Question

我还没有在网上找到解决方案，因为提出问题的正确问题并不容易。我有两个data.frames，x和y，并希望将它们组合起来得到z：

棘手的是z比较x和y的日期值，并采用最新的观察来更新A，B，C和D.因此“动态”更新/组合。

x=data.frame(c("2000-01-01","2000-06-01","2001-01-01"),c("100","100","100"),c("200","200","200"))
colnames(x)=c("Date","A","B")

y=data.frame(c("2000-01-05","2000-04-09"),c("10","0"),c("0","35"))
colnames(y)=c("Date","C","D")

z=data.frame(c("2000-01-01","2000-01-05","2000-04-09","2000-06-01","2001-01-01"),c("100","100","100","100","100"),c("200","200","200","200","200"),c("0","10","10","0","0"),c("0","0","35","0","0"))
colnames(z)=c("Date","A","B","C","D")

x$Date = as.Date(x$Date)
y$Date = as.Date(y$Date)

问题：如何通过有效的代码到达z？

举例说明：

> x
        Date   A   B
1 2000-01-01 100 200
2 2000-06-01 100 200
3 2001-01-01 100 200
> y
        Date  C  D
1 2000-01-05 10  0
2 2000-04-09  0 35
> z
        Date   A   B  C  D
1 2000-01-01 100 200  0  0
2 2000-01-05 100 200 10  0
3 2000-04-09 100 200 10 35
4 2000-06-01 100 200 10 35
5 2001-01-01 100 200 10 35
>

编辑：感谢下面的答案。解决方案似乎是一个简单的完全连接，然后是循环中的循环（我想出了第二步）：

x$Date = as.Date(x$Date)
y$Date = as.Date(y$Date)

tt=merge(x,y,by='Date',all=TRUE)

for (i in 2:(ncol(x)+ncol(y)-1)){
  for (j in 2:(nrow(x)+nrow(y))){
    if (is.na(tt[j,i])==TRUE & is.na(tt[j-1,i])==FALSE){
      tt[j,i]=tt[j-1,i]}
  }
}

EDIT2：其他人发布的解决方案似乎更有效率。仅仅为了完整性，如果y中的0被NA替换，则我的更长解决方案有效，即将y定义为：

y=data.frame(c("2000-01-05","2000-04-09"),c("10",NA),c(NA,"35"))
colnames(y)=c("Date","C","D")

然后在最后一步中替换z中的NA。

我从第一次编辑中学到了，我没有编辑上面的原始问题，以避免混淆。

非常感谢你的帮助！

Answer 1

可能的解决方案可能是使用data.table打包中的na.locf和zoo函数的组合：

# loading the needed packages
library(data.table)
library(zoo)

# converting x & y to datatables
setDT(x)
setDT(y)

# merge x & y into z
z <- merge(x, y, by="Date", all=TRUE) # this works in base R as well

# fill the NA's with the last observation
cols <- c("A","B","C","D") # in this specific case, you can also use: LETTERS[1:4]
z[, (cols) := lapply(.SD, na.locf, rule = 1, na.rm=FALSE), .SDcols= cols]

这给出了：

> z
         Date   A   B  C  D
1: 2000-01-01 100 200 NA NA
2: 2000-01-05 100 200 10  0
3: 2000-04-09 100 200  0 35
4: 2000-06-01 100 200  0 35
5: 2001-01-01 100 200  0 35

这个结果也可以在@Tensibai在评论中提到的基础R中实现（由于某种原因，起初我的系统没有工作）：

z <- merge(x, y, by="Date", all=TRUE)
z <- na.locf(z)

要获得准确的所需输出，您需要一些额外的步骤（省略第一步，因为它们是相同的）：

# merge x & y into z
z <- merge(x, y, by="Date", all=TRUE) # this works in base R as well

# replace the zero with NA
z[z==0] <- NA

# fill the NA's with the last observation
cols <- LETTERS[1:4]
z[, (cols) := lapply(.SD, na.locf, rule = 1, na.rm=FALSE), .SDcols= cols]

# replace the remaining NA's with zero's
z[is.na(z)] <- 0

这给出了：

> z
         Date   A   B  C  D
1: 2000-01-01 100 200  0  0
2: 2000-01-05 100 200 10  0
3: 2000-04-09 100 200 10 35
4: 2000-06-01 100 200 10 35
5: 2001-01-01 100 200 10 35

在基地R你会这样做：

z <- merge(x, y, by="Date", all=TRUE)
z[z==0] <- NA
z <- na.locf(z)
z[is.na(z)] <- 0

得到相同的结果。

Answer 2

使用dplyr和一些函数的替代方法：

library(lubridate)
library(dplyr)

# dataset
x=data.frame(c("2000-01-01","2000-06-01","2001-01-01"),
             c("100","100","100"),
             c("200","200","200"), stringsAsFactors = F)
colnames(x)=c("Date","A","B")

y=data.frame(c("2000-01-05","2000-04-09"),
             c("10","0"),
             c("0","35"), stringsAsFactors = F)
colnames(y)=c("Date","C","D")

# update date columns
x$Date = ymd(x$Date)
y$Date = ymd(y$Date)

# function that replaces NAs with 0s
ff = function(x){x[is.na(x)]=0 
                 return(as.numeric(x))}

# function that updates zero elements with the previous ones
ff2 = function(x){

  for (i in 2:length(x)){x[i] = ifelse(x[i]==0, x[i-1], x[i])}

  return(x)

}

# create the full dataset
xy =
    x %>% 
    full_join(y, by="Date") %>% 
    arrange(Date)

xy

#         Date    A    B    C    D
# 1 2000-01-01  100  200 <NA> <NA>
# 2 2000-01-05 <NA> <NA>   10    0
# 3 2000-04-09 <NA> <NA>    0   35
# 4 2000-06-01  100  200 <NA> <NA>
# 5 2001-01-01  100  200 <NA> <NA>


  xy %>%
  group_by(Date) %>% 
  mutate_each(funs(ff)) %>%
  ungroup %>% 
  select(-Date) %>%
  mutate_each(funs(ff2)) %>%
  bind_cols(data.frame(Date=xy$Date)) %>%
  select(Date,A,B,C,D)

#           Date   A   B  C  D
#   1 2000-01-01 100 200  0  0
#   2 2000-01-05 100 200 10  0
#   3 2000-04-09 100 200 10 35
#   4 2000-06-01 100 200 10 35
#   5 2001-01-01 100 200 10 35

在R中动态更新/组合两个data.frames

2 个答案: