我有一个 字典数组,如下所示:
(
{ key = 1, value = 40},
{ key = 4, value = 50},
{ key = 8, value = 60}
}
这些是这样的,
for >=1 项目成本是 40,
for >=4 项目成本是 50 和类似的。
我想获得 5 的值,在本例中为 50。
我试过的一段代码是:
NSMutableArray *wallpaperPriceArray = [[NSMutableArray alloc]init]; // Assuming this has all the dictionary values
float wallpaperPriceValue = 0;
int itemNumber = 0;
for (int i = 0; i<wallpaperPriceArray.count; i++) {
int check = 0;
if(itemNumber >= [[wallpaperPriceArray objectAtIndex:i] intValue])
{
wallpaperPriceValue = [[[wallpaperPriceArray objectAtIndex:i] objectForKey"value"] floatValue];
check++;
}
if(i + 1 <= wallpaperPriceArray.count)
{
if(itemNumber >= [[wallpaperPriceArray objectAtIndex:i+1] intValue] && itemNumber < [[wallpaperPriceArray objectAtIndex:i+1] intValue])
{
wallpaperPriceValue = [[[wallpaperPriceArray objectAtIndex:i+1] objectForKey"value"] floatValue];
check++;
if(check == 2)
{
break ;
}
}
}
if(i + 2 <= wallpaperPriceArray.count)
{
if(itemNumber >= [[wallpaperPriceArray objectAtIndex:i+2] intValue] && itemNumber < [[wallpaperPriceArray objectAtIndex:i+2] intValue])
{
wallpaperPriceValue = [[[wallpaperPriceArray objectAtIndex:i+2] objectForKey"value"] floatValue];
check++;
if(check == 2)
{
break ;
}
}
}
Best Answer-推荐答案 strong>
我不认为谓词是正确的,最好是枚举对象这里是一些示例代码:
NSArray *array = @[
@{ @"key" : @(1), @"value" : @(40)},
@{ @"key" : @(4), @"value" : @(50)},
@{ @"key" : @(8), @"value" : @(60)}
];
NSInteger searchedValue = 5; // <---
__block NSDictionary *closestDict = nil;
__block NSInteger closestValue = 0;
[array enumerateObjectsUsingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
NSDictionary *dict = obj;
NSInteger key = [dict[@"key"] integerValue];
// Check if we got better result
if(closestDict == nil || (key > closestValue && key <= searchedValue)){
closestDict = dict;
closestValue = key;
if(key == searchedValue) { *stop = YES; }
}
}];
NSLog(@"value %@", closestDict);
关于ios NSPredicate 在 NSDictionary 中查找最接近的值,我们在Stack Overflow上找到一个类似的问题:
https://stackoverflow.com/questions/25197307/
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