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What an "AquaButton", e.g. A Simple AquaButton[^]? They are rounded rectangles, but changing them into circles should not be too complicated.
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Hello All,
My Problem
Currently I am working on a project(C#,WinForm), where my requirement is to collect GPS data in my application like (Latitude,Longitude,Altitude,Speed,Satellites).
For getting above GPS data, I have a Garmin USB Device. Now problem is, How I can capture the GPS Device data into my application.
What I have done
I installed Garmin USB Driver, Franson GPS Client Gate.
Question
What library I should use? Has anyone faced same problem? please share.
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You should get the documetation for the device from Garmin; they are most likely to know the answer.
Use the best guess
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Do you need the data "online"? If not, you could configure the device to log the data to a file on its memory card and then import that file.
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my sttring bulilder string would be like this:
StringBuilder sb = new StringBuilder();
sb.Append("hi friends hi");
i want to check the count of "hi" in the sb
so my final output would be like this.
Count is 2.
how to do that.
thank u
SUBIN
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Probably not very efficient:
Get the string by calling ToString(). Split by space to get an arry of words. You can then use Distinct to get unique words and/or Select to get a specific word
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Something like this?
StringBuilder sb = new StringBuilder();
sb.Append("hi friends hi");
Console.WriteLine("Count is " + "".ToString().Split(new []{" "}, StringSplitOptions.RemoveEmptyEntries).
Where(w => w.Equals("hi", StringComparison.CurrentCultureIgnoreCase)).Count();
Andreas Johansson
Senior software developer at Tieto Sweden
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Yeah that's how I would do it, but copying your homework from Google is frowned upon. I think the OP should work to understand what the problem is, a method for solving it, and then the code for that method. Skipping to the last part won't help you pass the class.
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Jasmine2501 wrote: Skipping to the last part won't help you pass the class.
That's why more and more companies are giving a coding-test when applying for a job. If the OP wants to cheat, he'll find that he has been cheated soon enough, owning a degree and not being able to use it
Bastard Programmer from Hell
If you can't read my code, try converting it here[^]
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These days, I'm almost insulted when I haven't been asked to do a coding test. And, I did have a job recently where the interview should have clued me in to the fact they didn't know what the hell was going on... worked there for two months - everybody was real smart and legit with their skills - but I only did one productive thing the whole time. I should have known something was hinky in the interview when they did zero actual assessment of my skills.
modified 18-Apr-13 17:41pm.
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Jasmine2501 wrote: but I only did one productive thing the whole time. That's when you start updating that resume again
Bastard Programmer from Hell
If you can't read my code, try converting it here[^]
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Oh heck yeah, I had another gig lined up before I quit from there. It's hard to stay unemployed if you have programming skills and you're being honest about it
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Something like:
sb.ToString().Split(' ').ToList().Distinct().Count()
That does (or should be close to) what your subject line wants to do.
But that's different from what you wrote:
kanamala subin wrote: i want to check the count of "hi" in the sb
so my final output would be like this.
Count is 2.
Which is "count the number of instances of a word", for which you could do something like this:
Dictionary<string, int> wordCount = new Dictionary<string,int>();
sb.ToString().Split(' ').ToList().ForEach(w=>
{
if (wordCount.ContainsKey(w) ++wordCountCount[w];
else wordCount[w]=1;
}
Marc
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Hello,
me and a coworker had thought about optimzing the fibonacci calculation.
This is what we have ended up and is pretty fast. (It is fast enough, for our product, but we want to know what is possible)
private static ulong GetFib3(int n)
{
ulong a = 0;
ulong b = 1;
if (n == 0) return a;
if (n == 1) return b;
for (int i = 2; i <= n ; i++)
{
a ^= b;
b ^= a;
a ^= b;
b += a;
}
return b;
}
The method has following restrictions: Only up to n=200; n=201 result is bigger than ulong and no negative fibonaccis.
This method is actually pretty fast (20x faster than our first approach) and (as I think) it is well optimized.
My main question is: Is it possible to improve the execution speed even more? (I dont mean precalculating the values and then just access the values)
Regards,
Max
modified 18-Apr-13 8:58am.
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Yes.
Precalculate the values. Since you are only working with 200 items, run your existing code to output the list of values into a text file, then add those values as an array:
private static ulong[] fibValues = { 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...}; Then your routine is just a limit test and an array access.
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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Ofcourse precalculating the values and then just accessing the array is fastest.
But I want to calculate it the fastest way...
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That wasn't what you asked!
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
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Thought it will be understood correctly. I edited the inital post.
But thanks for your efforts answering my question
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There is a formula:
Private Function Fibonacci(ByVal n As Integer) As Double
Return (1 / Math.Sqrt(5)) * ((1 + Math.Sqrt(5)) / 2) ^ n - (1 / Math.Sqrt(5)) * ((1 - Math.Sqrt(5)) / 2) ^ n
End Function
And it seems to do the trick, mening no tables
http://en.wikipedia.org/wiki/Fibonacci_number[^]
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Yes, this implementation works great, BUT it is actually slower on my machine than my code I initially posted.
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Whats the slowdown? if it is Sqrt that could be fixed, and if its the power function you could make that better too.
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This is faster on my computer:
Private Function Fibonacci2(ByVal n As Integer) As Double
Dim gold As Double = 1.6180339887498949
Dim gold2 As Double = 0.6180339887498949
Dim sqrt1over5 As Double = 0.44721359549995793
Return sqrt1over5 * ((gold) ^ n - (gold2) ^ n)
End Function
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My Results so far:
Link to screenshot
Last one is your Goldratio implementation (second fast), second from bottom is my last implementation (and fastest I found)...
I ran it one time (n=50), speed is always a bit different, but differences are accurate.
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