I'm analysing a dataset with some data-mining tools.The response variable has ten levels and I'm trying to create a classifier.
Here comes the problem.When using nnet and bagging function,the result is not that good and the 5th level is even not in the prediction.
I want to use a confusion matrix to analyse the classifier.but as the 5th level is not shown in the prediction I can't get a well-formed matrix.So how can I get a well-formed matrix?i.e. I want a 10*10 matrix.
The confusion matrix:
library("mda")#This is where **confusion** comes from
> confusion(pre.bag$class,CLASS)#here confusion acts like table
true
predicted 1 2 3 4 6 7 8 9 10 5
1 338 9 6 0 5 12 10 1 15 46
2 9 549 1 59 18 0 3 0 0 6
3 18 1 44 0 0 0 2 0 0 4
4 0 1 0 21 0 0 0 0 0 0
6 2 13 0 1 299 2 9 0 0 0
7 5 2 1 0 10 231 6 0 1 0
8 0 0 0 0 0 5 76 0 0 0
9 5 1 0 0 0 0 0 62 0 0
10 7 3 1 0 0 2 1 6 181 16
attr(,"error")
[1] 0.1231743
attr(,"mismatch")
[1] 0.03386642
Try this:
pred <- factor(pre.bag$class, levels=levels(CLASS) )
confusion(pre.bag$class, CLASS)
(Tested with an fda-object.)
Related
I'm using a FOR loop to loop through an array to check a condition but it seems to duplicate the variables I want to print.
Here's my code:
/*Printing the matrix*/
for(row=0;row<15;row++)
{
for(col=0;col<8;col++)
{
for(a=0;a<sizeof(row_sel)/(sizeof(row_sel[0]));a++)
{
if(row==row_sel[a] && col==col_sel[a])
{
printf("|%d|",matrix[row][col]);
}
else
{
printf(" %d ",matrix[row][col]);
}
}
}
printf("\n");
}
What I'm trying to print is:
|0| 6 7 3 6 6 9 5
0 0 0 0 0 0 0 0
2 8 8 4 8 4 0 0
0 0 0 0 0 0 0 0
5 0 8 7 9 5 5 3
0 0 0 0 0 0 0 0
2 7 8 5 3 3 8 6
0 0 0 0 0 0 0 0
6 6 4 2 9 6 2 1
0 0 0 0 0 0 0 0
9 4 0 6 6 7 0 4
0 0 0 0 0 0 0 0
4 3 8 9 0 2 2 7
0 0 0 0 0 0 0 0
0 3 6 7 8 3 5 |2|
But instead I get this output, I don't know why the values are duplicated,the loop should've ended or skipped the statement:
|0| 0 0 1 1 1 2 2 2 7 7 7 7 7 7 3 3 3 6 6 6 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 6 6 6 3 3 3 2 2 2 0 0 0 4 4 4 2 2 2 9 9 9
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
5 5 5 4 4 4 6 6 6 3 3 3 5 5 5 8 8 8 6 6 6 5 5 5
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
5 5 5 7 7 7 1 1 1 9 9 9 1 1 1 2 2 2 0 0 0 4 4 4
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
9 9 9 7 7 7 3 3 3 6 6 6 8 8 8 0 0 0 3 3 3 8 8 8
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
5 5 5 2 2 2 6 6 6 5 5 5 2 2 2 7 7 7 5 5 5 5 5 5
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
5 5 5 7 7 7 8 8 8 3 3 3 2 2 2 1 1 1 1 1 1 6 6 6
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
9 9 9 6 6 6 9 9 9 7 7 7 2 2 2 6 6 6 2 2 2 5 |5| 5
Any tips to where I went wrong? Any help is quite appreciated.
Your problem:
You print each value of the matrix for every a, but you only want to print it once.
The solution:
You have to move the if-else statement out of the innermost loop:
for(row=0;row<15;row++)
{
for(col=0;col<8;col++)
{
bool foundMatch = false;
for(a=0;!foundMatch && a<sizeof(row_sel)/(sizeof(row_sel[0]));a++)
{
foundMatch = row==row_sel[a] && col==col_sel[a];
}
if(foundMatch)
{
printf("|%d|",matrix[row][col]);
}
else
{
printf(" %d ",matrix[row][col]);
}
}
printf("\n");
}
Why/How it works:
The inner loop will run until foundMatch is true, or a gets out of range.
foundMatch is set to true if there is any a for which row==row_sel[a] && col==col_sel[a] is true.
|%d| will print if foundMatch is true.
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Do you know if it is possible to get the following triangular matrix
[ N:-1:1; (N-1):-1:0; (N-2):-1:0 0; (N-3):-1:0 0 0; ....] without writing every line with horzcat and without using a loop?
thanks all
Fred
Is this what you want?
N = 8;
result = flipud(tril(toeplitz(1:N)));
This gives
result =
8 7 6 5 4 3 2 1
7 6 5 4 3 2 1 0
6 5 4 3 2 1 0 0
5 4 3 2 1 0 0 0
4 3 2 1 0 0 0 0
3 2 1 0 0 0 0 0
2 1 0 0 0 0 0 0
1 0 0 0 0 0 0 0
Maybe something like this:
N=10;
M=triu(gallery('circul',N)).'
M =
1 0 0 0 0 0 0 0 0 0
2 1 0 0 0 0 0 0 0 0
3 2 1 0 0 0 0 0 0 0
4 3 2 1 0 0 0 0 0 0
5 4 3 2 1 0 0 0 0 0
6 5 4 3 2 1 0 0 0 0
7 6 5 4 3 2 1 0 0 0
8 7 6 5 4 3 2 1 0 0
9 8 7 6 5 4 3 2 1 0
10 9 8 7 6 5 4 3 2 1
Or did you want this:
M=fliplr(triu(gallery('circul',N)))
M =
10 9 8 7 6 5 4 3 2 1
9 8 7 6 5 4 3 2 1 0
8 7 6 5 4 3 2 1 0 0
7 6 5 4 3 2 1 0 0 0
6 5 4 3 2 1 0 0 0 0
5 4 3 2 1 0 0 0 0 0
4 3 2 1 0 0 0 0 0 0
3 2 1 0 0 0 0 0 0 0
2 1 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
I couldn't really tell from your code sample which direction you wanted this to go.
The power of bsxfun compels you!
[[N:-1:1]' reshape(repmat([N-1:-1:1]',1,N).*bsxfun(#ge,[1:N-1]',1:N),N,[])]
Sample run -
>> N = 8;
>> [[N:-1:1]' reshape(repmat([N-1:-1:1]',1,N).*bsxfun(#ge,[1:N-1]',1:N),N,[])]
ans =
8 7 6 5 4 3 2 1
7 6 5 4 3 2 1 0
6 5 4 3 2 1 0 0
5 4 3 2 1 0 0 0
4 3 2 1 0 0 0 0
3 2 1 0 0 0 0 0
2 1 0 0 0 0 0 0
1 0 0 0 0 0 0 0
This is basically inspired by this another bsxfun-based solution to a very similar question - Replicate vector and shift each copy by 1 row down without for-loop. There you can see similar solutions and related benchmarks, as it seems performance is a concern here.
I'm learning a spread of programming languages in a class, and we're working on an APLX project at the moment. A restriction we have to work around is we cannot use If, For, While, etc. No loops or conditionals. I have to be able to take a plane of numbers, ranging 0-7, and replace each number 2 or greater into the depth of that number, and, ideally, change the 1's to 0's. For example:
0100230 => 0000560
I have no idea how I'm supposed to do the replacement with depth aspect, though the change from ones to zeros is quite simple. I'm able to produce the set of integers in a table and I understand how to replace specific values, but only with other specific values, not values that would have to be determined during the function. The depth should be the row depth, rather than the multi-dimensional depth.
For the record this is not the whole of the program, the program itself is a poker dealing and scoring program. This is a specific aspect of the scoring methodology that my professor recommended I use.
TOTALS„SCORE PHAND;TYPECOUNT;DEPTH;ISCOUNT;TEMPS;REPLACE
:If (½½PHAND) = 0
PHAND„DEAL PHAND
:EndIf
TYPECOUNT„CHARS°.¹PHAND
DEPTH„2Þ(½TYPECOUNT)
REPLACE „ 2 3 4 5 6 7
ISCOUNT „ +/ TYPECOUNT
ISCOUNT „ ³ISCOUNT
((1=,ISCOUNT)/,ISCOUNT)„0
©((2=,ISCOUNT)/,ISCOUNT)„1
©TEMPS „ ISCOUNT
Œ„ISCOUNT
Œ„PHAND
You may have missed the first lessons of your prof and it might help to look at at again to learn about vectors and how easy you can work with them - once you unlearned the ideas of other programming languages ;-)
Assume you have a vector A with numbers from 1 to 7:
A←⍳7
A
1 2 3 4 5 6 7
Now, if you wanted to search for values > 3, you'd do:
A>3
0 0 0 1 1 1 1
The result is a vector, too, and you can easily combine the two in lots of operations:
multiplication to only keep values > 0 and replace others with 0:
A×A>3
0 0 0 4 5 6 7
or add 500 to values >3
A+500×A>3
1 2 3 504 505 506 507
or, find the indices of values > 3:
(A>3)×⍳⍴A
0 0 0 4 5 6 7
Now, looking at your q again, the word 'depth' has a specific meaning in APL and I guess you meant something different. Do I understand correctly that you want to replace values > 2 with the ' indices' of these values?
Well, with what I've shown before, this is easy:
A←0 1 0 0 2 3 0
(A≥2)×⍳⍴A
0 0 0 0 5 6 0
edit: looking at multi-dimensional arrays:
let's look into this example:
A←(⍳5)∘.×⍳10
A
1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 12 14 16 18 20
3 6 9 12 15 18 21 24 27 30
4 8 12 16 20 24 28 32 36 40
5 10 15 20 25 30 35 40 45 50
Now, which numbers are > 20 and < 30?
z←(A>20)∧A<30
z
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 1 1 0
0 0 0 0 0 1 1 0 0 0
0 0 0 0 1 0 0 0 0 0
Then, you can multiply the values with that boolean result to filter out only the ones satisfying the condition:
A×z
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 21 24 27 0
0 0 0 0 0 24 28 0 0 0
0 0 0 0 25 0 0 0 0 0
Or, perhaps you're interested in the column-index of the values?
z×[2]⍳¯1↑⍴z
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 7 8 9 0
0 0 0 0 0 6 7 0 0 0
0 0 0 0 5 0 0 0 0 0
NB: this statement might not work in all APL-dialects. Here's another way to formulate this:
z×((1↑⍴z)⍴0)∘.+⍳¯1↑⍴z
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 7 8 9 0
0 0 0 0 0 6 7 0 0 0
0 0 0 0 5 0 0 0 0 0
I hope this gives you some ideas to play with. In general, using booleans to manipulate arrays in mathematical operations is an extremely powerful idea in APL which will take you loooooong ways ;-)
Also, if you'd like to see more of the same, have a look at the FinnAPL Idioms - some useful shorties grown over the years ;-)
edit re. "maintaining untouched values":
going back to example array A:
A←(⍳5)∘.×⍳10
A
1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 12 14 16 18 20
3 6 9 12 15 18 21 24 27 30
4 8 12 16 20 24 28 32 36 40
5 10 15 20 25 30 35 40 45 50
Replacing values between 20 and 30 with the power 2 of these values, keeping all others unchanged:
touch←(A>20)∧A<30
(touch×A*2)+A×~touch
1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 12 14 16 18 20
3 6 9 12 15 18 441 576 729 30
4 8 12 16 20 576 784 32 36 40
5 10 15 20 625 30 35 40 45 50
I hope you get the idea...
Or better: ask a new q, as otherwise this would truly take epic dimensions, whereas the idea of stackoverflow is more like "one issue - one question"...
I'm working on J.
How can I convert this matrix:
(i.10)*/(i.10)
0 0 0 0 0 0 0 0 0 0
0 1 2 3 4 5 6 7 8 9
0 2 4 6 8 10 12 14 16 18
0 3 6 9 12 15 18 21 24 27
0 4 8 12 16 20 24 28 32 36
0 5 10 15 20 25 30 35 40 45
0 6 12 18 24 30 36 42 48 54
0 7 14 21 28 35 42 49 56 63
0 8 16 24 32 40 48 56 64 72
0 9 18 27 36 45 54 63 72 81
in array?
0 0 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 9 . . .
I tried
(i.10)*/(i.10)"0
and then I've added
~.(i.10)*/(i.10)"0
to eliminate doubles, but it doesn't work.
If you want to turn a 2-dimensional table (matrix) into a 1-dimensional list (vector or "array", though in the J world "array" usually means "rectangle with any number [N] of dimensions"), you can use ravel (,):
matrix =: (i.10)*/(i.10)
list =: , matrix
list
0 0 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 ...
Now using nub (~.) to remove duplicates should work:
~. list
0 1 2 3 4 5 6 7 8 9 10 12 ...
Note that, in J, the shape of an array usually carries important information, so flattening a matrix like this would be fairly unusual. Still, nothing stopping you.
BTW, you can save yourself some keystrokes by using the adverb ~, which will copy the left argument of a dyad to the right side as well, so you could just say:
matrix =: */~ i. 10
and get the same result as (i.10) */ (i.10).
I am working on a minigame called 'Pogo Painter', and I need some mathematical solutions. Below is an image (made with Paint) to illustrate a bit what it's all about.
Four players, each of different color, must claim squares to gain points. The minigame will be similar to this: http://www.youtube.com/watch?v=rKCQfAlaRrc, but slightly different. The players will be allowed to run around the playground and claim any of the squares, and points are gathered when a pattern is closed. For example, claiming blue square on A3 will create a closed blue pattern.
What kind of variables should I declare and how do I check if the pattern is closed?
Please answer if you have a solution :)
Here’s another (Discrete Optimization) way to model your problem.
Notation
View your grid as a ‘graph’ with n^2 nodes, and edges of length 1 (Edges connect two neighboring nodes.) Let the nodes be numbered 1:n^2. (For ease of notation, you can use a double array (x,y) to denote each node if you prefer.)
Decision Variables
There are k colors, one for each player (1 through 4). 0 is an unclaimed cell (white)
X_ik = 1 if player k has claimed node i. 0 otherwise.
To start out
X_i0 = 1 for all nodes i.
All nodes start out as white (0).
Neighboring sets: Two nodes i and j are ‘neighbors’ if they are adjacent to each other. (Any given node i can have at most 4 neighbors: Up down right and left.)
Edge variables:
We can now define a new set of edge variables Y_ijk that connect two adjacent nodes (i and j) with a common color k.
Y_ijk = 1 if neighboring nodes i and j are both of color k. 0 Otherwise.
(That is, X_ik = X_jk) for non-zero k.
We now have an undirected graph. Checking for ‘closed patterns’ is the same as detecting cycles.
Detecting Cycles:
A simple DFS search will do, since we have undirected cycles. Start with each colored node i, and check for cycles. If a path leads you back to a visited node, cycles exist. You can award points accordingly.
Finally, one suggestion as you design the game. You can reward points according to the “longest cycle” you detect. The shortest cycle gets 4 points, one point for each edge (or one point for each node in the cycle) whichever works best for you.
1 1
1 1 scores 4 points
1 1 1
1 1 1 scores 6 points
1 1 1
1 1 1
1 1 scores 8 points
Hope that helps.
Okay,
This is plenty of text, but it's simple.
An N-by-N square will satisfy as the game-board.
Each time a player claims a square,
If the square is not attached to any square of that player, then you must give that square a unique ID.
If the square is attached,
Count how many neighbours of each ID it has.
( See the demos I put below, to see what this means)
For each group
patterns_count += group_size - 1
If the number of groups is more than 1
Change the ID of that group as well as every other square connected to it so they all share the same ID
You must remember which IDs belong to which players.
This is what you have in your example
1 1 1 0 0 0 0 2 2
1 0 0 0 1 3 3 0 0
1 1 0 0 3 3 0 0 0
0 1 0 0 4 5 0 0 0
0 0 0 6 4 0 0 0 0
7 7 0 0 0 0 8 8 8
0 7 7 0 9 8 8 0 8
A A 7 0 9 8 0 0 8
A 0 7 0 0 0 8 8 8
And this is what it would turn out like after blue grabs A-3
1 1 1 0 0 0 0 2 2
1 0 0 0 1 3 3 0 0
1 1 0 0 3 3 0 0 0
0 1 0 0 4 5 0 0 0
0 0 0 6 4 0 0 0 0
7 7 0 0 0 0 8 8 8
0 7 7 0 9 8 8 0 8
A A 7 0 9 8 0 0 8
A 0 7 0 0 8 8 8 8
More examples of the algorithm in use
1 1 1 0
1 0 1 0
1 1 0
0 0 0 0
2 neighbours. 2x'1'
1x closed pattern.
1 1 1 0
1 0 1 0
1 1 1 0
0 0 0 0
--
1 1 1 0 0
1 0 1 0 0
1 1 0 0
1 0 1 0 0
1 1 1 0 0
3 neighbours: 3x'1'
2x closed patterns
1 1 1 0 0
1 0 1 0 0
1 1 1 0 0
1 0 1 0 0
1 1 1 0 0
--
1 1 1 0 0
1 0 1 0 0
1 1 2 2
0 0 2 0 2
0 0 2 2 2
4 neighbours: 2x'1', 2x'2'
2 Closed patterns
1 1 1 0 0
1 0 1 0 0
1 1 1 1 1
0 0 1 0 1
0 0 1 1 1
But I also consider these a closed pattern. You haven't given any description as to what should be considered one and what shouldn't be.
1 1 0
1 1 0
0 0 0
1 1 1
1 1 1
0 0 0
1 1 1
1 1 1
1 1