这里有一个基本的解决方案，应该是相当快的：

acols <- grep("^a", colnames(df))
bcols <- grep("^b", colnames(df))

df[bcols] = lapply(
  df[bcols], 
  \(bb) {
    bb[rowSums(bb == df[acols], na.rm = TRUE) == 0] = NA_integer_
    bb
  })

df
#   row b1 b2 b3 a1 a2 a3
# 1   1 55 NA NA 55 NA NA
# 2  12 17 44 90 17 44 90
# 3  17 12 44 NA 12 44 NA
# 4  44 12 17 90 12 17 90
# 5  55  1 NA NA  1 NA NA
# 6  90 12 NA 44 12 28 44

我将你的数据扩展到50万行，在这个数据集上，这种方法的运行速度提高了50倍，只使用了1/50的内存。我怀疑随着列数的增加，差异会更大。

single_loop = function(df){
  df[bcols] = lapply(
  df[bcols], 
  \(bb) {
    bb[rowSums(bb == df[acols], na.rm = TRUE) == 0] = NA
    bb
  })
  df
}

double_loop = function(df){
  for (i in 1:nrow(df)) {
    for (j in bcols) {
      if (df[i, j] %in% df[i, acols]) {
        next
      } else {
        df[i, j] <- NA
      }
    }
  }
  df
}

df_big = df[sample(1:nrow(df), size = 5e5, replace = TRUE)]

bench::mark(
  op = double_loop(df_big),
  gregor = single_loop(df_big)
)
# # A tibble: 2 × 13
#   expression      min   median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc total_time result   
#   <bch:expr> <bch:tm> <bch:tm>     <dbl> <bch:byt>    <dbl> <int> <dbl>   <bch:tm> <list>   
# 1 op          624.8ms  624.8ms      1.60     381MB     9.60     1     6      625ms <df [6 ×…
# 2 gregor       12.8ms   13.4ms     48.5     7.63MB     9.33    26     5      536ms <df [6 ×…
# # … with 3 more variables: memory <list>, time <list>, gc <list>
# Warning message:
# Some expressions had a GC in every iteration; so filtering is disabled.

如果需要检查任何位置的值，可以对数据进行往返透视：

library(tidyr)
library(dplyr)

df %>%
  pivot_longer(-row) %>%
  mutate(grp = substr(name, 1L, 1L),
         value = replace(value, !(value[grp == "b"] %in% value[grp == "a"])[1:n()], NA), 
         grp = NULL,
         .by = row) %>%
  pivot_wider()
  
# A tibble: 6 × 7
    row    b1    b2    b3    a1    a2    a3
  <int> <int> <int> <int> <int> <int> <int>
1     1    55    NA    NA    55    NA    NA
2    12    17    44    90    17    44    90
3    17    12    44    NA    12    44    NA
4    44    12    17    90    12    17    90
5    55     1    NA    NA     1    NA    NA
6    90    12    NA    44    12    28    44

如果可以按列对进行检查，则可以执行以下操作：

df %>%
  mutate(replace(pick(b1:b3), pick(a1:a3) != pick(b1:b3) | is.na(pick(a1:a3)), NA))

  row b1 b2 b3 a1 a2 a3
1   1 55 NA NA 55 NA NA
2  12 17 44 90 17 44 90
3  17 12 44 NA 12 44 NA
4  44 12 17 90 12 17 90
5  55  1 NA NA  1 NA NA
6  90 12 NA 44 12 28 44