根据R中的一列日期扩展数据框

Question

我想根据我的日期列扩展我的数据框架，以便按时间顺序排列我当前日期之间有新的日期。我的日期专栏是按时间顺序排列的，持续了5年，并且包含我想忽略的重复日期。我希望相应的组并绘制新行的排为" na"。

zz <- "Date Group Draw
1  2006-05-11    bb     T
2  2006-05-11    bb     F
3  2006-05-14    aa     T
4  2006-05-16    aa     T
5  2006-05-20    cc     F
6  2006-05-20    bb     F
7  2006-05-21    aa     T"
Data <- read.table(text=zz, header = TRUE)

所以我希望我的新数据框架看起来像：

xx <- "Date Group Draw
1  2006-05-11    bb     T
2  2006-05-11    bb     F
3  2006-05-12    NA     NA
4  2006-05-13    NA     NA
5  2006-05-14    aa     T
6  2006-05-15    NA     NA
7  2006-05-16    aa     T
8  2006-05-17    NA     NA
9  2006-05-18    NA     NA
10 2006-05-19    NA     NA
11 2006-05-20    cc     F
12 2006-05-20    bb     F
13 2006-05-21    aa     T"
Output <- read.table(text=xx, header = TRUE)

任何帮助将不胜感激。我是R的新手，并且一直在尝试手动执行此操作。

Answer 1

我认为这应该很好：

merge(
    x = data.frame(
        Date = seq.Date(min(df$Date), max(df$Date), by = "day")
    ),
    y = df,
    all.x = TRUE
)
#          Date Group  Draw
# 1  2006-05-11    bb  TRUE
# 2  2006-05-11    bb FALSE
# 3  2006-05-12  <NA>    NA
# 4  2006-05-13  <NA>    NA
# 5  2006-05-14    aa  TRUE
# 6  2006-05-15  <NA>    NA
# 7  2006-05-16    aa  TRUE
# 8  2006-05-17  <NA>    NA
# 9  2006-05-18  <NA>    NA
# 10 2006-05-19  <NA>    NA
# 11 2006-05-20    cc FALSE
# 12 2006-05-20    bb FALSE
# 13 2006-05-21    aa  TRUE

---

所有这些都是创建一个日期序列，涵盖了实际数据范围，然后执行左联接。

---

和同一想法，使用data.table：

dt[dt[,.(Date = seq.Date(min(Date), max(Date), by = "day"))], on = .(Date)]
#           Date Group  Draw
#  1: 2006-05-11    bb  TRUE
#  2: 2006-05-11    bb FALSE
#  3: 2006-05-12    NA    NA
#  4: 2006-05-13    NA    NA
#  5: 2006-05-14    aa  TRUE
#  6: 2006-05-15    NA    NA
#  7: 2006-05-16    aa  TRUE
#  8: 2006-05-17    NA    NA
#  9: 2006-05-18    NA    NA
# 10: 2006-05-19    NA    NA
# 11: 2006-05-20    cc FALSE
# 12: 2006-05-20    bb FALSE
# 13: 2006-05-21    aa  TRUE

---

zz <- "Date Group Draw
1  2006-05-11    bb     T
2  2006-05-11    bb     F
3  2006-05-14    aa     T
4  2006-05-16    aa     T
5  2006-05-20    cc     F
6  2006-05-20    bb     F
7  2006-05-21    aa     T"
df <- read.table(
    text = zz, 
    header = TRUE
)
df$Date <- as.Date(df$Date) 
library(data.table)
dt <- data.table(read.table(text = zz, header = TRUE))[,Date := as.Date(Date)]

Answer 2

使用 @nrussell帖子中的数据，另一个选项是 complete tidyr

library(tidyr)
complete(df, Date = full_seq(Date, 1))
## A tibble: 13 × 3
#         Date  Group  Draw
#       <date> <fctr> <lgl>
#1  2006-05-11     bb  TRUE
#2  2006-05-11     bb FALSE
#3  2006-05-12     NA    NA
#4  2006-05-13     NA    NA
#5  2006-05-14     aa  TRUE
#6  2006-05-15     NA    NA
#7  2006-05-16     aa  TRUE
#8  2006-05-17     NA    NA
#9  2006-05-18     NA    NA
#10 2006-05-19     NA    NA
#11 2006-05-20     cc FALSE
#12 2006-05-20     bb FALSE
#13 2006-05-21     aa  TRUE

Answer 3

如果我正确理解您的问题，这是我的粗略示例：

date  <- format(seq.Date(from=as.Date(paste(2006, '05', '11', sep='-'), 
                                   '%Y-%m-%d'), 
                      to  =as.Date(paste(2006, 05, '21', sep='-'), 
                                   '%Y-%m-%d'), 
                      by  = "day"), '%Y-%m-%d')

以上生成日期列表。然后，您可以将上面的date的左连接到您的数据。Table。