Question

以下是两个df示例数据：

df1

ID      First.seen  Last.seen 
A10   2015-09-07  2015-09-16       
A11   2015-09-07  2015-09-19 



df2
 ID      First_seen  Last_seen
 A1      2015-09-07  0
A10      2015-09-07  0

如果df2$Last_seen在ID中都很常见，我想填写dfs。请注意，在实际数据中，我在两个dfs中都有几个ID。我尝试使用for循环，但我得到了数值：

for (i in 1:nrow(df2)){
  if (df2$ID[i] %in% df1$ID) {
    df2$Last_seen[i] <- df1$Last.seen[df1$ID == df2$ID[i]]
  }else{
    df2$Last_seen[i] <- 0
  }
}

我发现this回答了使用seq_along的同一问题，但是当我应用此代码时，我得到df1$Last_seen[i] == 1的结果：

 for (i in seq_along(1:nrow(df2))){
      if (df2$ID[i] %in% df1$ID) {
        df2$Last_seen[i] <- df1$Last.seen[df1$ID == df2$ID[i]]
      }else{
        df2$Last_seen[i] <- 0
      }
    }

有关如何正确使用它的任何建议吗？

Answer 1

你不需要循环来做到这一点。您需要在ID上加入表。这可以通过dplyr：

完成

df1 <- read.table(text="ID      First.seen  Last.seen
A10   2015-09-07  2015-09-16
A11   2015-09-07  2015-09-19",header=TRUE, stringsAsFactors=FALSE)

df2<- read.table(text="ID      First_seen  Last_seen
 A1      2015-09-07  0
A10      2015-09-07  0",header=TRUE, stringsAsFactors=FALSE)

library(dplyr)
left_join(df2,df1)
   ID First_seen Last_seen First.seen  Last.seen
1  A1 2015-09-07         0       <NA>       <NA>
2 A10 2015-09-07         0 2015-09-07 2015-09-16

如果你想要一个三列表：

left_join(df2,df1, by=c("ID" = "ID","First_seen"="First.seen")) %>%
mutate(Last_seen=ifelse(is.na(Last.seen),Last_seen,Last.seen)) %>%
select(-Last.seen)

   ID First_seen  Last_seen
1  A1 2015-09-07          0
2 A10 2015-09-07 2015-09-16

编辑要更改Last_seen为0的出现次数，您可以添加另一个ifelse：

left_join(df2,df1, by=c("ID" = "ID","First_seen"="First.seen")) %>%
mutate(Last_seen=ifelse(is.na(Last.seen),Last_seen,Last.seen),
       Last_seen=ifelse(Last_seen==0,format(as.Date(First_seen)+16,"%Y-%m-%d"),Last.seen))%>%
select(-Last.seen)

   ID First_seen  Last_seen
1  A1 2015-09-07 2015-09-23
2 A10 2015-09-07 2015-09-16

<强> EDIT2

left_join(df2,df1, by=c("ID" = "ID","First_seen"="First.seen")) %>%
mutate(Last_seen=ifelse(is.na(Last.seen),Last_seen,Last.seen),
       Last_seen=ifelse(Last_seen==0,format(as.Date(First_seen)+16,"%Y-%m-%d",origin = "1900-01-01"),Last.seen))%>%
select(-Last.seen)

   ID First_seen  Last_seen
1  A1 2015-09-07 2015-09-23
2 A10 2015-09-07 2015-09-16

使用seq_along（）处理for循环中的Date

1 个答案: