R: How to do fastest replacement in R?

I have a input dataframe like this (the real one is very large, so I want to do it faster):

df1 <- data.frame(A=c(1:5), B=c(5:9), C=c(9:13))

  A B  C
1 1 5  9
2 2 6 10
3 3 7 11
4 4 8 12
5 5 9 13

I have a dataframe with replacement code like this (the entries here maybe more than df1):

df2 <- data.frame(X=c(1:15), Y=c(101:115))

    X   Y
1   1 101
2   2 102
3   3 103
4   4 104
5   5 105
6   6 106
7   7 107
8   8 108
9   9 109
10 10 110
11 11 111
12 12 112
13 13 113
14 14 114
15 15 115

By matching df2$X with value in df开发者_运维技巧1$A and df1$B, I want to get a new_df1 by replace df1$A and df1$B with the corresponding values in df2$Y, i.e. resulting this new_df1

  A    B    C
1 101  105  9
2 102  106 10
3 103  107 11
4 104  108 12
5 105  109 13

Could you mind to give me some guidance how to do it faster in R, as my dataframe is very large? Many thanks.

---

As Thilo mentioned Nico's answer assumes that df2 is ordered by X and X contains every integer 1,2,3....

I would prefer to use match() as a more general case:

df1 <- data.frame(A=c(1:5), B=c(5:9), C=c(9:13))
df2 <- data.frame(X=c(1:15), Y=c(101:115))

new_df1 <- df1

new_df1$A <- df2$Y[match(df1$A,df2$X)]
new_df1$B <- df2$Y[match(df1$B,df2$X)]
    A   B  C
1 101 105  9
2 102 106 10
3 103 107 11
4 104 108 12
5 105 109 13

---

It's supereasy! You just need to get the proper offsets in the array.

So for instance, to get the Y column of df2 corresponding to the values in the A column of df1 you'll write df2$Y[df1$A]

Hence, your code will be:

df_new <- data.frame("A" = df2$Y[df1$A], "B" = df2$Y[df1$B], "C" = df1$C)

---

Here is another (one-liner) way of doing it.

> with(c(df2,df1),data.frame(A = Y[match(A,X)],B = Y[match(B,X)],C))
    A   B  C
1 101 105  9
2 102 106 10
3 103 107 11
4 104 108 12
5 105 109 13

However I am not sure whether it will be faster than the other suggestions