I have a matlab/octave for loop which gives me an inf error messages along with the incorrect data
I'm trying to get 240,120,60,30,15... every number is divided by two then that number is also divided by two
but the code below gives me the wrong value when the number hits 30 and 5 and a couple of others it doesn't divide by two.
ang=240;
for aa=2:2:10
ang=[ang;ang/aa];
end
240
120
60
30
40
20
10
5
30
15
7.5
3.75
5
2.5
1.25
0.625
24
12
6
3
4
2
1
0.5
3
1.5
0.75
0.375
0.5
0.25
0.125
0.0625
PS: I will be accessing these values from different arrays, that's why I used a for loop so I can access the values using their indexes
In addition to the divide-by-zero error you were starting with (fixed in the edit), the approach you're taking isn't actually doing what you think it is. if you print out each step, you'll see why.
Instead of that approach, I suggest taking more of a "matlab way": avoid the loop by making use of vectorized operations.
orig = 240;
divisor = 2.^(0:5); #% vector of 2 to the power of [0 1 2 3 4 5]
ans = orig./divisor;
output:
ans = [240 120 60 30 15 7.5]
Try the following:
ang=240;
for aa=1:5
% sz=size(ang,1);
% ang=[ang;ang(sz)/2];
ang=[ang;ang(end)/2];
end
You should be getting warning: division by zero if you're running it in Octave. That says pretty much everything.
When you divide by zero, you get Inf. Because of your recursion... you see the problem.
You can simultaneously generalise and vectorise by using logic:
ang=240; %Replace 240 with any positive integer you like
ang=ang*2.^-(0:log2(ang));
ang=ang(1:sum(ang==floor(ang)));
This will work for any positive integer (to make it work for negatives as well, replace log2(ang) with log2(abs(ang))), and will produce the vector down to the point at which it goes odd, at which point the vector ends. It's also faster than jitendra's solution:
octave:26> tic; for i=1:100000 ang=240; ang=ang*2.^-(0:log2(ang)); ang=ang(1:sum(ang==floor(ang))); end; toc;
Elapsed time is 3.308 seconds.
octave:27> tic; for i=1:100000 ang=240; for aa=1:5 ang=[ang;ang(end)/2]; end; end; toc;
Elapsed time is 5.818 seconds.
Related
So the task I have to solve is to calculate the binomial coefficient for 100>=n>k>=1 and then say how many solutions for n and k are over an under barrier of 123456789.
I have no problem in my formula of calculating the binomial coefficient but for high numbers n & k -> 100 the datatypes of c get to small to calculated this.
Do you have any suggestions how I can bypass this problem with overflowing the datatypes.
I thought about dividing by the under barrier straight away so the numbers don't get too big in the first place and I have to just check if the result is >=1 but i couldn't make it work.
Say your task is to determine how many binomial coefficients C(n, k) for 1 ≤ k < n ≤ 8 exceed a limit of m = 18. You can do this by using the recurrence C(n, k) = C(n − 1, k) + C(n − 1, k − 1) that can visualized in Pascal's triangle.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 (20) 15 6 1
1 7 (21 35 35 21) 7 1
1 8 (28 56 70 56 28) 8 1
Start at the top and work your way down. Up to n = 5, everything is below the limit of 18. On the next line, the 20 exceeds the limit. From now on, more and more coefficients are beyond 18.
The triangle is symmetric and strictly increasing in the first half of each row. You only need to find the first element that exceeds the limit on each line in order to know how many items to count.
You don't have to store the whole triangle. It is enough to keey the last and current line. Alternatively, you can use the algorithm detailed [in this article][ot] to work your way from left to right on each row. Since you just want to count the coefficients that exceed a limit and don't care about their values, the regular integer types should be sufficient.
First, you'll need a type that can handle the result. The larget number you need to handle is C(100,50) = 100,891,344,545,564,193,334,812,497,256. This number requires 97 bits of precision, so your normal data types won't do the trick. A quad precision IEEE float would do the trick if your environment provides it. Otherwise, you'll need some form of high/arbitrary precision library.
Then, to keep the numbers within this size, you'll want cancel common terms in the numerator and the denominator. And you'll want to calculate the result using ( a / c ) * ( b / d ) * ... instead of ( a * b * ... ) / ( c * d * ... ).
I'm working on typing speed app and I need to know what's the Formula of calculating WPM (Words Per Minute)
Edit:
indeed i know:
wpm = correct_characters_in_60_seconds / 5
but i have no idea what should i do with decimal numbers like 22.6 or 19.7 and...
for example if user typed 158 keystrokes in 60 seconds so, 158/5 = 31.6
so do the result should be 32 WPM or 31 WPM? How?
thanks.
WPM (Word per minute)
For the purpose of typing measurement, each word is standardized to be five characters or keystrokes long, including spaces and punctuation. For example, the phrase "I run" counts as one word, but "rhinoceros" and "let's talk" both count as two.
So the formula is:
Number_of_keystroke / time_in_minute * percentages_of_accurate_word
or
Number_of_keystroke / time_in_second * 60 * percentages_of_accurate_word
When dealing with decimals you should round down when the decimal is >.5 , else round down
Example:
5.5 -> 6
7.3 -> 7
3.49 -> 3
4.51 -> 5
Words per minute(WPM) should be rounded off to the closest decimal value.
In your case 158/5 = 31.6 should be reported as 32 and not 31.
However if the the value was 156/5 = 31.2 then it should be rounded off to 31 to approximate the closest decimal value and hence averaging the overall error.
if WPM<=x.5 then WPM = x else WPM = x+1
I have a very large dataset array with over a million values that looks like this:
Month Day Year Hour Min Second Line1 Line2 Power Dt
7 8 2013 0 1 54 1.91 4.98 826.8 0
7 8 2013 0 0 9 1.93 3.71 676.8 0
7 8 2013 0 1 15 1.92 5.02 832.8 0
7 8 2013 0 1 21 1.91 5.01 830.4 0
and so on.
When the measurement of seconds got to 60 it would start over again at 0 hence why the first number is bigger. I need to fill the delta t column (Dt) by taking the current rows seconds column and subtracting the previous rows seconds column and correcting for negatyive values. This opperation cannot preform this operation in a loop as the it would take ages to complete and needs to be completed in a simple, one-shot, vector subtraction operation.
You can try diff command to generate such results. Its very fast and should work wihout any for loop.
HTH
Dt=diff(datenum(A(:,1:6)))*60*60*24;
This gives the delta in seconds, but I'm not sure what you want you correction for negative differences to be. Could you give an example of the expected output?
Note that Dt will be one entry shorter than A, so you may have to pad it.
You can remove the negative values (I think) with the command
Dt(Dt<0)=Dt(Dt<0)+60;
If you need to pad the Dt vector so that it is the same length as the data set, try
Dt=[Dt;0];
I have a Sorted array .Lets assume
{4,7,9,12,23,34,56,78} Given min and max I want to find elements in array between min and max in efficient way.
Cases:min=23 and max is 78 op:{23,34,56,78}
min =10 max is 65 op:{12,23,34,56}
min 0 and max is 100 op:{4,7,9,12,23,34,56,78}
Min 30 max= 300:{34,56,78}
Min =100 max=300 :{} //empty
I want to find efficient way to do this?I am not asking code any algorithm which i can use here like DP exponential search?
Since it's sorted, you can easily find the lowest element greater than or equal to the minimum desired, by using a binary search over the entire array.
A binary search basically reduces the serch space by half with each iteration. Given your first example of 10, you start as follows with the midpoint on the 12:
0 1 2 3 4 5 6 7 <- index
4 7 9 12 23 34 56 78
^^
Since the element you're looking at is higher than 10 and the next lowest is lesser, you've found it.
Then, you can use a similar binary search but only over that section from the element you just found to the end. This time you're looking for the highest element less than or equal to the maximum desired.
On the same example as previously mentioned, you start with:
3 4 5 6 7 <- index
12 23 34 56 78
^^
Since that's less than 65 and the following one is also, you need to increase the pointer to the halfway point of 34..78:
3 4 5 6 7 <- index
12 23 34 56 78
^^
And there you have it, because that number is less and the following number is more (than 65)
Then you have the start at stop indexes (3 and 6) for extracting the values.
0 1 2 3 4 5 6 7 <- index
4 7 9 ((12 23 34 56)) 78
-----------
The time complexity of the algorithm is O(log N). Though keep in mind that this really only becomes important when dealing with larger data sets. If your data sets do consist of only about eight elements, you may as well use a linear search since (1) it'll be easier to write; and (2) the time differential will be irrelevant.
I tend not to worry about time complexity unless the operations are really expensive, the data set size gets into the thousands, or I'm having to do it thousands of times a second.
Since it is sorted, this should do:
List<Integer> subarray = new ArrayList<Integer>();
for (int n : numbers) {
if (n >= MIN && n <= MAX) subarray.add(n);
}
It's O(n) as you only look at every number once.
I have a matrix:
x = [0 0 0 1 1 0 5 0 7 0];
I need to remove all of the zeroes, like so:
x = [1 1 5 7];
The matrices I am using are large (1x15000) and I need to do this multiple times (5000+), so efficiency is key!
One way:
x(x == 0) = [];
A note on timing:
As mentioned by woodchips, this method seems slow compared to the one used by KitsuneYMG. This has also been noted by Loren in one of her MathWorks blog posts. Since you mentioned having to do this thousands of times, you may notice a difference, in which case I would try x = x(x~=0); first.
WARNING: Beware if you are using non-integer numbers. If, for example, you have a very small number that you would like to consider close enough to zero so that it will be removed, the above code won't remove it. Only exact zeroes are removed. The following will help you also remove numbers "close enough" to zero:
tolerance = 0.0001; % Choose a threshold for "close enough to zero"
x(abs(x) <= tolerance) = [];
Just to be different:
x=x(x~=0);
or
x=x(abs(x)>threshold);
This has the bonus of working on complex numbers too
Those are the three common solutions. It helps to see the difference.
x = round(rand(1,15000));
y = x;
tic,y(y==0) = [];toc
Elapsed time is 0.004398 seconds.
y = x;
tic,y = y(y~=0);toc
Elapsed time is 0.001759 seconds.
y = x;
tic,y = y(find(y));toc
Elapsed time is 0.003579 seconds.
As you should see, the cheapest way is the direct logical index, selecting out the elements to be retained. The find is more expensive, since matlab finds those elements, returning a list of them, and then indexes into the vector.
Here's another way
y = x(find(x))
I'll leave it to you to figure out the relative efficiency of the various approaches you try -- do write and let us all know.
Though my timing results are not conclusive to whether it is significantly faster, this seems to be the fastest and easiest approach:
y = nonzeros(y)
x = [0 0 0 1 1 0 5 0 7 0]
y = [0 2 0 1 1 2 5 2 7 0]
Then x2 and y2 can be obtained as:
x2=x(~(x==0 & y==0))
y2=y(~(x==0 & y==0))
x2 = [0 1 1 0 5 0 7]
y2 = [2 1 1 2 5 2 7]
Hope this helps!