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Generate Random numbers uniformly over entire range
How to use rand function to
Could anyone tell me how to use the rand() function in C programming with 1 = min and
7 = max, 1 and 7 included.
Thanks
This will do what you want:
rand() % 7 + 1
Explanation:
rand() returns a random number between 0 and a large number.
% 7 gets the remainder after dividing by 7, which will be an integer from 0 to 6 inclusive.
+ 1 changes the range to 1 to 7 inclusive.
Use the modulo operator. It returns the remainder when dividing. Generate a number in range 0 to 6 by using
rand() % 7. Add 1 to that to generate a number in range 1 to 7.
Related
How do I create a function (e.g. here, an anonymous one but I don't mind any) to get x elements from vec that are most centered (i.e. around the median)? In essence I want a function with same syntax as Matlab's randsample(n,k), but for non-random, with elements spanning around the center.
cntr=#(vec,x) vec(round(end*.5)+(-floor(x/2):floor(x/2))); %this function in question
cntr(1:10,3) % outputs 3 values around median 5.5 => [4 5 6];
cntr(1:11,5) % outputs => [4 5 6 7 8]
Note that vec is always sorted.
One part that I struggle with is not to output more than the limits of vec. For example, cntr(1:10, 10) should not throw an error.
edit: sorry to answer-ers for many updates of question
It's not a one-line anonymous function, but you can do this pretty simply with a couple calls to sort:
function vec = cntr(vec, x)
[~, index] = sort(abs(vec-median(vec)));
vec = vec(sort(index(1:min(x, end))));
end
The upside: it will still return the same set of values even if vec isn't sorted. Some examples:
>> cntr(1:10, 3)
ans =
4 5 6
>> cntr(1:11, 5)
ans =
4 5 6 7 8
>> cntr(1:10, 10) % No indexing errors
ans =
1 2 3 4 5 6 7 8 9 10
>> cntr([3 10 2 4 1 6 5 8 11 7 9], 5) % Unsorted version of example 2
ans =
4 6 5 8 7 % Same values, in their original order in vec
OLD ANSWER
NOTE: This applied to an earlier version of the question where a range of x values below and x values above the median were desired as output. Leaving it for posterity...
I broke it down into these steps (starting with a sorted vec):
Find the values in vec less than the median, get the last x indices of these, then take the first (smallest) of them. This is the starting index.
Find the values in vec greater than the median, get the first x indices of these, then take the last (largest) of them. This is the ending index.
Use the starting and ending indices to select the center portion of vec.
Here's the implementation of the above, using the functions find, min, and max:
cntr = #(vec, x) vec(min(find(vec < median(vec), x, 'last')):max(find(vec > median(vec), x)));
And a few tests:
>> cntr(1:10, 3) % 3 above and 3 below 5.5
ans =
3 4 5 6 7 8
>> cntr(1:11, 5) % 5 above and 5 below 6 (i.e. all of vec)
ans =
1 2 3 4 5 6 7 8 9 10 11
>> cntr(1:10, 10) % 10 above and 10 below 5.5 (i.e. all of vec, no indexing errors)
ans =
1 2 3 4 5 6 7 8 9 10
median requires sorting the array elements. Might as well sort manually, and pick out the middle block (edit: OP's comment indicates elements are already sorted, more justification for keeping it simple):
function data = cntr(data,x)
x = min(x,numel(data)); % don't pick more elements than exist
data = sort(data);
start = floor((numel(data)-x)/2) + 1;
data = data(start:start+x-1);
You could stick this into a single-line anonymous function with some tricks, but that just makes the code ugly. :)
Note that in the case of an uneven division (when we don't leave an even number of elements out), here we prioritize an element on the left. Here is what I mean:
0 0 0 0 0 0 0 0 0 0 0 => 11 elements, x=4
\_____/
picking these 4 values
This choice could be made more complex, for example shifting the interval left or right depending on which of those values is closest to the mean.
Given data (i.e. vec) is already sorted, the indexing operation can be kept to a single line:
cntr = #(data,x) data( floor((numel(data)-x)/2) + (1:x) );
The thing that is missing in that line is x = min(x,numel(data)), which we need to add twice becuase we can't change a variable in an anonymous function:
cntr = #(data,x) data( floor((numel(data)-min(x,numel(data)))/2) + (1:min(x,numel(data))) );
This we can simplify to:
cntr = #(data,x) data( floor(max(numel(data)-x,0)/2) + (1:min(x,numel(data))) );
I'm new to C and I'm reading a book about it. I just came across the rand() function. The book states that using rand() returns a random number from 0 to 32767. It also states that you can narrow the random numbers by using % (modulus operator) to do so.
Here is an example: the following expression puts a random number from 1 to 6 in the variable dice
dice = (rand() % 5) + 1;
I'm unable to get a remainder of 5 as any number from 0 to 33767 % 5 is equal to 0 to 4, but never 5.
Shouldn't it be % 6 in the above statement instead?
For example, if I choose randomly a number between 0 and 32767, let's say 75, then:
75 % 5 == 0
76 % 5 == 1
77 % 5 == 2
78 % 5 == 3
79 % 5 == 4
80 % 5 == 0
Etc.
So regardless of the random number between 0 and 32767, the remainder will never be 5, so it will not be possible to get a 6 number for the dice (as per the above statement).
Not sure if you will understand what I mean but your help would be much appreciated.
dice = (rand() % 5) + 1;
This will generate a random number between 1 to 5, inclusive, as you have analyzed. The % 5 in the book is probably just a typo. To get 1 to 6 it needs to be % 6.
First you will have to understand how modulo (%) works. If you have say 10 and divide it by 5 you get 2 with a remainder of 0, hence the 10 % 5. The possible range of remainders you would get when you mod(modulo) 5 is 0 - 4. Remember that the possible remainders would can get when you divide by x if from 0 to x-1. So in your case with the dice program you need numbers from the range of 1 to 6 (the faces of a die) hence you would mod 6 and add 1 to this number for give the necessary shift. (rand() % 6) + 1
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How to find the index of the n smallest elements in a vector
(2 answers)
Closed 7 years ago.
I would like to return five minimum integers from a vector consists of twenty integers in MATLAB. Any help? Thanks.
Example:
X = [6 7 8 3 5 6 7 2 5 1 0 6 6 2 9 6 3 3 4 77];
How to get the five minimum values from this vector?
You can use Sort function and then take the 5 in the end of the array
sorted_x = sort(X)
5Minimum = sorted_x(15:20)
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How to generate a random number from within a range - C
I saw the following code from programming pearls
int randint(int l, int u)
{ return l + (RAND_MAX*rand() + rand()) % (u-l+1);
}
Can anyone help me explain it?
Can we just use
return l + rand() % (u-l+1);
Thanks,
The problem with using rand() % n to get a number between 0 and n-1 is that it has some bias when n is not an exact divisor of RAND_MAX. The higher the value of n, the stronger this bias becomes.
To illustrate why this happens, let's imagine that rand() would be implemented with a six-sided die. So RAND_MAX would be 5. We want to use this die to generate random numbers between 0 and 3, so we do this:
x = rand() % 4
What's the value of x for each of the six outcomes of rand?
0 % 4 = 0
1 % 4 = 1
2 % 4 = 2
3 % 4 = 3
4 % 4 = 0
5 % 4 = 1
As you can see, the numbers 0 and 1 will be generated twice as often as the numbers 2 and 3.
When your use case doesn't permit bias, this is a better way to calculate a random number:
(int)((double)rand() / (double)RAND_MAX * (double)n)
yes that is ok, check that u>l and you can do only this:
return l + (RAND_MAX*rand()) % (u-l+1);
explaination:
if we would like to generate in union distribution a random integer number in [0,N] when N>0 we would use:
return (RAND_MAX*rand()) % (N+1);
since the range is shitted with a constant value l in your case we just have to add it to the final result.
python model:
>>> import random
>>> import sys
>>> for i in xrange(20):
int(random.random()*sys.maxint%4)
0
1
2
3
1
1
2
2
3
0
3
3
0
2
3
3
1
2
2
3
7 % 3 = 1 (remainder 1)
how does
3 % 7 (remainder ?)
work?
remainder of 3/7 is 3..since it went 0 times with 3 remainder so 3%7 = 3
7 goes into 3? zero times with 3 left over.
quotient is zero. Remainder (modulus) is 3.
Conceptually, I think of it this way. By definition, your dividend must be equal to (quotient * divisor) + modulus
Or, solving for modulus: modulus = dividend - (quotient * divisor)
Whenever the dividend is less than the divisor, the quotient is always zero which results in the modulus simply being equal to the dividend.
To illustrate with OP's values:
modulus of 3 and 7 = 3 - (0 * 7) = 3
To illustrate with other values:
1 % 3:
1 - (0 * 3) = 1
2 % 3:
2 - (0 * 3) = 2
The same way. The quotient is 0 (3 / 7 with fractional part discarded). The remainder then satisfies:
(a / b) * b + (a % b) = a
(3 / 7) * 7 + (3 % 7) = 3
0 * 7 + (3 % 7) = 3
(3 % 7) = 3
This is defined in C99 ยง6.5.5, Multiplicative operators.
7 divided by 3 is 2 with a remainder of 1
3 divided by 7 is 0 with a remainder of 3
As long as they're both positive, the remainder will be equal to the dividend. If one or both is negative, then you get reminded that % is really the remainder operator, not the modulus operator. A modulus will always be positive, but a remainder can be negative.
(7 * 0) + 3 = 3; therefore, the remainder is 3.
a % q = r means there is a x so that q * x + r = a.
So, 7 % 3 = 1 because 3 * 2 + 1 = 7,
and 3 % 7 = 3 because 7 * 0 + 3 = 3
It seems you forgot to mention the surprising case, if the divident is smaller and negative:
-3 % 7
result: 4
For my brain to understand this kind of question, I'm always converting it to a real world object, for example if I convert your question 3 % 7. I'm going to represent the "3" as a 3-inch wide metal hole, then the "7" as a 7 inch metal screw. Can you insert the 7 inch metal screw to a 3 inch wide metal hole? Of course not, therefore the answer should be the 3-inch metal hole, it doesn't matter even let say you have a 1000 or a million inch wide screw, it is still 3, because how many times can you insert the 1000 or a million inch wide screw to a 3 inch wide metal hole? Zero times, right?
The most simple and effective catch to remember would be:
Whenever dividend is less than the divisor, modulus is just that dividend.
Let's formulate this:
if x < y, then x % y = x