如何獲取字符串值爲1的元素的數量

Question

如何獲取數組中只包含字符串@「one」的元素的數量。

NSMutableArray *array = [[NSMutableArray alloc]initWithObject:@"one",@"one",@"two",@"one",@"five",@"one",nil]; 

如何獲取其中包含一個數組的數量。

Answer 1

很多路要走：

NSMutableArray *array = [[NSMutableArray alloc]initWithObject:@"one",@"one",@"two",@"one",@"five",@"one",nil]; 

使用塊：

NSInteger occurrenceCount = [[array indexesOfObjectsPassingTest:^(id obj, NSUInteger idx, BOOL *stop) {return [obj isEqual:@"one"];}] count]; 

使用循環：

int occurrenceCount = 0; 
for(NSString *str in array){ 
    occurrenceCount += ([string isEqualToString:@"one"]?1:0); 
} 

使用NSCountedSet：

NSCountedSet *countedSet = [[NSCountedSet alloc] initWithArray:array]; 
NSLog(@"Occurrences of one: %u", [countedSet countForObject:@"one"]); 

使用NSPredicate：（如EridB建議）

NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SELF contains %@", 
          @"one"]; 

NSInteger occurrenceCount = [array filteredArrayUsingPredicate:predicate].count; 

檢查答案here瞭解更多詳情。

Answer 2

有提到從那些另一溶液

// Query to find elements which match 'one' 
NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SELF contains %@", 
          @"one"]; 

// Use the above predicate on your array 
// The result will be a `NSArray` so from there we count the elements on this array 
NSInteger count = [array filteredArrayUsingPredicate:predicate].count; 

// Prints out number of elements 
NSLog(@"%li", (long)count); 

Answer 3

NSArray *array = @[@"one",@"one",@"two",@"one",@"five",@"one"]; 
    NSPredicate *searchCountString= [NSPredicate predicateWithFormat:@"SELF contains %@",@"one"]; 
    NSInteger count = [array filteredArrayUsingPredicate:searchCountString].count; 
    NSLog(@"%ld",(long)count);