使用基于匹配值的另一个数据帧更新一个数据帧中的列

I have a data frame "z"

我有一个数据框“z”

   letter color
1       a     0
2       e     0
3       b     0
4       b     0
5       d     0
6       d     0
7       a     0
8       b     0
9       c     0
10      d     0
11      c     0
12      c     0
13      c     0
14      c     0
15      e     0
16      e     0
17      a     0
18      d     0
19      e     0
20      b     0

and another data frame "y"

和另一个数据框“y”

  letter color
1      a   red
2      b  blue
3      c green

when the letter in z matches with a letter in y I would like to append the color from y into the corresponding color field in z but I do not want to remove any values from z. If a match doesn't occur, z$color should remain unchanged. I used"0" as a place holder in z$color, this could be text instead.

当z中的字母与y中的字母匹配时，我想将y中的颜色附加到z中的相应颜色字段中，但我不想从z中删除任何值。如果没有匹配，z $ color应保持不变。我使用“0”作为z $颜色的占位符，这可能是文本而不是。

I've been attempting things for loops, the match() command and statements with %in% but I'm not quite achieving the results I'm after.

我一直在尝试for循环，match（）命令和％in％的语句，但我还没有完全达到我想要的结果。

Any ideas?

有任何想法吗？

This is the code I used for the data frames

这是我用于数据帧的代码

set.seed(3)
z=data.frame(sample(c("a","b","c","d","e"),20,replace=T))
names(z)="letter"
z$color=rep(0,dim(z)[1])
z

y1=c("a","b","c")
y2=c("red","blue","green")
y=data.frame(cbind(y1,y2))
names(y)=c("letter","color")
y

3 个解决方案

#1

you don't need z$color in the first place if its just place holder, you can replace NA later with 0

如果它只是占位符，你首先不需要z $颜色，你可以稍后用0替换NA

z$color<-y[match(z$letter, y$letter),2]

#2

You can use merge:

你可以使用合并：

dat <- merge(z, y, by = "letter", all.x = TRUE)
transform(dat, color = ifelse(is.na(color.y), 
                              color.x, as.character(color.y)))[-(2:3)]

   letter color
1       a   red
2       a   red
3       a   red
4       b  blue
5       b  blue
6       b  blue
7       b  blue
8       c green
9       c green
10      c green
11      c green
12      c green
13      d     0
14      d     0
15      d     0
16      d     0
17      e     0
18      e     0
19      e     0
20      e     0

#3

sqldf/sqlite is very flexible:

sqldf / sqlite非常灵活：

library(sqldf)
z$color="0" # to avoid conflicts numeric/characters
z <- sqldf(c("UPDATE z
             SET color = (SELECT y.color
                          FROM y
                          WHERE z.letter = y.letter
                           )
             WHERE EXISTS (SELECT 1
                           FROM y
                           WHERE z.letter = y.letter
                           )"
             , "select * from main.z"
                  )
           )
z
   letter color
1       b  blue
2       a   red
3       d   0.0
4       d   0.0
5       e   0.0
6       a   red
7       a   red
8       c green
9       b  blue
10      c green
11      e   0.0
12      c green
13      b  blue
14      d   0.0
15      d   0.0
16      d   0.0
17      c green
18      e   0.0
19      a   red
20      c green

#1