Question

假设我有两个向量定义如下：

a <- c("a1", "a2", "a3")
b <- c("b1", "b2", "b3")

现在我想最终得到一个像这样的字符串向量：

c("a1","b1", "a2", "b2", "a3", "b3")

这可能吗？

Answer 1

你可以rbind他们，然后强制回到矢量。

a <- c("a1", "a2", "a3") 
b <- c("b1", "b2", "b3")
c(rbind(a,b))

# [1] "a1" "b1" "a2" "b2" "a3" "b3"

正如@Moody_Mudskipper指出的那样，as.vector(rbind(a,b)))更快

对于长度不同的情况，我在此链接中找到了Rolf Turner的以下解决方案：http://r.789695.n4.nabble.com/Interleaving-elements-of-two-vectors-td795123.html

riffle <- function (a,b) { 
  n  <- min(length(a),length(b)) 
  p1 <- as.vector(rbind(a[1:n],b[1:n])) 
  p2 <- c(a[-(1:n)],b[-(1:n)]) 
  c(p1,p2) 
} 

riffle(1:3, letters[1:5])

# [1] "1" "a" "2" "b" "3" "c" "d" "e"

Answer 2

那些也有效：

c(matrix(c(a,b),ncol=3,,T))
c(a,b)[rep(1:length(a),each=2)+c(0,length(a))]
c(mapply(c,a,b))
# [1] "a1" "b1" "a2" "b2" "a3" "b3"

但Ryan的解决方案更快：

a <- paste0("a",1:10000)
b <- paste0("b",1:10000)

microbenchmark::microbenchmark(
  ryan = c(rbind(a,b)),
  mm1 = c(matrix(c(a,b),ncol=length(a),,T)),
  mm2 = c(a,b)[rep(1:length(a),each=2)+c(0,length(a))],
  mm3 = c(mapply(c,a,b))
)

# Unit: microseconds
# expr    min      lq      mean median       uq     max neval
# ryan  172.9  192.10   218.366  200.7   222.00   643.5   100
#  mm1  258.5  288.10   308.359  300.6   316.05   685.6   100
#  mm2  749.9  772.05   787.783  780.9   795.70  1104.6   100
#  mm3 9203.0 9585.70 11491.192 9978.7 10209.80 35516.3   100

as.vector让它更快一点：

microbenchmark::microbenchmark(
  ryan = c(rbind(a,b)),
  ryan2 = as.vector(rbind(a,b)))

# Unit: microseconds
#  expr   min     lq    mean median     uq   max neval
#  ryan 169.7 185.65 198.797  194.2 208.10 302.0   100
# ryan2 108.8 121.05 135.465  128.3 142.85 297.7   100

隔行扫描两个向量

2 个答案: