Question

（链接的Vlookup线程没有回答这个问题）

我正在寻找一种方法来替换一个数据帧（DF2）中的值与来自另一个（DF1）的值，其中DF2包含重复的条目，但我想保留这些重复项。

作为一个例子：

假设我有2个数据帧。其中一个名为DF1，包含不同日期酒店遮阳伞的正确数字。

我们在5月20日，5月25日，6月1日以及相关联的保护伞上提供了希尔顿_A的订单项。与Hilton_B和Hilton_C相同。

这是DF1的参数，参考数据帧：

structure(list(Date = structure(c(15852, 15859, 15852, 15859, 
15852, 15859, 15852), class = "Date"), Hotel = structure(c(1L, 
1L, 2L, 2L, 3L, 3L, 4L), .Label = c("Hilton_A", "Hilton_B", "Hilton_C", 
"Hilton_D"), class = "factor"), Umbrellas = c(9340L, 6401L, 9089L, 
7716L, 5542L, 5565L, 8158L), datename = c("2013-05-27_Hilton_A", 
"2013-06-03_Hilton_A", "2013-05-27_Hilton_B", "2013-06-03_Hilton_B", 
"2013-05-27_Hilton_C", "2013-06-03_Hilton_C", "2013-05-27_Hilton_D"
)), .Names = c("Date", "Hotel", "Umbrellas", "datename"), row.names = c(NA, 
-7L), class = "data.frame")

DF2包含不同日期的许多其他酒店的信息，以及DF1中Hiltons的信息。问题是，DF2中的伞＃对于希尔顿来说是错误的，我需要用DF1中的＃替换它们。

这是DF2的输入，包括不正确的希尔顿号码，以及其他一些我不想触及的数据：

structure(list(Date = structure(c(15845, 15852, 15859, 15852, 
15859, 15845, 15859, 15845, 15845, 15852, 15845, 15845, 15882
), class = "Date"), Hotel = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 
2L, 2L, 3L, 4L, 5L, 6L, 7L), .Label = c("Hilton_A", "Hilton_B", 
"Hilton_C", "Hilton_D", "RedRoof_A", "RedRoof_D", "Sheraton_D"
), class = "factor"), Umbrellas = c(263L, 287L, 258L, 110L, 234L, 
212L, 265L, 542L, 81L, 51L, 162L, 232L, 493L), datename = c("2013-05-20_Hilton_A", 
"2013-05-27_Hilton_A", "2013-06-03_Hilton_A", "2013-05-27_Hilton_A", 
"2013-06-03_Hilton_A", "2013-05-20_Hilton_B", "2013-06-03_Hilton_B", 
"2013-05-20_Hilton_B", "2013-05-20_Hilton_C", "2013-05-27_Hilton_D", 
"2013-05-20_RedRoof_A", "2013-05-20_RedRoof_D", "2013-06-26_Sheraton_D"
)), .Names = c("Date", "Hotel", "Umbrellas", "datename"), row.names = c(NA, 
-13L), class = "data.frame")

通常这会起作用：

DF2$Umbrellas<- replace(DF2$Umbrellas, DF2$datename%in% DF1$datename, DF1$Umbrellas)

（其中“datename”只是酒店和日期的串联，因为同一家酒店有多个日期的信息（因此我们可以“唯一= ify”列表））

但DF2实际上对我想保留的每个酒店和日期有多个观察结果（即5/27的Hilton_A在DF2中显示2次）。

所以当我尝试将DF中的UAG＃替换为DF2时，我收到错误消息：

Warning message:
In replace(DF2$Umbrellas, DF2$hoteldatename %in% DF1$hoteldatename ,  :
  number of items to replace is not a multiple of replacement length

这些数字都错了。

有人知道这里发生了什么，以及我如何获取DF1中的数字来替换DF2中所有适用的观测值？

Answer 1

df3$Umbrellas<-df1$Umbrellas[match(df2$datename,df1$datename)]
> df3
         Date      Hotel Umbrellas              datename
1  2013-05-20   Hilton_A        NA   2013-05-20_Hilton_A
2  2013-05-27   Hilton_A      9340   2013-05-27_Hilton_A
3  2013-06-03   Hilton_A      6401   2013-06-03_Hilton_A
4  2013-05-27   Hilton_A      9340   2013-05-27_Hilton_A
5  2013-06-03   Hilton_A      6401   2013-06-03_Hilton_A
6  2013-05-20   Hilton_B        NA   2013-05-20_Hilton_B
7  2013-06-03   Hilton_B      7716   2013-06-03_Hilton_B
8  2013-05-20   Hilton_B        NA   2013-05-20_Hilton_B
9  2013-05-20   Hilton_C        NA   2013-05-20_Hilton_C
10 2013-05-27   Hilton_D      8158   2013-05-27_Hilton_D
11 2013-05-20  RedRoof_A        NA  2013-05-20_RedRoof_A
12 2013-05-20  RedRoof_D        NA  2013-05-20_RedRoof_D
13 2013-06-26 Sheraton_D        NA 2013-06-26_Sheraton_D

df3$Umbrellas<-ifelse(is.na(df3$Umbrellas),df2$Umbrellas,df3$Umbrellas)
> df3
         Date      Hotel Umbrellas              datename
1  2013-05-20   Hilton_A       263   2013-05-20_Hilton_A
2  2013-05-27   Hilton_A      9340   2013-05-27_Hilton_A
3  2013-06-03   Hilton_A      6401   2013-06-03_Hilton_A
4  2013-05-27   Hilton_A      9340   2013-05-27_Hilton_A
5  2013-06-03   Hilton_A      6401   2013-06-03_Hilton_A
6  2013-05-20   Hilton_B       212   2013-05-20_Hilton_B
7  2013-06-03   Hilton_B      7716   2013-06-03_Hilton_B
8  2013-05-20   Hilton_B       542   2013-05-20_Hilton_B
9  2013-05-20   Hilton_C        81   2013-05-20_Hilton_C
10 2013-05-27   Hilton_D      8158   2013-05-27_Hilton_D
11 2013-05-20  RedRoof_A       162  2013-05-20_RedRoof_A
12 2013-05-20  RedRoof_D       232  2013-05-20_RedRoof_D
13 2013-06-26 Sheraton_D       493 2013-06-26_Sheraton_D

用R中的重复观察替换多个数据帧之间的信息（寻找Vlookup类型函数）

1 个答案: