Hi All,
I have a dataset something similar to following
Name Location Nickname saleCost
rest1 east Reast1 [5, 7, 9 , 2, 5, 1, 9, 11, 1]
rest4 south Rsouth4 [12, 9, 2, 3, 5, 10, 11, 4, 4]
rest3 east Reast3 [20, 2, 2, 3, 20, 30, 1, 3, 7]
rest9 west Rwest9 [1, 9, 3, 33, 22, 10, 6, 6, 8]
Note that saleCost is a dataset as well and as number of columns which are multiple of 3.
I want to find max for every 3 columns so my resultant dataset is as follow
Name Location Nickname saleCostMax
rest1 east Reast1 [9, 9, 9, 5, 9, 11, 11, 11, 1]
rest4 south Rsouth4 [12, 9, 5, 10, 11, 11, 11, 4, 4]
rest3 east Reast3 [20, 3, 20, 30, 30, 30, 7, 7, 7]
rest9 west Rwest9 [9, 33, 33, 33, 22, 10, 8, 8, 8]
ANy other way so that I do not have to use for loop? WHat are my options?
Thanks

2 commentaires

Geoff Hayes
Geoff Hayes le 4 Nov 2015
Neesha - is your data set a cell array or something else? Why don't you want to use a for loop?
Neesha
Neesha le 4 Nov 2015
isn't for loop time consuming and not the best approach? I will have 100s of lines

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 Réponse acceptée

Image Analyst
Image Analyst le 4 Nov 2015

0 votes

If you have the Image Processing Toolbox, you can go down row by row and call imdilate() to get a moving max:
thisRow = [5, 7, 9 , 2, 5, 1, 9, 11, 1]
newRow = imdilate(thisRow, [1,1,1])
% Crop off the first element and tack on the last element
newRow = [newRow(2:end), thisRow(end)]
In the command window:
thisRow =
5 7 9 2 5 1 9 11 1
newRow =
7 9 9 9 5 9 11 11 11
newRow =
9 9 9 5 9 11 11 11 1

4 commentaires

Neesha
Neesha le 4 Nov 2015
seems it worked in this example for me. :) if i have to take max of every 6 instead of 3, do i have keep changing newRow? How?
Image Analyst
Image Analyst le 5 Nov 2015
The kernel in imdilate() defines what elements in a sliding window to look at. If you wanted to look at 6 use [1,1,1,1,1,1]. Use a zero if you want to ignore an element.
ok makes sense. I tried and this is what i get. It is not correct output
thisRow = [5, 7, 9 , 2, 5, 1, 9, 11, 1]
newRow = imdilate(thisRow, [1,1,1,1,1,1])
% Crop off the first element and tack on the last element
newRow = [newRow(2:end), thisRow(end)]
thisRow =
5 7 9 2 5 1 9 11 1
newRow =
9 9 9 9 9 11 11 11 11
newRow =
9 9 9 9 11 11 11 11 1
Image Analyst
Image Analyst le 5 Nov 2015
What would you want? A sliding window of 6 only fits completely in there in 4 spots. Do you want 4 output numbers, or do you want the window to slide off on both ends?

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