Question

我想从nsmutablearray中删除重复项。

数组结构： -

(
    {
        "act_end_date" = "";
        "act_entry_date" = "13/11/2014";
        "act_recurrrance_type" = Daily;
        "act_start_date" = "2014-11-13";
        "row_id" = 0;
        "act_id" = 2;
    }
    {
        "act_end_date" = "";
        "act_entry_date" = "13/11/2014";
        "act_recurrrance_type" = Daily;
        "act_start_date" = "2014-11-13";
        "row_id" = 1;
        "act_id" = 2;
    }
    {
        "act_end_date" = "";
        "act_entry_date" = "16/11/2014";
        "act_recurrrance_type" = Daily1;
        "act_start_date" = "2014-11-15";
        "row_id" = 2;
        "act_id" = 3;
    }
)

我想要的是检查数组包含任何具有相同act_id的字典。如果是，则删除第一个字典。即上述数组的预期结果： -

(
        {
            "act_end_date" = "";
            "act_entry_date" = "13/11/2014";
            "act_recurrrance_type" = Daily;
            "act_start_date" = "2014-11-13";
            "row_id" = 1;
            "act_id" = 2;
        }
        {
            "act_end_date" = "";
            "act_entry_date" = "16/11/2014";
            "act_recurrrance_type" = Daily1;
            "act_start_date" = "2014-11-15";
            "row_id" = 2;
            "act_id" = 3;
        }
    )

他们有什么办法吗？是否可以使用NSPREDICATE？

我试图用以下方式做到： -

  NSMutableArray *filteredArray = [NSMutableArray array];

    for (NSMutableDictionary* E1 in Event_Array)
    {

        BOOL hasDuplicate = [[filteredArray filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:@"act_id == %@", [E1 objectForKey:@"act_id"]]] count] > 0;

        if (!hasDuplicate)
        {
            [filteredArray addObject:E1];
        }
    }

因为它返回数组：---

(
    {
        "act_end_date" = "";
        "act_entry_date" = "13/11/2014";
        "act_recurrrance_type" = Daily;
        "act_start_date" = "2014-11-13";
        "row_id" = 0;
        "act_id" = 2;
    }
)

我想要的是

(
    {
        "act_end_date" = "";
        "act_entry_date" = "13/11/2014";
        "act_recurrrance_type" = Daily;
        "act_start_date" = "2014-11-13";
        "row_id" = 1;
        "act_id" = 2;
    }
)

Answer 1

    NSMutableSet *keysSet = [[NSMutableSet alloc] init];
    NSMutableArray *filteredArray = [[NSMutableArray alloc]init];
    for (NSDictionary *msg in germanMakes) {
    NSString *key = [NSString stringWithFormat:@"%@", msg[@"act_id"]];
    if (![keysSet containsObject:key]) {
        [filteredArray addObject:msg];
        [keysSet addObject:key];
    }
    }
    NSLog(@"filteredResults %@ keyset%@",filteredArray , keysSet);

germanMakes是你的初始数组，keysSet包含唯一的act_id（例如1,2,3,4 ..），filteredArray是由act_id过滤的数组。请记住，这不会按升序排序，但这部分很容易。

Answer 2

假设您希望最新的对象位于列表中。

解决方案1：（简单）

反转你的数组

NSMutableArray *filteredArray = [NSMutableArray array];
for (int i = Event_Array.count-1; i>= 0; i--)
{
    NSMutableDictionary* E1 = [Event_Array objectAtIndex:i];
    BOOL hasDuplicate = [[filteredArray filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:@"act_id == %@", [E1 objectForKey:@"act_id"]]] count] > 0;

    if (!hasDuplicate)
    {
        [filteredArray addObject:E1];
    }
}

解决方案2：

使用NSMutableDictionary而不是数组。

//Keys are act_id(s)
"2" = {
    "act_end_date" = "";
    "act_entry_date" = "13/11/2014";
    "act_recurrrance_type" = Daily;
    "act_start_date" = "2014-11-13";
    "row_id" = 1;
    "act_id" = 2;
};
"3" = {
    "act_end_date" = "";
    "act_entry_date" = "16/11/2014";
    "act_recurrrance_type" = Daily1;
    "act_start_date" = "2014-11-15";
    "row_id" = 2;
    "act_id" = 3;
}

从包含nsdictionary的nsmutablearray中删除重复项

2 个答案: