I found this hard to put a title on so please feel free to modify it.
I have an array that contains 5 CGSizes that I use in cellForRowAtIndexPath.
NSArray *mElements = [NSArray arrayWithObjects:[NSValue valueWithCGSize:
CGSizeMake(600.0, 900.0)],
[NSValue valueWithCGSize:CGSizeMake(300.0, 800.0)],
[NSValue valueWithCGSize:CGSizeMake(546.0, 1032.0)],
[NSValue valueWithCGSize:CGSizeMake(300.0, 700.0)],
[NSValue valueWithCGSize:CGSizeMake(300.0, 800.0)],
nil];
I currently have it that a random size is taken from the array and applied to each cell, giving it a staggered layout. However the problem is that if I reload the UITableView all the rows will have different sizes, which is obvious because it will pick a random value each time.
int random = arc4random_uniform((uint32_t) mElements.count);
CGSize size =[[mElements objectAtIndex:random] CGSizeValue];
However the effect I want is that row 0 will always have the same size, possibly the index 0 of the array. Then row 1 to have the value at index 1, and then row 2 to have the value at index 2, and so on. I would then like the row at index 4 to have the value at index 0 of the array (so the array starts again). I have tried this, but it requires too much hardcoding.
if (indexPath.item%2 && !indexPath.item%4) {
size =[[mElements objectAtIndex:1] CGSizeValue];
}
else if(indexPath.item%3 && !indexPath.item%6){
size =[[mElements objectAtIndex:2] CGSizeValue];
}
else if(indexPath.item%4){
size =[[mElements objectAtIndex:2] CGSizeValue];
}
//... and so on
Im hoping to get help in creating a dynamic conditional state where I do not have to hard code every possible row index in the table. I can't seem to get my head around it though.
size = [[mElements objectAtIndex:indexPath.row % [mElements count]] CGSizeValue];
That'd give you a constant repeating pattern all the way down.
Number of rows could be
return [mElements count]*ROW_MULTIPLYER;
where
#define ROW_MULTIPLYER 10000
Related
I have a 2D array PointAndTangent of dimension 8500 x 5. The data is row-wise with 8500 data rows and 5 data values for each row. I need to extract the row index of an element in 4th column when this condition is met, for any s:
abs(PointAndTangent[:,3] - s) <= 0.005
I just need the row index of the first match for the above condition. I tried using the following:
index = np.all([[abs(s - PointAndTangent[:, 3])<= 0.005], [abs(s - PointAndTangent[:, 3]) <= 0.005]], axis=0)
i = int(np.where(np.squeeze(index))[0])
which doesn't work. I get the follwing error:
i = int(np.where(np.squeeze(index))[0])
TypeError: only size-1 arrays can be converted to Python scalars
I am not so proficient with NumPy in Python. Any suggestions would be great. I am trying to avoid using for loop as this is small part of a huge simulation that I am trying.
Thanks!
Possible Solution
I used the following
idx = (np.abs(PointAndTangent[:,3] - s)).argmin()
It seems to work. It returns the row index of the nearest value to s in the 4th column.
You were almost there. np.where is one of the most abused functions in numpy. Half the time, you really want np.nonzero, and the other half, you want to use the boolean mask directly. In your case, you want np.flatnonzero or np.argmax:
mask = abs(PointAndTangent[:,3] - s) <= 0.005
mask is a 1D array with ones where the condition is met, and zeros elsewhere. You can get the indices of all the ones with flatnonzero and select the first one:
index = np.flatnonzero(mask)[0]
Alternatively, you can select the first one directly with argmax:
index = np.argmax(mask)
The solutions behave differently in the case when there are no rows meeting your condition. Three former does indexing, so will raise an error. The latter will return zero, which can also be a real result.
Both can be written as a one-liner by replacing mask with the expression that was assigned to it.
I know mapping 2D array into 1D array has been asked many times, but I did not find a solution that would fit a where the column count varies.
So I want get a 1-dimensional index from this 2-dimensional array
Col> _0____1____2__
Row 0 |_0__|_1__|_2__|
V 1 |_3__|_4__|
2 |_5__|_6__|_7__|
3 |_8__|_9__|
4 |_10_|_11_|_12_|
5 |_13_|_14_|
The normal formula index = row * columns + column does not work, since after the 2nd row the index is out of place.
What is the correct formula here?
EDIT:
The specific issue is that I have a list of items in with the layout like in the grid, but a one dimensional array for the data. So while looping through the elements in the UI, I need to get the correct data, but can only get the row and column for that element. I need to find a way to turn a row/column value into an index for the data-array
Bad picture trying to explain it
A truly optimal answer (or even a provably correct one) will depend on the language you are using and how it lays out memory for such arrays.
However, taking your question simply at face value, you have to know what the actual length of each row is in order to calculate a 1D index.
So either the row length follows some pattern that can be inferred from the data, or you have (or can write) a rlen = rowLength( 2dTable, RowNumber) function.
Then, depending on how big the tables are and how fast you need to run, you can calculate a 1D index from the 2d table by adding all the previous row lengths until the current row length is less than the 2d column index.
or build a 1d table of the row lengths (or commulative rowlengths) so you can scan it and so only call your rowlength function for each row only once.
With a better description of your problem, you might get a better answer...
For your example which alternates between 3 and 2 columns you can construct a formula:
index = (row / 2) * (3 + 2) + (row % 2 ? 3 : 0) + column
(C-like syntax, assuming integer division)
In general though, the one and only way to implement what you're doing here, jagged arrays, is to make an array of arrays, a.k.a. an Iliffe vector. That means, use the row number as index into an array of pointers which point to the individual row arrays containing the actual data.
You can have an additional 1D array having the length of the columns say "length". Then your formula is index=sum {length(i)}+column. i runs from 0 to row.
There is an integer array, for eg.
{3,1,2,7,5,6}
One can move forward through the array either each element at a time or can jump a few elements based on the value at that index. For e.g., one can go from 3 to 1 or 3 to 7, then one can go from 1 to 2 or 1 to 2(no jumping possible here), then one can go 2 to 7 or 2 to 5, then one can go 7 to 5 only coz index of 7 is 3 and adding 7 to 3 = 10 and there is no tenth element.
I have to only count the number of possible paths to reach the end of the array from start index.
I could only do it recursively and naively which runs in exponential time.
Somebody plz help.
My recommendation: use dynamic programming.
If this key word is sufficient and you want the challenge to find a possible solution on your own, dont read any further!
Here a possible DP-algorithm on the example input {3,1,2,7,5,6}. It will be your job to adjust on the general problem.
create array sol length 6 with just zeros in it. the array will hold the number of ways.
sol[5] = 1;
for (i = 4; i>=0;i--) {
sol[i] = sol[i+1];
if (i+input[i] < 6 && input[i] != 1)
sol[i] += sol[i+input[i]];
}
return sol[0];
runtime O(n)
As for the directed graph solution hinted in the comments :
Each cell in the array represents a node. Make an directed edge from each node to the node accessable. Basically you can then count more easily the number of ways by just looking at the outdegrees on the nodes (since there is no directed cycle) however it is a lot of boiler plate to actual program it.
Adjusting the recursive solution
another solution would be to pruning. This is basically equivalent to the DP-algorithm. The exponentiel time comes from the fact, that you calculate values several times. Eg function is recfunc(index). The initial call recFunc(0) calls recFunc(1) and recFunc(3) and so on. However recFunc(3) is bound to be called somewhen again, which leads to a repeated recursive calculation. To prune this you add a Map to hold all already calculated values. If you make a call recFunc(x) you lookup in the map if x was already calculated. If yes, return the stored value. If not, calculate, store and return it. This way you get a O(n) too.
In the code below "G" returns a (10*4) matrix which is in the correct orientation.
All I want then is to be able to view/call these (10*4) matrices indexed by (j,k). However when I store the matrix "G" in the 4-D matrix "test" the data is displayed in a way that is a little counter intuitive? When I look at test in the variable editor I get:
val(:,:,1,1) =
1
val(:,:,2,1) =
0
val(:,:,3,1) =
0
val(:,:,4,1) =
0
.
.
.
val(:,:,1,10) =
1
val(:,:,2,10) =
0
val(:,:,3,10) =
0
val(:,:,4,10) =
0
So all the data is there but I want it displayed at a 10*4 matrix?
Also as you will see I had to change the code for "Correl_betas" and transpose "G" to get to where I am above. However I felt I did this by playing around rather than what I thought the code should be doing. Why does the original code not work? I am having to change the order of the 3rd and 4th dimensions when declaring "Correl_betas" and then pass the transpose of G, but this seems totally counter intuitive as the last two dimensions in the original "Correl_betas" and the original (un-transposed) "G" also match? But when I did it this way the ordering seemed even further from the 10*$ matrix I want.
So I have 2 questions?
1.) How can I get to the 2-dimentional (10*4) matrices I want indexed by j,k from where I am?
2.) How come the original code above doesn't result in the last two columns producing (10,4) matrices?
A large part of the problem is that I have very little experience working with matrices that have more than 2 dimensions so sorry if this question shows a lack of understanding. Maybe a pointer to a god tutorial on how to interpret manipulate higher dimensional matrices would help too.
%Correl_betas=zeros(50,50,10,4);
Correl_betas=zeros(50,50,4,10);
mats=[1:10]';
L1=-1;
for j=1:51
L1=L1+1;
L2=-1;
for k=1:51
L2=L2+1;
lambda=[ L1; L2 ];
nObs=size(mats,1);
G= [ones(nObs,1) (1-exp(-mats./lambda(1)))./(mats./lambda(1)) ((1-exp(-mats./lambda(1)))./(mats./lambda(1))-exp(-mats./lambda(1))) ((1-exp(-mats./lambda(2)))./(mats./lambda(2))-exp(-mats./lambda(2)))];
%Correl_betas(j,k,:,:)=G;
Correl_betas(j,k,:,:)=G';
test=Correl_betas(j,k,:,:);
temp1=corrcoef(Correl_betas(j,k,:,2),Correl_betas(j,k,:,3),'rows','complete');
temp2=corrcoef(Correl_betas(j,k,:,2),Correl_betas(j,k,:,4),'rows','complete');
temp3=corrcoef(Correl_betas(j,k,:,3),Correl_betas(j,k,:,4),'rows','complete');
F2_F3(j,k)=temp1(1,2);
F2_F4(j,k)=temp2(1,2);
F3_F4(j,k)=temp3(1,2);
end
end
To reshape the matrix as desired,
val2 = permute(val,[4 3 2 1]);
This brings the 4th dimension (size of 10) onto the first, and the the 3rd dimension (size of 4) onto the second.
In your loop, both j and k cycle through 1:51 so the first two dimensions of Correl_betas will end up being length 51 too.
In my program I need to work with arrays roughly 500x500 to 1500x1500 within a function that is looped over 1000's of times. In each iteration, I need to start with an array that has the same form (whose dimensions are fixed across all iterations). The initial values will be:
[0 0 0 ... 1]
[0 0 0 ... 1]
....
However, the contents of the array will be modified within the loop. What is the most efficient way to "reset" the array to this format so I can pass the same array to the function every time without having to allocate a new set of memory every time? (I know the range of rows that were modified)
I have tried:
a[first_row_modified:last_row_modified,:] = 0.
a[first_row_modified:last_row_modified,:-1] = 1.
but it takes roughly the same amount of time as just creating a new array every time with the following:
a = zeros((sizeArray, sizeArray))
a[:,-1] = 1.
Is there a faster way to effectively "erase" the array and change the last column to ones? I think this is similar to this question, clearing elements of numpy array , although my array doesn't change sizes and i didn't see the definitive answer to the previously asked question.
No; I think the way you are doing it is about as fast as it gets.