在具有多列

时间:2015-11-16 18:31:18

标签: r date padding

如何在超过2列的数据框中插入缺少的日期? 在我的数据中,每个日期都有sp1和sp2之间的观察。如果一天中sp1和sp2之间没有观察,则缺少该日期。

这是我的df的一部分:

the_date    sp1 sp2 win     loss    sp1_name    sp2_name
4/1/13      A   B   8       8         A_name    B_name
4/2/13      A   B   6       10        A_name    B_name
4/3/13      A   B   7       5         A_name    B_name
4/5/13      A   B   7       5         A_name    B_name
4/6/13      A   B   6       2         A_name    B_name
4/7/13      A   B   15      10        A_name    B_name
4/1/13      A   C   3       8         A_name    C_name
4/2/13      A   C   2       12        A_name    C_name
4/3/13      A   C   9       7         A_name    C_name
4/4/13      A   C   14      8         A_name    C_name
4/6/13      A   C   9       10        A_name    C_name
4/1/13      A   D   13      13        A_name    D_name
4/2/13      A   D   13      5         A_name    D_name
4/3/13      A   D   7       1         A_name    D_name
4/4/13      A   D   15      11        A_name    D_name
4/5/13      A   D   3       11        A_name    D_name
4/6/13      A   D   12      11        A_name    D_name
4/7/13      A   D   9       9         A_name    D_name

例如,A-B的4/4/13缺失。在我的输出中我想要的是插入所有相应列的缺少日期,并将0分配给胜负。所以我的输出看起来像是添加*的行:

the_date    sp1 sp2 win     loss    sp1_name    sp2_name
4/1/13      A   B   8       8         A_name    B_name
4/2/13      A   B   6       10        A_name    B_name
4/3/13      A   B   7       5         A_name    B_name
*4/4/13     A   B   0       0         A_name    B_name
4/5/13      A   B   7       5         A_name    B_name
4/6/13      A   B   6       2         A_name    B_name
4/7/13      A   B   15      10        A_name    B_name
4/1/13      A   C   3       8         A_name    C_name
4/2/13      A   C   2       12        A_name    C_name
4/3/13      A   C   9       7         A_name    C_name
4/4/13      A   C   14      8         A_name    C_name
*4/5/13     A   C   0       0         A_name    C_name
4/6/13      A   C   9       10        A_name    C_name
*4/7/13     A   C   0       0         A_name    C_name
4/1/13      A   D   13      13        A_name    D_name
4/2/13      A   D   13      5         A_name    D_name
4/3/13      A   D   7       1         A_name    D_name
4/4/13      A   D   15      11        A_name    D_name
4/5/13      A   D   3       11        A_name    D_name
4/6/13      A   D   12      11        A_name    D_name
4/7/13      A   D   9       9         A_name    D_name

我知道如果我们有一个2列数据帧(值,日期),我们可以通过将数据帧与全范围时间合并来填充缺少日期的数据帧。但是,我的数据框有两列以上。

此外,这只是我数据的一部分,所以我还有其他日期的其他组合:

sp1 sp2 
B    C
B    A
B    D
C    A
C    B
C    D
D    B
D    C
D    A

任何线索?

2 个答案:

答案 0 :(得分:4)

这是dplyr方法。考虑到您拥有大型数据集,您可能需要考虑data.table方法。

d <- read.table(textConnection("the_date    sp1 sp2 win     loss    sp1_name    sp2_name
4/1/13      A   B   8       8         A_name    B_name
4/2/13      A   B   6       10        A_name    B_name
4/3/13      A   B   7       5         A_name    B_name
4/5/13      A   B   7       5         A_name    B_name
4/6/13      A   B   6       2         A_name    B_name
4/7/13      A   B   15      10        A_name    B_name
4/1/13      A   C   3       8         A_name    C_name
4/2/13      A   C   2       12        A_name    C_name
4/3/13      A   C   9       7         A_name    C_name
4/4/13      A   C   14      8         A_name    C_name
4/6/13      A   C   9       10        A_name    C_name
4/1/13      A   D   13      13        A_name    D_name
4/2/13      A   D   13      5         A_name    D_name
4/3/13      A   D   7       1         A_name    D_name
4/4/13      A   D   15      11        A_name    D_name
4/5/13      A   D   3       11        A_name    D_name
4/6/13      A   D   12      11        A_name    D_name
4/7/13      A   D   9       9         A_name    D_name"),
stringsAsFactors = FALSE, header = TRUE)

d$the_date <- as.Date(d$the_date, "%m/%d/%y")

更新

我意识到我下面的原始答案并不完全正确。例如,它没有使用4/7/13填充A组。考虑到这一点,我提出了一个更好的,我认为更快的方法。

#Step one combine sp1 and sp2 into one group
d$group <- paste0(d$sp1,d$sp2)

#Step two find min and max date in the database

min_d <- min(d$the_date)
max_d <- max(d$the_date)

#Step three use dplyr
d %>%
  do(expand.grid(unique(.$group), seq(min_d, max_d, 1))) %>% 
  rename(group = Var1, the_date = Var2) %>%
  left_join(d) %>%
  arrange(group) %>%
  select(-group)

原始

#Step one combine sp1 and sp2 into one group
d$group <- paste0(d$sp1,d$sp2)

#Step two use dplyr.  
d %>%
  group_by(group) %>%
  summarise(min = min(the_date), max = max(the_date)) %>%
  rowwise() %>%
  do(data.frame(group = .$group, the_date = seq(.$min, .$max, 1))) %>%
  left_join(d) %>%
  select(-group)

一般来说,您的问题与this类似。请查看更多信息/想法。

答案 1 :(得分:1)

以下是使用pad中的fill_by_valuepadr的解决方案:

library(dplyr)
library(tidyr)
library(padr)

df %>%
  mutate(the_date = as.Date(the_date, "%m/%d/%y")) %>%
  group_by(sp1, sp2) %>%
  pad() %>%              
  fill(sp1_name:sp2_name) %>%      
  fill_by_value(win, loss)

<强>结果:

# A tibble: 20 x 7
# Groups:   sp1, sp2 [3]
     the_date    sp1    sp2   win  loss sp1_name sp2_name
       <date> <fctr> <fctr> <dbl> <dbl>   <fctr>   <fctr>
 1 2013-04-01      A      B     8     8   A_name   B_name
 2 2013-04-02      A      B     6    10   A_name   B_name
 3 2013-04-03      A      B     7     5   A_name   B_name
 4 2013-04-04      A      B     0     0   A_name   B_name
 5 2013-04-05      A      B     7     5   A_name   B_name
 6 2013-04-06      A      B     6     2   A_name   B_name
 7 2013-04-07      A      B    15    10   A_name   B_name
 8 2013-04-01      A      C     3     8   A_name   C_name
 9 2013-04-02      A      C     2    12   A_name   C_name
10 2013-04-03      A      C     9     7   A_name   C_name
11 2013-04-04      A      C    14     8   A_name   C_name
12 2013-04-05      A      C     0     0   A_name   C_name
13 2013-04-06      A      C     9    10   A_name   C_name
14 2013-04-01      A      D    13    13   A_name   D_name
15 2013-04-02      A      D    13     5   A_name   D_name
16 2013-04-03      A      D     7     1   A_name   D_name
17 2013-04-04      A      D    15    11   A_name   D_name
18 2013-04-05      A      D     3    11   A_name   D_name
19 2013-04-06      A      D    12    11   A_name   D_name
20 2013-04-07      A      D     9     9   A_name   D_name

数据:

df = structure(list(the_date = structure(c(1L, 2L, 3L, 5L, 6L, 7L, 
1L, 2L, 3L, 4L, 6L, 1L, 2L, 3L, 4L, 5L, 6L, 7L), .Label = c("4/1/13", 
"4/2/13", "4/3/13", "4/4/13", "4/5/13", "4/6/13", "4/7/13"), class = "factor"), 
    sp1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "A", class = "factor"), 
    sp2 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 
    2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("B", "C", "D"
    ), class = "factor"), win = c(8L, 6L, 7L, 7L, 6L, 15L, 3L, 
    2L, 9L, 14L, 9L, 13L, 13L, 7L, 15L, 3L, 12L, 9L), loss = c(8L, 
    10L, 5L, 5L, 2L, 10L, 8L, 12L, 7L, 8L, 10L, 13L, 5L, 1L, 
    11L, 11L, 11L, 9L), sp1_name = structure(c(1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "A_name", class = "factor"), 
    sp2_name = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 
    2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("B_name", 
    "C_name", "D_name"), class = "factor")), .Names = c("the_date", 
"sp1", "sp2", "win", "loss", "sp1_name", "sp2_name"), class = "data.frame", row.names = c(NA, 
-18L))