Question

我有两个data.frames，（df1，df2），我想将P1-P10列中的值替换为df1$V2的值，但保留df2的前两列。

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))

这给了我我的愿望输出，我的真实数据有10,000列，有没有办法避免cbind步骤或使过程快速？

> 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

由于

Answer 1

您可match df2[3:12] df1[[1]]的值df1[2]。这些行号用于从df2[3:12] <- df1[match(as.character(unlist(df2[3:12])), as.character(df1[[1]])), 2]中提取值。

df2

结果（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）：

df2

如果您不想替换df4中的值，可以使用

创建新数据框

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

{{1}}

Answer 2

尝试一些*pply魔术：

lookup<-tapply(df1$V2, df1$V1, unique) #Creates a lookup table
lookup.function<-function(x) as.numeric(lookup[as.character(x)]) #The function
df4<-data.frame(df2[,1:2], apply(df2[,3:12], 2,lookup.function )) #Builds the output

<强>更新：

*pply系列比merge快得多，至少一个数量级。看看这个

num<-1000
df1 = data.frame(V1=LETTERS, V2=rnorm(26))
df2<-data.frame(cbind(first=1:num,second=1:num, matrix(sample(LETTERS, num^2, replace=T), nrow=num, ncol=num)))


start<-Sys.time()
lookup<-tapply(df1$V2, df1$V1, unique)
lookup.function<-function(x) as.numeric(lookup[as.character(x)])
df4<-data.frame(cbind(df2[,1:2], data.frame(apply(df2[,3:(num+2)], 2, lookup.function ))))
(difftime(Sys.time(),start))


start<-Sys.time()
df4.merge <- "[<-"(df2, 3:num, value = df1[match(as.character(unlist(df2[3:num])), as.character(df1[[1]])), 2])
(difftime(Sys.time(),start))

sum(df4==df4.merge)==num^2

对于3000列和行，*pply组合需要4.3s，而merge在我的慢速英特尔上需要大约22秒。它很好地扩展。对于4000列和行，相应的时间分别为7.4秒和118秒。

根据另一个data.frame替换data.frame中的某些列值

2 个答案: