用R中的特定列填写缺少值的缺少日期

Question

I have the following table:

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

You can see that there are some time gaps. I would like to fill in those time gaps by name and replace the quiz, homework scores for those missing dates with zero. Thus, the final outcome I want would be the following

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

Is there a fast way of doing it? What I did was the following:

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

but that was rather slow. I was wondering if there's a fast way of doing it.

Answer 1

A solution using data.table package which should be fast:

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)
        )]

Explanation:

1. Create the sequence of dates using allDates <- DT[, .(Date=seq(min(Date), max(Date), by="1 day")), by=.(Name)]
2. Join with original dataset using DT[allDates, on=.(Name, Date)]
3. Finally, replace NAs with 0

Answer 2

A tidyverse solution:

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