How to remove duplicity from array of dictionaries in IOS? [closed] - iphone

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(
{
color = blue;
},
{
color = blue;
},
{
color = red;
},
{
color = white;
}
)
This is an Array of dictionary, i have to remove duplicate dictionary from array corresponding to key color.

Answer 1

NSSet is to save you in this case. Use:
NSSet *set = [NSSet setWithArray:duplicateArray];
NSArray *uniqueArray = [set allObjects];
Avoid using loop for this because if you have more object loop is a consuming process. You can directly use NSSet and it will work for sure.

Answer 2

Working Code :
NSArray *html = #[#{#"color": #("blue")},#{#"color": #("blue")},#{#"color": #("red")},#{#"color": #("yellow")}];
NSMutableArray *finalArray = [NSMutableArray array];
NSMutableSet *mainSet = [NSMutableSet set];
for (NSDictionary *item in html) {
//Extract the part of the dictionary that you want to be unique:
NSDictionary *dict = [item dictionaryWithValuesForKeys:#[#"color"]];
if ([mainSet containsObject:dict]) {
continue;
}
[mainSet addObject:dict];
[finalArray addObject:item];
}
NSLog(#"%#", finalArray);

Answer 3

An alternative to Vin's solution which I believe would work. But this one does not create a resulting array. It manipulates the existing one. For doing so it creates temporary copies to drive the iterations.
NSArray workingCopy = [NSArray arrayWithArray:yourArray];
for (int i = 0; i < [workingCopy count] - 1; i++) { // count - 1 just saves time. Works nicely without.
for (int j = i+1; j < [workingCopy count]; j++) {
if ([[[workingCopy objectAtIndex:i] objectForKey:#"color"] isEqualToString: [[workingCopy objectAtIndex:j] objectForKey:#"color"]] {
[yourArray removeOjbect:[[workingCopy objectAtIndex:i] objectForKey:#"color"]] // yourArray must be mutable for this.
}
}
}
This algo creates a copy of the original array before. That is to avoid hasseling with changes to the very array that is used for iterations/enumerations. Then it iterates though the copy in a 2-dimensional loop by avoiding to compare the same object with itself (i is never qual to j) and it avoids compaing A with B when B was already compared with A. Both is achieved by stating the j loop with i+1.
The very last iteration would be i = [workingCopy count]. Then j would start off with i+1 and therefore already be larger than [workingCopy count]. The loop's body would not be executed a single time. That's why the i loop can already finish with [workingCopy count] - 1.
The same can be achieved without a copy of the original array. But that does require rather smart manipulations of the running idices i and j, which is no good programming style, rather complex and error prone.

Answer 4

if arr is the array from which you want to remove duplicates
for(int index = 0;index<arr.count;index++){
NSDictionary *dict = [arr objectAtIndex:index];
for(int i = index-1 ; i>=0 ;i++){
NSDictionary *dictToComp = [arr objectAtIndex:i];
if([[dict objectForKey:#"color"] isEqualToString:[dictToComp objectForKey:#"color"]]){
[arr removeObject:dict];
}
}
}