Anyone know how to get Engineering Notation in a cell in a datagridview similar to how excel displays with a ##0.0E+0 custom format?
myDGV.Columns(0).DefaultCellStyle.Format = "E3"
I got to here, but don't find anything to work: https://learn.microsoft.com/en-us/dotnet/standard/base-types/custom-numeric-format-strings
I've tried every combination of E, 3, and ### I can think of.
Engineering notation examples:
0.1 => 100e-3 or 100m (not 1e-1)
1,000 => 1e3 or 1k
10,000 => 10e3 or 10k (not 1e4)
i.e. the powers of 10 always follow multiples of three (https://en.wikipedia.org/wiki/Engineering_notation)
Related
This is more of an open discussion topic than anything else. Currently I'm storing 50 Float32 values in my NSMutableArray *voltageArray before I refresh my CPTPlot *plot. Every time I obtain 50 values, I remove the previous 50 from the voltageArray and repeat the process....always displaying the 50 values in "real time" on my plot.
However, the data I'm receiving (which is voltage coming from a Cypress BLE module equipped with a pressure transducer) is so quick that any variation (0.4 V to 4.0 V; no pressure to lots of pressure) cannot be seen on my graph. It just shows up as a straight line, varying up and down without showing increased or decreased slopes.
To show overall change, I wanted to take those 50 values, store them in the first index of another NSMutableArray *stampArray and use the index of stampArray to display information. Meanwhile, the numberOfRecordsForPlot: method would look like this:
- (NSUInteger)numberOfRecordsForPlot:(CPTPlot *)plotnumberOfRecords {
return (DATA_PER_STAMP * _stampCount);
}
This would initially be 50, then after 50 pieces of data are captured from the BLE module, _stampCount would increase by one, and the number of records for plot would increase by 50 (till about 2500-10000 range, then I'd refresh the whole the thing and restart the process.)
Is this the right approach? How would I be able to make the first 50 points stay on the graph, while building the next 50, etc.? Imagine an y = x^2 graph, and what the graph looks like when applying integration (the whole breaking the area under the curve into rectangles).
Look at the "Real Time Plot" demo in the Plot Gallery example app included with Core Plot. It starts off with an empty plot, adding a new point each cycle until reaching the maximum number of points. After that, one old point is removed for each new one added so the total number stays constant. The demo uses a timer to pass random data to the plot, but your app can of course collect data from anywhere. Be sure to always interact with the graph from the main thread.
I doubt you'll be able to display 10,000 data points on one plot (does your display have enough pixels to resolve that many points?). If not, you'll get much better drawing performance if you filter and/or smooth the data to remove some of the points before sending them to the plot.
I am currently working on a project which has some requirements for VBA. The data is found in excel. What I need to ask/bounce ideas off of is for a way to write some code that will abide to the following conditions:
where Xmw and Ymw are in megawatt, and X and Y are generation plants
Xmw=<1000 -always true
and
Ymw=<1000 -always true
2000=Xmw+Ymw -Equation 1
and
10=X+Y -Equation 2
Essentially, since the maximum absolute value of generation to increase and decrease is 2000, and the maximum amount of plants that can be used is 10. I'm stuck at this point because I can't find the relation between the 2 equations. Additionally, the existing program identifies generation to use, but it doesn't follow it to the 2 provided equations.
Existing programming identifies which generation plants are in the "pools" of X and Y.
Any help would be greatly appreciated.
I am posting this in an answer because it would be too much to post in a comment (please comment and let me know if this hits the mark)
You could use a conditional that looks something like this:
power_generated=Ymw+Xmw
if power_generated<>2000 then
'do stuff here to add another power plant to generation
'
'
if X+Y>10 then
'deal with the fact that you have not enough power plants to deal with your draw
end if
end if
I was trying to solve Google Code Jam problems and there is one of them that I don't understand. Here is the question (World Finals 2013 - problem C): https://code.google.com/codejam/contest/2437491/dashboard#s=p2&a=2
And here follows the problem analysis: https://code.google.com/codejam/contest/2437491/dashboard#s=a&a=2
I don't understand why we can use binary search. In order to use binary search the elements have to be sorted. In order words: for a given element e, we can't have any element less than e at its right side. But that is not the case in this problem. Let me give you an example:
Suppose we do what the analysis tells us to do: we start with a left bound angle of 90° and a right bound angle of 0°. Our first search will be at angle of 45°. Suppose we find that, for this angle, X < N. In this case, the analysis tells us to make our left bound 45°. At this point, we can have discarded a viable solution (at, let's say, 75°) and at the same time there can be no more solutions between 0° and 45°, leading us to say that there's no solution (wrongly).
I don't think Google's solution is wrong =P. But I can't figure out why we can use a binary search in this case. Anyone knows?
I don't understand why we can use binary search. In order to use
binary search the elements have to be sorted. In order words: for a
given element e, we can't have any element less than e at its right
side. But that is not the case in this problem.
A binary search works in this case because:
the values vary by at most 1
we only need to find one solution, not all of them
the first and last value straddle the desired value (X .. N .. 2N-X)
I don't quite follow your counter-example, but here's an example of a binary search on a sequence with the above constraints. Looking for 3:
1 2 1 1 2 3 2 3 4 5 4 4 3 3 4 5 4 4
[ ]
[ ]
[ ]
[ ]
*
I have read the problem and in the meantime thought about the solution. When I read the solution I have seen that they have mostly done the same as I would have, however, I did not thought about some minor optimizations they were using, as I was still digesting the task.
Solution:
Step1: They choose a median so that each of the line splits the set into half, therefore there will be two provinces having x mines, while the other two provinces will have N - x mines, respectively, because the two lines each split the set into half and
2 * x + 2 * (2 * N - x) = 2 * x + 4 * N - 2 * x = 4 * N.
If x = N, then we were lucky and accidentally found a solution.
Step2: They are taking advantage of the "fact" that no three lines are collinear. I believe they are wrong, as the task did not tell us this is the case and they have taken advantage of this "fact", because they assumed that the task is solvable, however, in the task they were clearly asking us to tell them if the task is impossible with the current input. I believe this part is smelly. However, the task is not necessarily solvable, not to mention the fact that there might be a solution even for the case when three mines are collinear.
Thus, somewhere in between X had to be exactly equal to N!
Not true either, as they have stated in the task that
You should output IMPOSSIBLE instead if there is no good placement of
borders.
Step 3: They are still using the "fact" described as un-true in the previous step.
So let us close the book and think ourselves. Their solution is not bad, but they assume something which is not necessarily true. I believe them that all their inputs contained mines corresponding to their assumption, but this is not necessarily the case, as the task did not clearly state this and I can easily create a solvable input having three collinear mines.
Their idea for median choice is correct, so we must follow this procedure, the problem gets more complicated if we do not do this step. Now, we could search for a solution by modifying the angle until we find a solution or reach the border of the period (this was my idea initially). However, we know which provinces have too much mines and which provinces do not have enough mines. Also, we know that the period is pi/2 or, in other terms 90 degrees, because if we move alpha by pi/2 into either positive (counter-clockwise) or negative (clockwise) direction, then we have the same problem, but each child gets a different province, which is irrelevant from our point of view, they will still be rivals, I guess, but this does not concern us.
Now, we try and see what happens if we rotate the lines by pi/4. We will see that some mines might have changed borders. We have either not reached a solution yet, or have gone too far and poor provinces became rich and rich provinces became poor. In either case we know in which half the solution should be, so we rotate back/forward by pi/8. Then, with the same logic, by pi/16, until we have found a solution or there is no solution.
Back to the question, we cannot arrive into the situation described by you, because if there was a valid solution at 75 degrees, then we would see that we have not rotated the lines enough by rotating only 45 degrees, because then based on the number of mines which have changed borders we would be able to determine the right angle-interval. Remember, that we have two rich provinces and two poor provinces. Each rich provinces have two poor bordering provinces and vice-versa. So, the poor provinces should gain mines and the rich provinces should lose mines. If, when rotating by 45 degrees we see that the poor provinces did not get enough mines, then we will choose to rotate more until we see they have gained enough mines. If they have gained too many mines, then we change direction.
I have a Fortran program which I need to modify, so I'm reading it and trying to understand. Can you please explain what the formatting string in the following statement means:
write(*,'(1p,(5x,3(1x,g20.10)))') x(jr,1:ncols)
http://www.fortran.com/F77_std/rjcnf0001-sh-13.html
breifly, you are writing three general (g) format floats per line. Each float has a total field width of 20 characters and 10 places to the right of the decimal. Large magnitude numbers are in exponential form.
The 1xs are simply added spaces (which could as well have been accomplished by increasing the field width ie, g21.10 since the numbers are right justified. The 5x puts an additional 5 spaces at the beginning of each line.
The somewhat tricky thing here is tha lead 1p which is a scale factor. It causes the mantissa of all exponential form numbers produced by the following g format to be multiplied by 10, and the exponent changed accordingly, ie instead of the default,
g17.10 -> b0.1234567890E+12
you get:
1p,g17.10 -> b1.2345678900E+11
b denotes a blank in the output. Be sure to allow room for a - in your field width count...
for completeness in the case of scale greater than one the number of decimal places is reduced (preserving the total precision) ie,
3p,g17.10 -> b123.45678900E+09 ! note only 8 digits after the decimal
that is 1p buys you a digit of precision over the default, but you don't get any more. Negative scales cost you precision, preserving the 10 digits:
-7p,g17.10 -> b0.0000000123E+19
I should add, the p scale factor edit descriptor does something completely different on input. Read the docs...
I'd like to add slightly to George's answer. Unfortunately this is a very nasty (IMO) part of Fortran. In general, bear in mind that a Fortran format specification is automatically repeated as long as there are values remaining in the input/output list, so it isn't necessary to provide formats for every value to be processed.
Scale factors
In the output, all floating point values following kP are multiplied by 10k. Fields containing exponents (E) have their exponent reduced by k, unless the exponent format is fixed by using EN (engineering) or ES (scientific) descriptors. Scaling does not apply to G editing, unless the value is such that E editing is applied. Thus, there is a difference between (1P,G20.10) and (1P,F20.10).
Grouping
A format like n() repeats the descriptors within parentheses n times before proceeding.
I'm attempting to gauge the percentage difference between two images.
Having done a lot of reading I seem to have a number of options but I'm not sure what the best method to follow for:
Ease of coding
Performance.
The methods I've seen are:
Non language specific - academic Image comparison - fast algorithm and Mac specific direct pixel access http://www.markj.net/iphone-uiimage-pixel-color/
Does anyone have any advice about what solutions make most sense for the above two cases and have code samples to show how to apply them?
I've had success calculating the difference between two images using the histogram technique mentioned here. redmoskito's answer in the SO question you linked to was actually my inspiration!
The following is an overview of the algorithm I used:
Convert the images to grayscale—compare one channel instead of three.
Divide each image into an n * n grid of "subimages". Then, for subimage pair:
Calculate their colour composition histograms.
Calculate the absolute difference between the two histograms.
The maximum difference found between two subimages is a measure of the two images' difference. Other metrics could also be used (e.g. the average difference betwen subimages).
As tskuzzy noted in his answer, if your ultimate goal is a binary "yes, these two images are (roughly) the same" or "no, they're not", you need some meaningful threshold value. You could produce such a value by passing images into the algorithm and tweaking the threshold based on its output and how similar you think the images are. A form of machine learning, I suppose.
I recently wrote a blog post on this very topic, albeit as part of a larger goal. I also created a simple iPhone app to demonstrate the algorithm. You can find the source on GitHub; perhaps it will help?
It is really difficult to suggest something when you don't tell us more about the images or the variations. Are they shapes? Are they the different objects and you want to know what class of objects? Are they the same object and you want to distinguish the object instance? Are they faces? Are they fingerprints? Are the objects in the same pose? Under the same illumination?
When you say performance, what exactly do you mean? How large are the images? All in all it really depends. With what you've said if it is only ease of coding and performance I would suggest to just find the absolute value of the difference of pixels. That is super easy to code and about as fast as it gets, but really unlikely to work for anything other than the most synthetic examples.
That being said I would like to point you to: DHOG, GLOH, SURF and SIFT.
You can use fairly basic subtraction technique that the lads above suggested. #carlosdc has hit the nail on the head with regard to the type of image this basic technique can be used for. I have attached an example so you can see the results for yourself.
The first shows a image from a simulation at some time t. A second image was subtracted away from the first which was taken some (simulation) time later t + dt. The subtracted image (in black and white for clarity) then shows how the simulation has changed in that time. This was done as described above and is very powerful and easy to code.
Hope this aids you in some way
This is some old nasty FORTRAN, but should give you the basic approach. It is not that difficult at all. Due to the fact that I am doing it on a two colour pallette you would do this operation for R, G and B. That is compute the intensities or values in each cell/pixal, store them in some array. Do the same for the other image, and subtract one array from the other, this will leave you with some coulorfull subtraction image. My advice would be to do as the lads suggest above, compute the magnitude of the sum of the R, G and B componants so you just get one value. Write that to array, do the same for the other image, then subtract. Then create a new range for either R, G or B and map the resulting subtracted array to this, the will enable a much clearer picture as a result.
* =============================================================
SUBROUTINE SUBTRACT(FNAME1,FNAME2,IOS)
* This routine writes a model to files
* =============================================================
* Common :
INCLUDE 'CONST.CMN'
INCLUDE 'IO.CMN'
INCLUDE 'SYNCH.CMN'
INCLUDE 'PGP.CMN'
* Input :
CHARACTER fname1*(sznam),fname2*(sznam)
* Output :
integer IOS
* Variables:
logical glue
character fullname*(szlin)
character dir*(szlin),ftype*(3)
integer i,j,nxy1,nxy2
real si1(2*maxc,2*maxc),si2(2*maxc,2*maxc)
* =================================================================
IOS = 1
nomap=.true.
ftype='map'
dir='./pictures'
! reading first image
if(.not.glue(dir,fname2,ftype,fullname))then
write(*,31) fullname
return
endif
OPEN(unit2,status='old',name=fullname,form='unformatted',err=10,iostat=ios)
read(unit2,err=11)nxy2
read(unit2,err=11)rad,dxy
do i=1,nxy2
do j=1,nxy2
read(unit2,err=11)si2(i,j)
enddo
enddo
CLOSE(unit2)
! reading second image
if(.not.glue(dir,fname1,ftype,fullname))then
write(*,31) fullname
return
endif
OPEN(unit2,status='old',name=fullname,form='unformatted',err=10,iostat=ios)
read(unit2,err=11)nxy1
read(unit2,err=11)rad,dxy
do i=1,nxy1
do j=1,nxy1
read(unit2,err=11)si1(i,j)
enddo
enddo
CLOSE(unit2)
! substracting images
if(nxy1.eq.nxy2)then
nxy=nxy1
do i=1,nxy1
do j=1,nxy1
si(i,j)=si2(i,j)-si1(i,j)
enddo
enddo
else
print *,'SUBSTRACT: Different sizes of image arrays'
IOS=0
return
endif
* normal finishing
IOS=0
nomap=.false.
return
* exceptional finishing
10 write (*,30) fullname
return
11 write (*,32) fullname
return
30 format('Cannot open file ',72A)
31 format('Improper filename ',72A)
32 format('Error reading from file ',72A)
end
! =============================================================
Hope this is of some use. All the best.
Out of the methods described in your first link, the histogram comparison method is by far the simplest to code and the fastest. However key point matching will provide far more accurate results since you want to know a precise number describing the difference between two images.
To implement the histogram method, I would do the following:
Compute the red, green, and blue histograms of each image
Add up the differences between each bucket
If the difference is above a certain threshold, then the percentage is 0%
Otherwise the colors found in the images are similar. So then do a pixel by pixel comparison and convert the difference into a percentage.
I don't know any precise algorithms for finding the key points of an image. However once you find them for each image you can do a pixel by pixel comparison for each of the key points.