使用公用值组合R中的透视行
使用公用值组合R中的透视行
提问于 2017-03-03 06:17:14
我有一个数据框，看起来像这样
Name Visit Arrival Departure
Jack week 1 8:00 NA
Jack week 1 NA 8:30
Sally week 5 9:00 NA
Sally week 5 NA 9:30
Adam week 2 2:00 NA
Adam week 2 NA 3:00到达和离开时间最初是行的，我将其转换为列，这就是为什么会有空值的原因。我想根据姓名和访问来合并行，这样到达和出发就在同一行中，如下所示
Name Visit Arrival Departure
Jack week 1 8:00 8:30
Sally week 5 9:00 9:30
Adam week 2 2:00 3:00任何解决方案都将受到赞赏，因为在尝试合并时会遇到困难。
回答 4
Stack Overflow用户
回答已采纳
发布于 2017-03-03 06:31:27
实际上，如果您能够在pivot之前返回数据，tidyr::spread将完成一项出色的工作。
Name <- c("Jack", "Jack","Sally", "Sally", "Adam", "Adam")
Visit <- c("week1", "week1", "week5", "week5", "week2", "week2")
Itenary <- rep(c("Arrival", "Departure"), 3)
Time <- c("8:00", "8:30", "9:00", "9:30", "2:00", "2:30")
df <- data.frame(Name, Visit, Itenary, Time)
df
Name Visit Itenary Time
1 Jack week1 Arrival 8:00
2 Jack week1 Departure 8:30
3 Sally week5 Arrival 9:00
4 Sally week5 Departure 9:30
5 Adam week2 Arrival 2:00
6 Adam week2 Departure 2:30
df %>%
spread(key = Itenary, value = Time)
Name Visit Arrival Departure
1 Adam week2 2:00 2:30
2 Jack week1 8:00 8:30
3 Sally week5 9:00 9:30Stack Overflow用户
发布于 2017-03-03 06:31:19
只需使用na.omit作为聚合函数进行aggregate即可：
aggregate(dat[c("Arrival","Departure")], dat[c("Name","Visit")], FUN=na.omit)
# or
aggregate(cbind(Arrival,Departure) ~ ., data=dat, FUN=na.omit, na.action=na.pass)
# Name Visit Arrival Departure
#1 Jack week1 8:00 8:30
#2 Adam week2 2:00 3:00
#3 Sally week5 9:00 9:30同样的逻辑也适用于data.table
dat[, lapply(.SD,na.omit), by=.(Name,Visit)]...or dplyr
dat %>% group_by(Name,Visit) %>% summarise_all(na.omit)Stack Overflow用户
发布于 2017-03-03 06:26:46
这里有一种方法，假设访问的人将恰好有两行数据：
library(dplyr)
df = readr::read_table("Name Visit Arrival Departure
Jack week 1 8:00 NA
Jack week 1 NA 8:30
Sally week 5 9:00 NA
Sally week 5 NA 9:30
Adam week 2 2:00 NA
Adam week 2 NA 3:00", col_types="cccc")
df %>%
group_by(Name, Visit) %>%
mutate(Arrival = ifelse(is.na(Arrival), lag(Arrival), Arrival),
Departure = ifelse(is.na(Departure), lead(Departure), Departure)) %>%
ungroup() %>%
distinct(Name, Visit, .keep_all=TRUE)
# A tibble: 3 × 4
Name Visit Arrival Departure
<chr> <chr> <chr> <chr>
1 Jack week 1 8:00 8:30
2 Sally week 5 9:00 9:30
3 Adam week 2 2:00 3:00页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/42567075
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