Question

我有一个像这样的数据框：

Name   Position   Value
a         1        0.2
a         3        0.4
a         4        0.3
b         1        0.5
b         2        0.4
b         5        0.3
c         2        0.3
c         3        0.4
c         5        0.1
d         1        0.2
d         2        0.4
d         3        0.5

我想这样做，以便每个名字的位置总是从1到5，并将NAs填入Value中，如下所示：

Name   Position   Value
a         1        0.2
a         2        NA
a         3        0.4
a         4        0.3
a         5        NA
b         1        0.5
b         2        0.4
b         3        NA
b         4        NA
b         5        0.3
c         1        NA
c         2        0.3
c         3        0.4
c         4        NA
c         5        0.1
d         1        0.2
d         2        0.4
d         3        0.5
d         4        NA
d         5        NA

有没有办法在没有创建前两列的虚拟数据框的情况下执行此操作，然后使用合并进行某种外连接？

感谢。

Answer 1

我会使用data.table，但以不同的方式@akrun强调：

library(data.table)
dt = as.data.table(df)
setkey(dt, Name, Position)
dt[CJ(unique(Name),unique(Position))]

Answer 2

您可以使用reshape2包：

# make sample data frame
df <- read.table(text = "Name   Position   Value
a         1        0.2
a         3        0.4
a         4        0.3
b         1        0.5
b         2        0.4
b         5        0.3
c         2        0.3
c         3        0.4
c         5        0.1
d         1        0.2
d         2        0.4
d         3        0.5", header = TRUE, stringsAsFactors = FALSE)

library('reshape2')
df2 <- dcast(df, Name ~ Position)
df3 <- melt(df2, value.name = "Value", variable.name = "Position")
df3[order(df3$Name), ]
#    Name Position Value
# 1     a        1   0.2
# 5     a        2    NA
# 9     a        3   0.4
# 13    a        4   0.3
# 17    a        5    NA
# 2     b        1   0.5
# 6     b        2   0.4
# 10    b        3    NA
# 14    b        4    NA
# 18    b        5   0.3
# 3     c        1    NA
# 7     c        2   0.3
# 11    c        3   0.4
# 15    c        4    NA
# 19    c        5   0.1
# 4     d        1   0.2
# 8     d        2   0.4
# 12    d        3   0.5
# 16    d        4    NA
# 20    d        5    NA

Answer 3

您可以使用data.table

 library(data.table)
 DT <- data.table(df)
 setkey(DT, Position)
 DT[, .SD[J(1:5), roll=FALSE], by=Name][order(Name, Position),]
 #   Name Position Value
 #1:    a        1   0.2
 #2:    a        2    NA
 #3:    a        3   0.4
 #4:    a        4   0.3
 #5:    a        5    NA
 #6:    b        1   0.5
 #7:    b        2   0.4
 #8:    b        3    NA
 #9:    b        4    NA
#10:    b        5   0.3
#11:    c        1    NA
#12:    c        2   0.3
#13:    c        3   0.4
#14:    c        4    NA
#15:    c        5   0.1
#16:    d        1   0.2
#17:    d        2   0.4
#18:    d        3   0.5
#19:    d        4    NA
#20:    d        5    NA

或者您可以使用tidyr/dplyr

 library(dplyr)
 library(tidyr)

  df %>% 
      spread(Position, Value) %>%
      gather(Position, Value, `1`:`5`) %>%
      arrange(Name, Position)

数据

 df <- structure(list(Name = c("a", "a", "a", "b", "b", "b", "c", "c", 
 "c", "d", "d", "d"), Position = c(1L, 3L, 4L, 1L, 2L, 5L, 2L, 
 3L, 5L, 1L, 2L, 3L), Value = c(0.2, 0.4, 0.3, 0.5, 0.4, 0.3, 
 0.3, 0.4, 0.1, 0.2, 0.4, 0.5)), .Names = c("Name", "Position", 
 "Value"), class = "data.frame", row.names = c(NA, -12L))

Answer 4

也许它有点矫枉过正，但我认为您可以使用sqldf来执行此操作：

library(sqldf)
# Your data frame:
df <- data.frame(
  name = c('a', 'a', 'a', 'b', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'd'),
  position = c(1, 3, 4, 1, 2, 5, 2, 3, 5, 1, 2, 3),
  value = c(0.2, 0.4, 0.3, 0.5, 0.4, 0.3, 0.3, 0.4, 0.1, 0.2, 0.4, 0.5)
)
# A data frame to hold the positions you want to fill:
pos = data.frame(pos = 1:5)
# SQLdf let's you write SQL sentences that use data frames like SQL tables:
df2 <- sqldf(
  "select a.*, b.value as value
  from (
    select a.name, p.pos as position 
    from (select distinct name from df) as a, pos as p
  ) as a
  left join df as b on a.name = b.name and a.position = b.position"
)
df2
## Result:
##   name position value
##1     a        1   0.2
##2     a        2    NA
##3     a        3   0.4
##4     a        4   0.3
##5     a        5    NA
##6     b        1   0.5
##7     b        2   0.4
##8     b        3    NA
##9     b        4    NA
##10    b        5   0.3
##11    c        1    NA
##12    c        2   0.3
##13    c        3   0.4
##14    c        4    NA
##15    c        5   0.1
##16    d        1   0.2
##17    d        2   0.4
##18    d        3   0.5
##19    d        4    NA
##20    d        5    NA

当然，您可以将sqldf()的结果直接分配到df以覆盖原始数据框

Answer 5

以下是几个基本解决方案：

as.data.frame.table(tapply(df[[3]], df[2:1], c))

和

merge(df, 
      expand.grid(Position = unique(df$Position), Name = unique(df$Name)), 
      all = TRUE)

R：在数据帧中添加NA

5 个答案:

数据