how can i find same values in an array?
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I have an array a, where
a=[303.1093
94.5581
86.6591
108.1389
303.1093
107.4338
138.6919
106.8294
108.1389
149.3491
187.8940
125.0922
96.7580
303.1093
75.9192
96.2085
118.8788
303.1093
303.1093
303.1093]
then i wanna do
b=round(a);
In b I want to find the positions who are having same values and I want to mark those value as 1 and others 0.
please help me.
Thank you
-- Suchi
2 个评论
Meghana Dinesh
2016-2-5
编辑：Meghana Dinesh
2016-2-5
How much "similar"? Or do you mean "same"?
suchismita
2016-2-5
编辑：suchismita
2016-2-5
采纳的回答
Star Strider
2016-2-5
编辑：Star Strider
2016-2-5
The unique function will obviously provide the information you want, but ‘the positions who are having similar values’ is less clear. I believe the au (unique values in ‘a’) and ‘ic’ values (the indices in ‘a’ that correspond to them) is what you want:
[au,ia,ic] = unique(round(a),'stable')
EDIT —
‘I wanna mark those same values as 1 and others 0’
I believe this does what you want:
[au,ia] = unique(round(a),'stable');
Same = ones(size(a));
Same(ia) = 0;
Result = [a Same]
Result =
303.11 0
94.558 0
86.659 0
108.14 0
303.11 1
107.43 0
138.69 0
106.83 1
108.14 1
149.35 0
187.89 0
125.09 0
96.758 0
303.11 1
75.919 0
96.209 0
118.88 0
303.11 1
303.11 1
303.11 1
-
4 个评论
suchismita
2016-2-5
It seems that this solution is not correct, as it misses 107.43 and 108.14 (rows 6 and 4). Surely these should be marked with 1's as (rounded) they match 106.83 and 108.14 respectively (rows 8 and 9, which may or may not be marked with 1's, depending on which results you read).
See my Answer for a solution that correctly identifies all matching values.
Star Strider
2016-2-5
I wasn’t certain what you wanted. I thought you only wanted to flag the repeated values, other than the first occurrence. Since this now seems to be resolved, I’ll not pursue it further.
Stephen23
2016-2-5
Good point about the "first occurrence". This did not occur to me, and the original question is ambiguous on this point.
更多回答（3 个）
Stephen23
2016-2-5
This identifies all matching rounded values, unlike the accepted answer:
b = round(a(:));
c = bsxfun(@eq,b,b.');
x = any(tril(c,-1)|triu(c,1),1);
[a(:),+x(:)]
303.1093 1
94.5581 0
86.6591 0
108.1389 1
303.1093 1
107.4338 1
138.6919 0
106.8294 1
108.1389 1
149.3491 0
187.8940 0
125.0922 0
96.7580 0
303.1093 1
75.9192 0
96.2085 0
118.8788 0
303.1093 1
303.1093 1
303.1093 1
Bhavini Sarkar
2016-9-17
2 个投票
To find the duplicate elements of an array, first you should sort array . After sorting all duplicate elements will aggregate in adjacent positions. Now using a for loop , traverse sorted array and compare adjacent elements. If adjacent elements are equal then you found one duplicate element of array . We can also find duplicate elements of an array in linear time using some extra space.
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