1

我有一个NSMutableArray包含 70 个对象的对象，它们是NSMutableDictionary，在每个字典中我有 2 个值：“距离”和“索引路径”。我想NSMutableArray从小到大排序。距离以 KM 为单位，使用我在此处粘贴的代码后数组如下所示：

NSSortDescriptor *descriptor = [NSSortDescriptor sortDescriptorWithKey:@"distance" ascending:YES];
[branchDistance sortedArrayUsingDescriptors:[NSArray arrayWithObjects:descriptor,nil]];

这是此方法的结果：

(lldb) po [branchDistance objectAtIndex:0]
(id) $1 = 0x0d6ac240 {
    "<NSIndexPath 0xd6cf180> 1 indexes [0]" = indexPath;
    "8.078547" = distance;
}
(lldb) po [branchDistance objectAtIndex:1]
(id) $2 = 0x0d6a64a0 {
    "<NSIndexPath 0xd6cf3b0> 1 indexes [1]" = indexPath;
    "45.044069" = distance;
}
(lldb) po [bran(lldb) po [branchDistance objectAtIndex:2]
(id) $4 = 0x0d6cf550 {
    "<NSIndexPath 0xd69b3f0> 1 indexes [2]" = indexPath;
    "9.992081" = distance;
}
(lldb) po [branchDistance objectAtIndex:3]
(id) $5 = 0x0d6cf750 {
    "283.356727" = distance;
    "<NSIndexPath 0xd6cf480> 1 indexes [3]" = indexPath;
}

我想NSMutableArray使用NSMutableDictionary. 我尝试了一些来自网络和 stackoverflow 的代码，但找不到适合我的东西。

请帮我！

谢谢！

4

1 回答 1

5

我假设距离值使用NSNumber对象存储在字典中：

NSArray *sortedArray = [branchDistance sortedArrayUsingComparator:^(id obj1, id obj2) {
    NSNumber *distance1 = [(NSDictionary *)obj1 objectForKey:@"distance"];
    NSNumber *distance2 = [(NSDictionary *)obj2 objectForKey:@"distance"];
    return [distance1 compare:distance2];
}];
于 2012-11-08T10:36:53.703 回答