怎么样：

rowMeans(cbind(a, b), na.rm=TRUE)

或

colMeans(rbind(a, b), na.rm=TRUE)

我不太清楚你想要什么，但是

apply(rbind(a,b),2,mean,na.rm = TRUE)

做你想做的事

tidyverse解决方案使用purrr：

library(purrr)
a <- c(1, 2, 3, 4)
b <- c(NA, 6, 7, 8)

# expected:
c(1, 4, 5, 6) 
#> [1] 1 4 5 6

# actual:
map2_dbl(a,b, ~mean(c(.x,.y), na.rm=T)) # actual
#> [1] 1 4 5 6

对于任意数量的向量：

>  pmap_dbl(list(a,b, a, b), compose(partial(mean, na.rm = T), c))
 [1] 1 4 5 6

另一个选项是collapse::fmean，它默认为矩阵的列平均值和na.rm = TRUE：

fmean(rbind(a, b))
#[1] 1 4 5 6

基准

向量a和b（长度= 4）：

microbenchmark::microbenchmark(
  collapse = fmean(rbind(a, b)),
  rowMeans = rowMeans(cbind(a, b), na.rm=TRUE),
  colMeans = colMeans(rbind(a, b), na.rm=TRUE),
  purrr = purrr::map2_dbl(a,b, ~mean(c(.x,.y), na.rm=T)),
  apply = apply(rbind(a,b),2,mean,na.rm = TRUE)
)

# Unit: microseconds
#      expr    min       lq      mean   median       uq      max neval
#  collapse  6.501   7.9020  10.72705   9.7010  10.8010   56.101   100
#  rowMeans  4.601   6.0505   9.21504   7.8010   9.4515   28.102   100
#  colMeans  4.700   5.7010   7.76410   6.8515   8.2015   27.301   100
#     purrr 94.101 104.4505 140.36694 108.8010 121.9510 2120.901   100
#     apply 50.301  55.1010  65.37305  59.9005  65.6510  156.700   100

大向量（大小1e6）：

a = sample(1e6)
b = sample(1e6)

# Unit: milliseconds
#      expr       min        lq     mean    median       uq     max neval
#  collapse  8.384401  9.621752 13.02568 10.160101 18.83060 34.2746   100
#  rowMeans 18.504201 21.513251 27.88083 23.509051 31.28925 94.2124   100
#  colMeans  8.117601  9.344551 12.69392  9.897702 12.50430 54.1703   100