What the best way to position an element say 2 columns in. Is there a short hand syntax like
span (6 at 2)
Which i've tried but doesn't work.
Or is the correct way to apply a left margin. Also can you get the span count to start from the right?
Thanks
span (6 at 2) works if you are using output: isolate — because isolation calculates position from the left edge for all elements. With the standard float output, you have to push or pull elements where you want them. Those push and pull the element from their default place in the flow (using margins).
#include span(6);
#inlude push(2);
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I am learning how to use formulas in spreadsheets, I do use libre office.
I need to sort out data in a quite huge messy spreadsheet.
Each column contains mixed data, the sheet is huge, dozens of columns and thousands of rows, if the spreadsheet does not contain errors each cell in a row either contains a different keyword or is empty, there should not be two cells in the same row containing the same keyword.
The problem to solve is to sort out all the data so to reach to have a new spreadsheet in which each cell marked with a given specific keyword is kept in the same position but placed in one column dedicated to that same keyword.
the kind of spreadsheet with mixed up cells to be sorted out
the data in the spreadsheet has to be fixed so to appear in this way
A formula that can be used to extract sorted out data from a cell is the following:
=IF(SEARCH("Text1";B2;1);B2;0)
The formula can be dragged to each cell below to hit the proper cell next to it. The result is correct.
The results are correct, but I do not know why the expected 0 is not printed, there is #VALUE! instead
The logic is very simple, if the cell contains the keyword or any other text that contains that keyword the result is the full content of that cell, otherwise the result is 0.
Here comes the first question, why do I get #VALUE! as a result for those cells that do not contain the keyword? I expected to get 0 instead, just as indicated in the formula,
I tried to leave this filed empty and also to put the 0 result in quotes, the actual result is always the same, #VALUE!...
However, of course this formula extracts only the information contained in one column, so for each other column the process must be repeated.
In order to avoid to create a column with the formula for each column in the spreadsheet or anyway to process each column one by one and more importantly to have then to merge all the results to form one columns containing only cells with a given keyword I thought to use the same formula extending the parsing to each next cell in the row as follows:
=IF(SEARCH("text";B2;1);B2;IF(SEARCH("text";C2;1);C2;IF(SEARCH("text";D2;1);D2;0)))
The logic is very simple and should output in one go a column containing all the cells containing the keyword that are found in the row, check if the first cell in the row contains a word using the search function, if does then the result is the content of that cell, otherwise perform the next test, the next test is the same, check if the next cell contains a certain word using the search function, if does then the result is the content of that cell, otherwise proceed to the next test…. and so on until last test, if no test gave a true result then print 0 (but we get #VALUE!, OK I could live with that...).
In theory should work for a any number of cells, but in the practice does not at all, in fact does work only for the first IF test and cell indicated in the formula.
WHY?
The result using the extended version of the formula to parse N cells in sequence is the same obtained with the simple formula to parse only one cell
Finally, how do I resolve this problem using IF and Search?
Is there any other better approach and way to solve this kind of problems and sort out data in huge spreadsheets of this kind?
Thank you for any hint and help.
I have a list that should display 7 items that each look like this:
Date Weekday Distance Time
Long text that may span many lines
two column text Distance Time
two column text Distance Time
two column text Distance Time
The last lines repeat in a number depending on the data, i e there may be different amounts of such lines for each list item.
I have tried implementing this with a ListCellRenderer that creates a table according to the requirements above, but I have a few problems with it:
The long text that may span many lines is implemented in a SpanLabel. But this text will not display more than one line anyway
Each item in the list will get space for the same number of lines below the first two..
So it seems that items in a list must be of the same size.
Later I also want to be able to detect selection on the entire list item, not just individual fields of it.
Is there a better way to do this?
How do I ensure that the SpanLabel actually gets as much space as it needs?
How do I ensure that the unknown number of lines gets the space they need, depending on how many they are?
Don't use a list: https://www.codenameone.com/blog/deeper-in-the-renderer.html
Lists in Codename One assume every entry is exactly the same height and provide no flexibility here.
I suggest doing something like the property cross demo: https://www.udemy.com/learn-mobile-programming-by-example-with-codename-one/
Where we use a Container with components within to provide a list like behavior with the full flexibility that arbitrary components allow.
I’m not specialist in signal processing. I’m doing simple processing on 1D signal using c++. I want really to know how I can determine the part that have the highest zero cross rate (highest frequency!). Is there a simple way or method to tell the beginning and the end of this part.
This image illustrate the form of my signal, and this image is what I need to do (two indexes of beginning and end)
Edited:
Actually I have no prior idea about the width of the beginning and the end, it's so variable.
I could calculate the number of zero crossing, but I have no idea how to define it's range
double calculateZC(vector<double> signals){
int ZC_counter=0;
int size=signals.size();
for (int i=0; i<size-1; i++){
if((signals[i]>=0 && signals[i+1]<0) || (signals[i]<0 && signals[i+1]>=0)){
ZC_counter++;
}
}
return ZC_counter;
}
Here is a fairly simple strategy which might give you some point to start. The outline of the algorithm is as follows
Input: Vector of your data points {y0,y1,...}
Parameters:
Window size sigma.
A threshold 0<p<1 defining when to start looking for a region.
Output: The start- and endpoint {t0,t1} of the region with the most zero-crossings
I won't give any C++ code, but the method should be easy to implement. As example let us use the following function
What we desire is the region between about 480 and 600 where the zero density higher than in the front. First step in the algorithm is to calculate the positions of zeros. You can do this by what you already have but instead of counting, you store the values for i where you met a zero.
This will give you a list of zero positions
From this list (you can do this directly in the above for-loop!) you create a list having the same size as your input data which looks like {0,0,0,...,1,0,..,1,0,..}. Every zero-crossing position in your input data is marked with a 1.
The next step is to smooth this list with a smoothing filter of size sigma. Here, you can use what you like; in the simplest case a moving average or a Gaussian filter. The higher you choose sigma the bigger becomes your look around window which measures how many zero-crossings are around a certain point. Let me give the output of this filter together with the original zero positions. Note that I used a Gaussian filter of size 10 here
In a next step, you go through the filtered data find the maximum value. In this case it is about 0.15. Now you choose your second parameter which is some percentage of this maximum. Lets say p=0.6.
The final step is to go through the filtered data and when the value is greater than p you start to remember a new region. As soon as the value drops below p, you end this region and remember start and endpoint. Once you are finished walking through the data, you are left with a list of regions, each defined by a start and an endpoint. Now you choose the region with the biggest extend and you are done.
(Optionally, you could add the filter size to each end of the final region)
For the above example, I get 11 regions as follows
{{164,173},{196,205},{220,230},{241,252},{259,271},{278,290},
{297,309},{318,327},{341,350},{458,468},{476,590}}
where the one with the biggest extend is the last one {476,590}. The final result looks (with 1/2 filter region padding)
Conclusion
Please don't be discouraged by the length of my answer. I tried to explain everything in detail. The implementation is really just some loops:
one loop to create the zero-crossings list {0,0,..,1,0,...}
one nested loop for the moving average filter (or you use some library Gaussian filter). Here you can at the same time extract the maximum value
one loop to extract all regions
one loop to extract the largest region if you haven't already extracted it in the above step
I am writing a code to do some template matching using cv::matchTemplate but I have run into some problems with the 2-dimensional vector of vectors (vov) I created which I have called vvABC. At the moment, my vov has 10 elements which can change based on the values I pass while running the code.
My problem is moving from one column in my vov to the next so I can calculate the size. From my understanding of how vov works, if I have my elements stored in my vov as:
C_A C_B
0 0
1 1
2 2
3
4
5
6
To calculate the size of the first column, I should simply do something like:
vvABC[0].size() to get the size of the first column (which would give 3 in this case) and vvABC[1].size() to get the size of the second column (which would give 7). The problem I am now faced with is both of them give '3' in both cases which is obviously wrong.
Can someone please help me out on how I can get the correct size of the next column?
I stored my detections in my vvABC, now I want to match them one at a time.
It seems like you made a mistake here:
for (uint iCaTemplate = iCa + 1; iCaTemplate < vvABC[iCa].size(); ++iCaTemplate) {
iCa is an index on the 'first level' of vector (of size 2 in your example above), i.e. columns, and you use it to go through the elements of the 'second level' of vector, i.e. rows.
Thanks a lot guys, esp. JGab, after several debug outputs, I finally found that my vector of vectors wasn't being filled up the way I thought it was...thanks once more and my apologies for my belated response.
In OpenOffice.org Calc, I would like to apply a formula to a column that references a cell from the same row but in a different column. I.e., =C1*48 in cell D1, but I want all cells in column D to do this without having to copy the formula to each one manually. Or another way of saying it, for each cell in C that has a number, I want to fill in the corresponding row-cell in D with a formula value based on that number. So far, all I can see from the documentation is that I'd have to copy/adjust the formula for every new row in the spreadsheet. And since I have 4 such columns that need calculation, this will get to be tiresome. I have little experience with spreadsheets at all, so I'm hoping that my problem is actually very simple, and that I just am looking at the wrong parts of the documentation.
I don't have OpenOffice in front of me, but it tries really hard to be Excel like in many ways, so usually assumptions about Calc based on Excel are fairly close to reality.
In Excel, a formula in cell D1, that points to a cell in C1, is treated as a relative reference - that is, one column back from where I am now. So when that formula is filled into other cells (either by Fill Down, or dragging the little handle in the corner of the cell outline, or by copy-pasting the formula into a range of selected cells) the new formulas are similarly treated, by default, as referring to the cell that is one column back from them, in the same row.
To force a formula to use an absolute reference, one specifies the cell address with a dollar sign - $C$1 will always point to (use) the contents of cell C1, regardless of where the formula ends up. ($C1 and C$1 are alternates that allow one parameter of the address to change in the usual relative sense while fixing the other half in place... this probably isn't important to you yet).
In other words, I'd expect that you can type the formulas in the first row of your OpenOffice Calc sheet and copy them down to the rest of the row, and things will just work.
If you want to extend your range down you can do this by calculating the first 2 cells with your formula and then highlighting them. You now grab the little square on the bottom right of the highlighted area and drag that down, across or both.
If you have a specific cell (e.g. D2) which you wish to have remain in all the cells you extend your range to, then in your 2 initial cell calculations use the following:
Extending Down
=G2*D$2
=G3*D$2
Extending Across
=G2*$D2
=H2*$D2
Extending Down and Across
Use $D$2
From your question it seems that you are trying to apply the same formula on whole row (or column) of cells and show the result on another row (or column).
The best way to do this is to use something called array formulas or array functions (in libre office). This link has very good explanation - https://help.libreoffice.org/Calc/Array_Functions
The way you would implement this in your case is
Type this formula in cell D1
=C1:C30 *48
and press ctrl + shift + enter
The formula now looks like this with the flower braces {..}
={C1:C30 *48}
and the formula gets applied all the way from D1 to D30.
You can even extrapolate the same concept to a whole matrix (for example: A1:C10)
(The way to tell excel/open office/ libre office that you wrote an array formula is to press ctrl + shift + enter. Never press enter as that will break the array function and convert it to a regular function)