用NA替换矩阵中的0

Ale*_*nda 5 performance r matrix

用NA替换矩阵中的所有零的最有效方法是什么?

我所做的:

my_matrix[my_matrix==0] <- NA
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我需要它用于推荐系统(recommenderlab).填充NA与构建推荐系统的时间相同.

编辑1:

昏暗(my_matrix)~500000x500

零为~90%.

Pie*_*une 10

答案和基准

my_matrix <- matrix(1:5e5, ncol=50)
my_matrix[4000:5000, 3:10] <- 0

library(microbenchmark)
microbenchmark(
  insubset     = my_matrix[my_matrix %in% 0],
  replace1     = replace(my_matrix, my_matrix %in% 0, NA),
  replace2     = replace(my_matrix, which( my_matrix==0), NA),
  Aleksandro   = my_matrix[my_matrix==0] <- NA,
  excloperator = my_matrix[!my_matrix] <- NA,
  is.na        = is.na(my_matrix) <- which(my_matrix == 0)
)

Unit: milliseconds
         expr       min        lq      mean    median        uq        max neval
     insubset 22.579762 22.890431 26.197510 23.453346 25.210976 151.957848   100
     replace1 21.630386 23.621707 27.573375 25.643425 26.225683 104.389554   100
     replace2  3.979487  4.069095  4.872796  4.159493  6.449839   8.887427   100
   Aleksandro 12.787962 13.100210 14.837055 13.689376 14.098338  96.258866   100
 excloperator 11.894246 12.275969 13.541593 13.011391 15.144429  17.307862   100
        is.na  7.642823  8.901978   15.7352  9.342954  10.13166   68.31235   100
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  • @adilapapaya感谢您的测试.我认为通过使用`which`我们获得的位置而不是具有巨大的逻辑矩阵 (2认同)