Question

我有一个看起来像这样的数据框

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

抵达和离开时间最初是行，我转入colums，这就是为什么有空值。我想基于名称和访问合并行，所以到达和离开在同一行，如

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

任何解决方案都会受到赞赏，在那里尝试合并时会遇到困难。

Answer 1

只需aggregate na.omit作为聚合函数：

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

相同的逻辑适用于data.table：

dat[, lapply(.SD,na.omit), by=.(Name,Visit)]

...或dplyr：

dat %>% group_by(Name,Visit) %>% summarise_all(na.omit)

Answer 2

这是一种方法，假设访问者将有两行数据：

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

Answer 3

我确信这可能有更漂亮的方法，但这对我有用：

 library(data.table)
library(reshape2)

test <- data.table(Name = c("Jack", "Jack", "Sally", "Sally", "Adam", "Adam"), Visit = c("week 1", "week 1", "week 5", "week 5", "week 2", "week 2"), Arrival = c("8:00", NA, "9:00", NA, "2:00", NA), Departure = c(NA, "8:30", NA, "9:30", NA, "3:00"))

test_m <- melt(test,id.vars = c("Name", "Visit"))
test_m <- test_m[!is.na(value),]
test_c <- dcast(test_m, Name + Visit ~ variable)

> test_c
   Name  Visit Arrival Departure
1  Adam week 2    2:00      3:00
2  Jack week 1    8:00      8:30
3 Sally week 5    9:00      9:30

希望有所帮助

Answer 4

实际上，如果你能够在转移之前回到数据，tidyr :: spread将会做得很漂亮。

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      9:30

通过公共值组合R中的旋转行

4 个答案: