根据另一个 data.frame 替换 data.frame 中的一些列值

Question

我有两个 data.frames (df1, df2)，我想用 df1$V2 的值替换 P1-P10 列中的字母值，但保留前两列df2.

df1 = data.frame(V1=LETTERS, V2=rnorm(26))

df2 <- data.frame(Name=sample(LETTERS, 6), bd=sample(1:6), P1=sample(LETTERS,6), P2=sample(LETTERS, 6), P3=sample(LETTERS, 6), P4=sample(LETTERS, 6), P5=sample(LETTERS, 6), P6=sample(LETTERS, 6), P7=sample(LETTERS, 6), P8=sample(LETTERS, 6), P9=sample(LETTERS, 6), P10=sample(LETTERS, 6))

我的方法如下:

df3 <- matrix(setNames(df1[,2], df1[,1])[as.character(unlist(df2[,3:12]))], nrow=6, ncol=10)
df4 <- data.frame(cbind(df2[,1:2], df3))

这给了我想要的输出，我的真实数据有 10,000 列，有什么方法可以避免 cbind 步骤或使过程更快？

> df4
Name bd         X1          X2         X3         X4         X5         X6        X7         X8         X9        X10
1    V  6 -1.8991102  0.40269050 -0.1517500 -2.5297829  1.5315622  1.4897071  1.364071 -1.2443708 -1.3197276 -0.4917057
2    T  1 -2.5297829 -0.44614123 -0.1894970 -0.6693774 -0.1517500 -1.0650962 -0.151750 -0.4461412 -0.6693774 -1.1351770
3    R  5 -0.6693774  0.09059365 -2.5297829  0.3233827 -0.9383348 -0.4461412  1.281797  1.5315622  1.4897071 -0.4461412
4    B  4 -0.4461412 -0.93833476 -1.2443708 -0.4461412 -0.1894970 -0.9383348 -1.135177 -1.8991102 -0.1894970  0.4026905
5    K  2 -1.0180271 -1.06509624 -0.1939600 -0.1894970  1.4897071 -0.6693774 -1.899110 -1.3197276  1.5315622 -0.1517500
6    Y  3  1.5315622 -0.19396005 -0.4917057 -0.4664239 -1.8991102  0.4026905 -1.065096 -0.9383348 -1.2443708 -0.4664239

谢谢

Answer 1

您可以在 df1[[1]] 中匹配 df2[3:12] 的值。这些行号用于从 df1[2] 中提取值。

df2[3:12] <- df1[match(as.character(unlist(df2[3:12])), 
                       as.character(df1[[1]])), 2]

结果(df2):

  Name bd         P1         P2         P3         P4         P5         P6         P7         P8         P9        P10
1    H  5  0.1199355  0.3752010 -0.3926061 -1.1039548 -0.1107821  0.9867373 -0.3360094 -0.7488000 -0.3926061  2.0667704
2    U  4  0.1168599  0.1168599  0.9867373  1.3521418  0.9867373 -0.3360094 -0.7724007 -0.3926061 -0.3360094 -1.2543480
3    R  3 -1.2337890 -0.1107821 -0.7724007  2.0667704  0.3752010  0.4645504  0.9867373  0.1168599 -0.0981773 -0.3926061
4    G  2 -0.3926061  0.3199261 -0.0981773 -0.1107821  2.0667704 -1.1039548 -1.2337890  0.3199261 -1.2337890 -2.1534678
5    C  6 -2.1534678 -1.1039548 -1.1039548 -0.7488000  0.4645504  0.3199261 -2.1534678 -0.3360094  0.9867373  0.8771467
6    I  1  0.6171634  0.6224091  1.8011711  0.7292998  0.8771467  2.0667704  0.3752010  0.4645504 -2.1534678 -0.7724007

如果你不想替换df2里面的值，你可以创建一个新的data frame df4 with

df4 <- "[<-"(df2, 3:12, value = df1[match(as.character(unlist(df2[3:12])), 
                                          as.character(df1[[1]])), 2])