5

我有矩阵

m <- matrix(1:9, nrow = 3, ncol = 3, byrow = TRUE,dimnames = list(c("s1", "s2", "s3"),c("tom", "dick","bob")))

   tom dick bob
s1   1    2   3
s2   4    5   6
s3   7    8   9

#and the data frame

current<-c("tom", "dick","harry","bob")
replacement<-c("x","y","z","b")
df<-data.frame(current,replacement)

  current replacement
1     tom           x
2    dick           y
3   harry           z
4     bob           b

#I need to replace the existing names i.e. df$current with df$replacement if 
#colnames(m) are equal to df$current thereby producing the following matrix


m <- matrix(1:9, nrow = 3, ncol = 3, byrow = TRUE,dimnames = list(c("s1", "s2", "s3"),c("x", "y","b")))

   x y b
s1 1 2 3
s2 4 5 6
s3 7 8 9

有什么建议吗？我应该使用“if”循环吗？谢谢。

4

2 回答 2

7

您可以使用将from与 中的值which进行匹配。然后，当您拥有索引时，您可以从.colnamesmdf$currentdf$replacement

colnames(m) = df$replacement[which(df$current %in% colnames(m))]

在上面：

  1. %in%测试被比较的对象之间的任何匹配TRUE。FALSE
  2. which(df$current %in% colnames(m))标识匹配名称的索引（在本例中为行号）。
  3. df$replacement[...]df$replacement是对仅返回与步骤 2 匹配的行的列进行子集化的基本方法。
于 2012-07-17T08:48:13.880 回答
7

查找索引的一种更直接的方法是使用match：

> id <- match(colnames(m), df$current)
> id
[1] 1 2 4
> colnames(m) <- df$replacement[id]
> m
   x y b
s1 1 2 3
s2 4 5 6
s3 7 8 9

如下所述%in%，通常使用起来更直观，效率差异很小，除非集合相对较大，例如

> n <- 50000 # size of full vector
> m <- 10000 # size of subset
> query <- paste("A", sort(sample(1:n, m)))
> names <- paste("A", 1:n)
> all.equal(which(names %in% query), match(query, names))
[1] TRUE
> library(rbenchmark)
> benchmark(which(names %in% query))
                     test replications elapsed relative user.self sys.self user.child sys.child
1 which(names %in% query)          100   0.267        1     0.268        0          0         0
> benchmark(match(query, names))
                 test replications elapsed relative user.self sys.self user.child sys.child
1 match(query, names)          100   0.172        1     0.172        0          0         0
于 2012-07-17T09:01:52.313 回答