我有下表：
Name Date Quiz Homework
John 11-01-02 40 10
John 11-01-03 47 20
John 11-01-04 41 10
John 11-01-08 35 10
John 11-01-10 43 15
John 11-01-13 40 10
Adam 11-01-05 41 10
Adam 11-01-08 41 15
Adam 11-01-14 49 10
Adam 11-01-19 40 20
Adam 11-01-21 40 10
你可以看到有一些时间差距。我想按名称填写这些时间差距，并将那些缺失日期的测验和作业分数替换为零。因此，我想要的最终结果将是以下
Name Date Quiz Homework
John 11-01-02 40 10
John 11-01-03 47 20
John 11-01-04 41 10
John 11-01-05 0 0
John 11-01-06 0 0
John 11-01-07 0 0
John 11-01-08 35 10
John 11-01-09 0 0
John 11-01-10 43 15
John 11-01-11 0 0
John 11-01-12 0 0
John 11-01-13 40 10
Adam 11-01-05 41 10
Adam 11-01-06 0 0
Adam 11-01-07 0 0
Adam 11-01-08 41 15
Adam 11-01-09 0 0
Adam 11-01-10 0 0
Adam 11-01-11 0 0
Adam 11-01-12 0 0
Adam 11-01-13 0 0
Adam 11-01-14 49 10
Adam 11-01-15 0 0
Adam 11-01-16 0 0
Adam 11-01-17 0 0
Adam 11-01-18 0 0
Adam 11-01-19 40 20
Adam 11-01-20 0 0
Adam 11-01-21 40 10
有没有快速的方法呢？我做的是以下内容：
1) Find a minimum, maximum dates by name
2) For each name, create a sequence of dates from minimum, maximum dates found in step 1)
3) Join the table created in step 2) with the original table.
4) replace NA values in Quiz, Homework by zero
但这很慢。我想知道是否有快速的方法。
答案 0 :(得分：1)
使用data.table包的解决方案应该很快：
library(data.table)
DT <- fread("Name Date Quiz Homework
John 11-01-02 40 10
John 11-01-03 47 20
John 11-01-04 41 10
John 11-01-08 35 10
John 11-01-10 43 15
John 11-01-13 40 10
Adam 11-01-05 41 10
Adam 11-01-08 41 15
Adam 11-01-14 49 10
Adam 11-01-19 40 20
Adam 11-01-21 40 10")
DT[, Date := as.Date(Date, "%y-%m-%d")]
DT[DT[, .(Date=seq(min(Date), max(Date), by="1 day")), by=.(Name)],
on=.(Name, Date)][,
':=' (
Quiz = ifelse(is.na(Quiz), 0, Quiz),
Homework = ifelse(is.na(Homework), 0, Homework)
)]
说明：
allDates <- DT[,
.(Date=seq(min(Date), max(Date), by="1 day")), by=.(Name)] DT[allDates, on=.(Name, Date)] 答案 1 :(得分：1)
tidyverse解决方案：
library(dplyr)
library(tidyr)
library(lubridate) # for easier year conversion
df1 <- structure(list(Name = c("John", "John", "John", "John", "John",
"John", "Adam", "Adam", "Adam", "Adam", "Adam"),
Date = c("11-01-02", "11-01-03", "11-01-04",
"11-01-08", "11-01-10", "11-01-13",
"11-01-05", "11-01-08", "11-01-14",
"11-01-19", "11-01-21"),
Quiz = c(40L, 47L, 41L, 35L, 43L, 40L, 41L, 41L, 49L, 40L, 40L),
Homework = c(10L, 20L, 10L, 10L, 15L, 10L,
10L, 15L, 10L, 20L, 10L)),
.Names = c("Name", "Date", "Quiz", "Homework"),
class = "data.frame",
row.names = c(NA, -11L))
df1 %>%
mutate(Date = as_date(Date, "%C-%m-%d")) %>%
group_by(Name) %>%
complete(Date = seq(min(Date), max(Date), by = "1 day"),
fill = list(Quiz = 0, Homework = 0))
Name Date Quiz Homework
1 Adam 2011-01-05 41 10
2 Adam 2011-01-06 0 0
3 Adam 2011-01-07 0 0
4 Adam 2011-01-08 41 15
5 Adam 2011-01-09 0 0
6 Adam 2011-01-10 0 0
7 Adam 2011-01-11 0 0
8 Adam 2011-01-12 0 0
9 Adam 2011-01-13 0 0
10 Adam 2011-01-14 49 10
11 Adam 2011-01-15 0 0
12 Adam 2011-01-16 0 0
13 Adam 2011-01-17 0 0
14 Adam 2011-01-18 0 0
15 Adam 2011-01-19 40 20
16 Adam 2011-01-20 0 0
17 Adam 2011-01-21 40 10
18 John 2011-01-02 40 10
19 John 2011-01-03 47 20
20 John 2011-01-04 41 10
21 John 2011-01-05 0 0
22 John 2011-01-06 0 0
23 John 2011-01-07 0 0
24 John 2011-01-08 35 10
25 John 2011-01-09 0 0
26 John 2011-01-10 43 15
27 John 2011-01-11 0 0
28 John 2011-01-12 0 0
29 John 2011-01-13 40 10