date 以及一个 start_loc 都有行。我想知道哪些人有
end_date start_date、- 如果有两对或更多对符合此条件，则优先考虑
end_loc 与他们的 start_loc 匹配的那些
- 否则取最早。
因此，所需的输出类似于：
person start_loc start_date end_date.1 end_date.2 end_date.3 end_loc.1 end_loc.2 end_loc.3 end_date end_loc
1 1 a 2021-02-10 2021-02-17 <NA> <NA> g 2021-02-17 g
2 2 a 2021-01-30 2020-09-29 2020-12-12 2021-02-04 a a g <NA>
3 3 g 2020-12-04 <NA> <NA> <NA> <NA>
4 4 r 2020-12-09 2020-12-12 2020-12-14 2021-01-05 c c g 2020-12-12 c
5 5 t 2021-03-22 2021-03-25 2021-03-29 <NA> b t 2021-03-29 t
6 6 b 2021-04-04 2021-04-07 2021-04-09 <NA> b t 2021-04-07 b
我在上一个问题中遵循了一些技巧，例如使用 c_across、across 和 rowwise，但我似乎无法让 R 返回单个输出。这可能吗？我是否需要再次对数据进行纵向结构化？
您可以使用 dplyr 和 tidyr:
library(dplyr)
library(tidyr)
tmp0 <- example %>%
pivot_longer(cols = starts_with("end_date"), names_to=c("id"), names_pattern="end_date.(.)", values_to="end_date", values_drop_na = TRUE) %>%
pivot_longer(cols = starts_with("end_loc"), names_to=c("id2"), names_pattern="end_loc.(.)", values_to="end_loc", values_drop_na = TRUE) %>%
filter(id==id2, end_date <= start_date + 7 & end_date >= start_date) %>%
select(-id, -id2) %>%
group_by(person)
tmp1 <- tmp0 %>%
mutate(match = case_when(end_loc == start_loc ~ end_date,
TRUE ~ NA_real_)) %>%
filter(end_date == match) %>%
select(-match)
tmp2 <- tmp0 %>%
filter(end_date == min(end_date)) %>%
anti_join(tmp1, by=c("person"))
tmp1 %>%
bind_rows(tmp2) %>%
right_join(example, by=c("person", "start_loc", "start_date")) %>%
arrange(person, start_loc, start_date)
returns
# A tibble: 6 x 11
# Groups: person [6]
person start_loc start_date end_date end_loc end_date.1 end_date.2 end_date.3 end_loc.1 end_loc.2
<int> <chr> <date> <date> <chr> <date> <date> <date> <chr> <chr>
1 1 a 2021-02-10 2021-02-17 g 2021-02-17 NA NA "g" ""
2 2 a 2021-01-30 2021-02-04 g 2020-09-29 2020-12-12 2021-02-04 "a" "a"
3 3 g 2020-12-04 NA NA NA NA NA "" ""
4 4 r 2020-12-09 2020-12-12 c 2020-12-12 2020-12-14 2021-01-05 "c" "c"
5 5 t 2021-03-22 2021-03-29 t 2021-03-25 2021-03-29 NA "b" "t"
6 6 b 2021-04-04 2021-04-07 b 2021-04-07 2021-04-09 NA "b" "t"
# ... with 1 more variable: end_loc.3 <chr>
我正在建立三个临时表来获取所需的信息。 tmp1看的是start_loc和end_loc相同的数据，tmp2看的是不同位置的数据。最后，我们合并这两个表并创建所需的输出。
抱歉延迟回复，但你可以这样做
- 人应该有一个结果（可能是你输入的错字）
- 用数据中的 NA 替换空字符串 ''
example <- structure(list(person = 1:6, start_loc = c("a", "a", "g", "r",
"t", "b"), start_date = structure(c(18668, 18657, 18600, 18605,
18708, 18721), class = "Date"), end_date.1 = structure(c(18675,
18534, NA, 18608, 18711, 18724), class = "Date"), end_date.2 = structure(c(NA,
18608, NA, 18610, 18715, 18726), class = "Date"), end_date.3 = structure(c(NA,
18662, NA, 18632, NA, NA), class = "Date"), end_loc.1 = c("g",
"a", NA, "c", "b", "b"), end_loc.2 = c(NA, "a", NA, "c", "t",
"t"), end_loc.3 = c(NA, "g", NA, "g", NA, NA)), class = "data.frame", row.names = c(NA,
-6L))
library(tidyverse)
example %>% left_join(example %>% pivot_longer(cols = !c(person, start_loc, start_date), names_sep = '\.',
names_to = c('.value', 'number'),
values_drop_na = T) %>%
group_by(person) %>%
mutate(diff = end_date - start_date,
cond2 = diff <= 7 & diff >= 0,
cond1 = start_loc == end_loc) %>%
filter(cond2) %>%
arrange(person, -cond1, diff) %>%
summarise(end_date = first(end_date),
end_loc = first(end_loc)), by = 'person')
#> person start_loc start_date end_date.1 end_date.2 end_date.3 end_loc.1
#> 1 1 a 2021-02-10 2021-02-17 <NA> <NA> g
#> 2 2 a 2021-01-30 2020-09-29 2020-12-12 2021-02-04 a
#> 3 3 g 2020-12-04 <NA> <NA> <NA> <NA>
#> 4 4 r 2020-12-09 2020-12-12 2020-12-14 2021-01-05 c
#> 5 5 t 2021-03-22 2021-03-25 2021-03-29 <NA> b
#> 6 6 b 2021-04-04 2021-04-07 2021-04-09 <NA> b
#> end_loc.2 end_loc.3 end_date end_loc
#> 1 <NA> <NA> 2021-02-17 g
#> 2 a g 2021-02-04 g
#> 3 <NA> <NA> <NA> <NA>
#> 4 c g 2020-12-12 c
#> 5 t <NA> 2021-03-29 t
#> 6 t <NA> 2021-04-07 b
实际上，left_join 中的语法完成了总结
的工作
example %>% pivot_longer(cols = !c(person, start_loc, start_date), names_sep = '\.',
names_to = c('.value', 'number'),
values_drop_na = T) %>%
group_by(person) %>%
mutate(diff = end_date - start_date,
cond2 = diff <= 7 & diff >= 0,
cond1 = start_loc == end_loc) %>%
filter(cond2) %>%
arrange(person, -cond1, diff) %>%
summarise(end_date = first(end_date),
end_loc = first(end_loc))
# A tibble: 5 x 3
person end_date end_loc
<int> <date> <chr>
1 1 2021-02-17 g
2 2 2021-02-04 g
3 4 2020-12-12 c
4 5 2021-03-29 t
5 6 2021-04-07 b
与我在这里问的一个问题相关：
示例初始数据：
person start_loc start_date end_date.1 end_date.2 end_date.3 end_loc.1 end_loc.2 end_loc.3
1 1 a 2021-02-10 2021-02-17 <NA> <NA> g
2 2 a 2021-01-30 2020-09-29 2020-12-12 2021-02-04 a a g
3 3 g 2020-12-04 <NA> <NA> <NA>
4 4 r 2020-12-09 2020-12-12 2020-12-14 2021-01-05 c c g
5 5 t 2021-03-22 2021-03-25 2021-03-29 <NA> b t
6 6 b 2021-04-04 2021-04-07 2021-04-09 <NA> b t
example <- structure(list(person = 1:6, start_loc = c("a", "a", "g", "r",
"t", "b"), start_date = structure(c(18668, 18657, 18600, 18605,
18708, 18721), class = "Date"), end_date.1 = structure(c(18675,
18534, NA, 18608, 18711, 18724), class = "Date"), end_date.2 = structure(c(NA,
18608, NA, 18610, 18715, 18726), class = "Date"), end_date.3 = structure(c(NA,
18662, NA, 18632, NA, NA), class = "Date"), end_loc.1 = c("g",
"a", "", "c", "b", "b"), end_loc.2 = c("", "a", "", "c", "t",
"t"), end_loc.3 = c("", "g", "", "g", "", "")), class = "data.frame", row.names = c(NA,
-6L))
我的数据是这样排列的，每个 person 和一个 start_date 以及一个 start_loc 都有行。我想知道哪些人有
end_datestart_date、 7 天内
- 如果有两对或更多对符合此条件，则优先考虑
end_loc与他们的start_loc 匹配的那些
- 否则取最早。
因此，所需的输出类似于：
person start_loc start_date end_date.1 end_date.2 end_date.3 end_loc.1 end_loc.2 end_loc.3 end_date end_loc
1 1 a 2021-02-10 2021-02-17 <NA> <NA> g 2021-02-17 g
2 2 a 2021-01-30 2020-09-29 2020-12-12 2021-02-04 a a g <NA>
3 3 g 2020-12-04 <NA> <NA> <NA> <NA>
4 4 r 2020-12-09 2020-12-12 2020-12-14 2021-01-05 c c g 2020-12-12 c
5 5 t 2021-03-22 2021-03-25 2021-03-29 <NA> b t 2021-03-29 t
6 6 b 2021-04-04 2021-04-07 2021-04-09 <NA> b t 2021-04-07 b
我在上一个问题中遵循了一些技巧，例如使用 c_across、across 和 rowwise，但我似乎无法让 R 返回单个输出。这可能吗？我是否需要再次对数据进行纵向结构化？
您可以使用 dplyr 和 tidyr:
library(dplyr)
library(tidyr)
tmp0 <- example %>%
pivot_longer(cols = starts_with("end_date"), names_to=c("id"), names_pattern="end_date.(.)", values_to="end_date", values_drop_na = TRUE) %>%
pivot_longer(cols = starts_with("end_loc"), names_to=c("id2"), names_pattern="end_loc.(.)", values_to="end_loc", values_drop_na = TRUE) %>%
filter(id==id2, end_date <= start_date + 7 & end_date >= start_date) %>%
select(-id, -id2) %>%
group_by(person)
tmp1 <- tmp0 %>%
mutate(match = case_when(end_loc == start_loc ~ end_date,
TRUE ~ NA_real_)) %>%
filter(end_date == match) %>%
select(-match)
tmp2 <- tmp0 %>%
filter(end_date == min(end_date)) %>%
anti_join(tmp1, by=c("person"))
tmp1 %>%
bind_rows(tmp2) %>%
right_join(example, by=c("person", "start_loc", "start_date")) %>%
arrange(person, start_loc, start_date)
returns
# A tibble: 6 x 11
# Groups: person [6]
person start_loc start_date end_date end_loc end_date.1 end_date.2 end_date.3 end_loc.1 end_loc.2
<int> <chr> <date> <date> <chr> <date> <date> <date> <chr> <chr>
1 1 a 2021-02-10 2021-02-17 g 2021-02-17 NA NA "g" ""
2 2 a 2021-01-30 2021-02-04 g 2020-09-29 2020-12-12 2021-02-04 "a" "a"
3 3 g 2020-12-04 NA NA NA NA NA "" ""
4 4 r 2020-12-09 2020-12-12 c 2020-12-12 2020-12-14 2021-01-05 "c" "c"
5 5 t 2021-03-22 2021-03-29 t 2021-03-25 2021-03-29 NA "b" "t"
6 6 b 2021-04-04 2021-04-07 b 2021-04-07 2021-04-09 NA "b" "t"
# ... with 1 more variable: end_loc.3 <chr>
我正在建立三个临时表来获取所需的信息。 tmp1看的是start_loc和end_loc相同的数据，tmp2看的是不同位置的数据。最后，我们合并这两个表并创建所需的输出。
抱歉延迟回复，但你可以这样做
- 人应该有一个结果（可能是你输入的错字）
- 用数据中的 NA 替换空字符串 ''
example <- structure(list(person = 1:6, start_loc = c("a", "a", "g", "r",
"t", "b"), start_date = structure(c(18668, 18657, 18600, 18605,
18708, 18721), class = "Date"), end_date.1 = structure(c(18675,
18534, NA, 18608, 18711, 18724), class = "Date"), end_date.2 = structure(c(NA,
18608, NA, 18610, 18715, 18726), class = "Date"), end_date.3 = structure(c(NA,
18662, NA, 18632, NA, NA), class = "Date"), end_loc.1 = c("g",
"a", NA, "c", "b", "b"), end_loc.2 = c(NA, "a", NA, "c", "t",
"t"), end_loc.3 = c(NA, "g", NA, "g", NA, NA)), class = "data.frame", row.names = c(NA,
-6L))
library(tidyverse)
example %>% left_join(example %>% pivot_longer(cols = !c(person, start_loc, start_date), names_sep = '\.',
names_to = c('.value', 'number'),
values_drop_na = T) %>%
group_by(person) %>%
mutate(diff = end_date - start_date,
cond2 = diff <= 7 & diff >= 0,
cond1 = start_loc == end_loc) %>%
filter(cond2) %>%
arrange(person, -cond1, diff) %>%
summarise(end_date = first(end_date),
end_loc = first(end_loc)), by = 'person')
#> person start_loc start_date end_date.1 end_date.2 end_date.3 end_loc.1
#> 1 1 a 2021-02-10 2021-02-17 <NA> <NA> g
#> 2 2 a 2021-01-30 2020-09-29 2020-12-12 2021-02-04 a
#> 3 3 g 2020-12-04 <NA> <NA> <NA> <NA>
#> 4 4 r 2020-12-09 2020-12-12 2020-12-14 2021-01-05 c
#> 5 5 t 2021-03-22 2021-03-25 2021-03-29 <NA> b
#> 6 6 b 2021-04-04 2021-04-07 2021-04-09 <NA> b
#> end_loc.2 end_loc.3 end_date end_loc
#> 1 <NA> <NA> 2021-02-17 g
#> 2 a g 2021-02-04 g
#> 3 <NA> <NA> <NA> <NA>
#> 4 c g 2020-12-12 c
#> 5 t <NA> 2021-03-29 t
#> 6 t <NA> 2021-04-07 b
实际上，left_join 中的语法完成了总结
的工作example %>% pivot_longer(cols = !c(person, start_loc, start_date), names_sep = '\.',
names_to = c('.value', 'number'),
values_drop_na = T) %>%
group_by(person) %>%
mutate(diff = end_date - start_date,
cond2 = diff <= 7 & diff >= 0,
cond1 = start_loc == end_loc) %>%
filter(cond2) %>%
arrange(person, -cond1, diff) %>%
summarise(end_date = first(end_date),
end_loc = first(end_loc))
# A tibble: 5 x 3
person end_date end_loc
<int> <date> <chr>
1 1 2021-02-17 g
2 2 2021-02-04 g
3 4 2020-12-12 c
4 5 2021-03-29 t
5 6 2021-04-07 b