Question

用NAs替换矩阵中的所有零的最有效方法是什么？

我的所作所为：

- (void)registerSettingsAndCategories {
// Create a mutable set to store the category definitions.
NSMutableSet* categories = [NSMutableSet set];

// Define the actions for a meeting invite notification.
UIMutableUserNotificationAction* acceptAction = [[UIMutableUserNotificationAction alloc] init];
acceptAction.title = NSLocalizedString(@"Repondre", @"Repondre commentaire");
acceptAction.identifier = @"respond";
acceptAction.activationMode = UIUserNotificationActivationModeForeground; //UIUserNotificationActivationModeBackground if no need in foreground.
acceptAction.authenticationRequired = NO;

// Create the category object and add it to the set.
UIMutableUserNotificationCategory* inviteCategory = [[UIMutableUserNotificationCategory alloc] init];
[inviteCategory setActions:@[acceptAction]
                forContext:UIUserNotificationActionContextDefault];
inviteCategory.identifier = @"respond";

[categories addObject:inviteCategory];

// Configure other actions and categories and add them to the set...

UIUserNotificationSettings* settings = [UIUserNotificationSettings settingsForTypes:
                                        (UIUserNotificationTypeAlert | UIUserNotificationTypeBadge | UIUserNotificationTypeSound)
                                                                         categories:categories];

[[UIApplication sharedApplication] registerUserNotificationSettings:settings];}

我需要它用于推荐系统（recommenderlab）。填充NAs与构建推荐系统的时间相同。

编辑1：

dim（my_matrix）~500000x500

零为~90％。

Answer 1

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

用NA替换矩阵中的0

1 个答案: