Generate a random number, excluding a single number - random
I'm writing a Monty Hall simulator, and found the need to generate a number within a range, excluding a single number.
This seemed easy, so I naively wrote up:
(The g/... functions are part of my personal library. Their use should be fairly clear):
(defn random-int-excluding
"Generates a random number between min-n and max-n; excluding excluding-n.
min-n is inclusive, while max-n is exclusive."
[min-n max-n excluding-n rand-gen]
(let [rand-n (g/random-int min-n max-n rand-gen)
rand-n' (if (= rand-n excluding-n) (inc rand-n) rand-n)]
(g/wrap rand-n' min-n (inc max-n))))
This generates a random number within the range, and if it equals the excluded number, adds one; wrapping if necessary. Of course this ended up giving the number after the excluded number twice the chance of being picked since it would be picked either if it or the excluded number are chosen. Sample output frequencies for a range of 0 to 10 (max exclusive), excluding 2:
([0 0.099882]
[1 0.100355]
[3 0.200025]
[4 0.099912]
[5 0.099672]
[6 0.099976]
[7 0.099539]
[8 0.100222]
[9 0.100417])
Then I read this answer, which seemed much simpler, and based on it, wrote up:
(defn random-int-excluding
"Generates a random number between min-n and max-n; excluding excluding-n.
min-n is inclusive, while max-n is exclusive."
[min-n max-n excluding-n rand-gen]
(let [r1 (g/random-int min-n excluding-n rand-gen)
r2 (g/random-int (inc excluding-n) max-n rand-gen)]
(if (g/random-boolean rand-gen) r1 r2)))
Basically, it splits the range into 2 smaller ranges: from the min to the excluded number, and from excluded number + 1 to the max. It generates random number from these ranges, then randomly chooses one of them. Unfortunately though, as I noted under the answer, this gives skewed results unless both the partitions are of equal size. Sample output frequencies; same conditions as above:
([0 0.2499497]
[1 0.2500795]
[3 0.0715849]
[4 0.071297]
[5 0.0714366]
[6 0.0714362]
[7 0.0712715]
[8 0.0715285]
[9 0.0714161])
Note the numbers part of the smaller range before the excluded number are much more likely. To fix this, I'd have to skew it to pick numbers from the larger range more frequently, and really, I'm not proficient enough in maths in general to understand how to do that.
I looked at the accepted answer from the linked question, but to me, it seems like a version of my first attempt that accepts more than 1 number to exclude. I'd expect, against what the answerer claimed, that the numbers at the end of the exclusion range would be favored, since if a number is chosen that's within the excluded range, it just advances the number past the range.
Since this is going to be one of the most called functions in the simulation, I'd really like to avoid the "brute-force" method of looping while the generated number is excluded since the range will only have 3 numbers, so there's a 1/3 chance that it will need to try again each attempt.
Does anyone know of a simple algorithm to chose a random number from a continuous range, but exclude a single number?
To generate a number in the range [a, b] excluding c, simply generate a number in the range [a, b-1], and if the result is c then output b instead.
Just generate a lazy sequence and filter out items you don't want:
(let [ignore #{4 2}]
(frequencies
(take 2000
(remove ignore (repeatedly #(rand-int 5))))))
Advantage to the other approach of mapping to different new values: This function will also work with different discrete random number distributions.
If the size of the collection of acceptable answers is small, just put all values into a vector and use rand-nth:
http://clojuredocs.org/clojure.core/rand-nth
(def primes [ 2 3 5 7 11 13 17 19] )
(println (rand-nth primes))
(println (rand-nth primes))
(println (rand-nth primes))
~/clj > lein run
19
13
11
Update
If some of the values should include more than the others, just put them in the array of values more than once. The number of occurrances of each value determines its relative weight:
(def samples [ 1 2 2 3 3 3 4 4 4 4 ] )
(def weighted-samples
(repeatedly #(rand-nth samples)))
(println (take 22 weighted-samples))
;=> (3 4 2 4 3 2 2 1 4 4 3 3 3 2 3 4 4 4 2 4 4 4)
If we wanted any number from 1 to 5, but never 3, just do this:
(def samples [ 1 2 4 5 ] )
(def weighted-samples
(repeatedly #(rand-nth samples)))
(println (take 22 weighted-samples))
(1 5 5 5 5 2 2 4 2 5 4 4 5 2 4 4 4 2 1 2 4 1)
Just to show the implementation I wrote, here's what worked for me:
(defn random-int-excluding
"Generates a random number between min-n and max-n; excluding excluding-n.
min-n is inclusive, while max-n is exclusive."
[min-n max-n excluding-n rand-gen]
(let [rand-n (g/random-int min-n (dec max-n) rand-gen)]
(if (= rand-n excluding-n)
(dec max-n)
rand-n)))
Which gives a nice even distribution:
([0 0.111502]
[1 0.110738]
[3 0.111266]
[4 0.110976]
[5 0.111162]
[6 0.111266]
[7 0.111093]
[8 0.110815]
[9 0.111182])
Just to make Alan Malloy's answer explicit:
(defn rand-int-range-excluding [from to without]
(let [n (+ from (rand-int (dec (- to from))))]
(if (= n without)
(dec to)
n)))
(->> #(rand-int-range-excluding 5 10 8)
repeatedly
(take 100)
frequencies)
;{6 28, 9 22, 5 29, 7 21}
No votes required :).
Related
Scheme making matrix
I am new to scheme and I'm having problems with matrices in Scheme. I need to create a function that takes one big and one small square matrices (with the condition: the small's length should be divisor of big one) and make a new matrix with doing an operation on the big one with small one. I've successfully split the big matrix to size that I wanted and I’m successfully operating on it to get the result. Here is how I did it: (define (matrix-op big small x y) (if (< y (/ (length big) (length small)))) (if (< x (/ (length big) (length small))) (cons (calculate (split-y (split-x big small x) small y) small) (matrix-op big small (+ x 1) y)) (matrix-op big small 0 (+ y 1)) ; <- this is where i need to split ) '() ) ) My calculate function returns only 1 atomic value so when I run the function like this it gives me an output like '(val val val val), but what i want is formatting the output like '((val val) (val val)). How can I do it? Thanks in advance. I realized that I couldn't explain the problem properly. What i want to have is a function that takes two different square matrices one big and one small, Splits the big one to same size as smaller one, operates on them to create a new matrix that has the size m/n if the big one is mxm and small one is nxn. Example: big '( small '( (8 0 3 1 5 3 2 2) (8 4) (7 1 1 4 3 7 1 4) (9 5) (1 3 7 4 3 6 6 3) ) (0 9 8 6 5 6 4 3) (1 7 6 9 6 6 7 2) (5 7 1 0 2 9 5 3) (0 5 4 6 6 6 3 0) (3 6 2 7 7 5 7 0) ) I need to split big over the same size as small and calculate results like: for x=0 y=0 part is '( calculate result is 5 (8 0) (7 1) ) for x=1 y=0 part is '( calculate result is 2 (3 1) (1 4) ) I actually did returned the results calculated but with the method i gave above my return was like '(5 2 4 2 2 6 4 4 4 3 5 4 2 4 6 3) but I wanted to return as: '( (5 2 4 2) (2 6 4 4) (4 3 5 4) (2 4 6 3) ) So how can I manage to split the return list where i want to split?
I think you are trying to do too much at once. It is always OK to split a bigger problem into a smaller problem. If I understand yours, the idea is to take two square matrics, one of which may be some multiple of the other’s dimensions, and perform a pair-wise operation on the elements. For example: '((1 2 3) '((1 2 3) '((7 7 7) '(( 8 9 10) (4 5 6) + '((7)) --> (4 5 6) + (7 7 7) --> (11 12 13) (7 8 9)) (7 8 9)) (7 7 7)) (14 15 16)) I will continue with the assumption that this is what is desired. Notice that if the two matrices were the same size, a simple nested map would easily combine all elements. What is left is the problem of the different sizes. Solve that and you are golden. Recap: (define (f op small-M big-M) (f-apply-pairwise-op op (f-biggify small-M (/ (length big-M) (length small-M))) big-M)) Now you have broken the problem into two smaller pieces: (define (f-apply-pairwise-op op A B) ...) ; produces pairwise 'A op B' (define (f-biggify M n) ...) ; tile M n times wider and taller Good luck!
Principal component function Incanter
I have been trying to use the principal-components function from Incanter to do PCA and seem to be off track in using it. I found some sample data online from a PCA tutorial and wanted to practice on it: (def data [[0.69 0.49] [-1.31 -1.21] [0.39 0.99] [0.09 0.29] [1.29 1.09] [0.49 0.79] [0.19 (- 0 0.31)] [(- 0 0.81) (- 0 0.81)] [(- 0 0.31) (- 0 0.31)] [(- 0 0.71) (- 0 1.01)]]) Upon first attempt to implement PCA I tried passing vectors to Incanter's matrix function, but found myself passing it too many arguments. At this point I decided to try a nested vector structure as defined above, but would like to avoid this route. How would I turn data into a matrix (Incanter) such that it will be accepted as input into Incanter's function principal-components. For simplicity let's call the new matrix fooMatrix. Once this matrix, fooMatrix, has been constructed the following code should work to extract the first two principal components (def pca (principal-components fooMatrix)) (def components (:rotation pca)) (def pc1 (sel components :cols 0)) (def pc2 (sel components :cols 1)) and then the data can be projected on the principal components by (def principal1 (mmult fooMatrix pc1)) (def principal2 (mmult fooMatrix pc2))
Check out the Incanter API. I believe you just want (incanter.core/matrix data). These are your options for Incanter's matrix function. Maybe A2 is what you're interested in.
(def A (matrix [[1 2 3] [4 5 6] [7 8 9]])) ; produces a 3x3 matrix
(def A2 (matrix [1 2 3 4 5 6 7 8 9] 3)) ; produces the same 3x3 matrix
(def B (matrix [1 2 3 4 5 6 7 8 9])) ; produces a 9x1 column vector
Example using your data:
user=> (use '[incanter core stats charts datasets])
nil
user=>(def data [0.69 0.49 -1.31 -1.21 0.39 0.99 0.09 0.29 1.29
1.09 0.49 0.79 0.19 (- 0 0.31) (- 0 0.81) (- 0 0.81)
(- 0 0.31) (- 0 0.31) (- 0 0.71) (- 0 1.01)])
user=>(def fooMatrix (matrix data 2))
user=>(principal-components fooMatrix)
{:std-dev (1.3877785387777999 0.27215937850413047), :rotation A 2x2 matrix
-------------
-7.07e-01 -7.07e-01
-7.07e-01 7.07e-01
}
Voilà. Nested vector structure gone.
Sum of numbers in a list using Scheme
I want to sum the numbers in a list without using recursion. I know you can sum a list of numbers like this (+ num1 num2 ... numN) but what if you have a list L which equals to '(num1 num2 ... numN) is there a way to make + take the numbers in this list as arguments. I need to do this without recursion or helper functions.
Sure, just use apply: (apply + '(1 2 3 4 5 6)) ; same as (+ 1 2 3 4 5 6) (apply + 1 2 3 '(4 5 6)) ; ditto (apply + 1 2 3 4 5 '(6)) ; ditto (apply + 1 2 3 4 5 6 '()) ; ditto
The general answer to the question you seem to be asking -- how to take a list and use it as the arguments -- is apply, as Chris Jester-Young answered. However, for this particular question, there might some other considerations. You may want to sum lists of arbitrary size. However, implementations often have some limit of the number of arguments you can call a function with. A more reliable solution may be to use some kind of fold function (various implementations have different fold functions) to fold + over the list.
Sort two list with the elements in an increasing order
The question requires me to Complete the Scheme function merge, which consumes two lists of sorted numbers (in an increasing order) and produces a list of numbers which consists of all the two consumed lists in sorted order. For example, (merge (list 1 4 5 9) (list -1 2 4)) => (list -1 1 2 4 4 5 9) (merge (list 1 4 5 9) empty) => (list 1 4 5 9) (merge empty (list 1 4 5 9)) => (list 1 4 5 9) (merge empty empty) => empty Thanks for helping out!!
Since this smells like homework, I won't write any code, but I will tell you that what are doing is part of the merge sort algorithm. Remember these two things: In functional languages like Scheme, you are asking the question what value do I need to produce rather than what do I need to do In Scheme, you often write more than one procedure to accomplish a single task If you remember these two things and figure out which part of merge sort you need to implement, it should become fairly easy to figure out.
Scheme Understanding
I am supposed to write a scheme function (digit-count n) that accepts a positive integer n and evaluates to the number of digits of n that are 6, 4, or 9. I am having trouble understanding what exactly I am supposed to do, I am confused about the "digits of n that are 6, 4 or 9", what does this mean?
This is just an interpretation question, but I would say that you would take the decimal representation of a number, and count the total number of digits that are 6, 4, or 9. For example: 100 --> 0 4 --> 1 469 --> 3 444 --> 3 Get it now?
One interpretation - example: Given 678799391, the number of digits would be 0 for 4, 1 for 6 and 3 for 9. The sum of the occurences would be 0 + 1 + 3 = 4.
Convert the whole number to a list and check each one individually. (define (number->list x) (string->list (number->string x)) (define (6-4-or-9 x) (cond ((= x 6) true)) ((= x 4) true)) ((= x 9) true)))) (define (count-6-4-9 x) (cond ((6-4-or-9 (car (number->list x))))....... I'm sure you can see where that's going. It's pretty crude, and I'm not sure it's really idiomatic, but it should work. The idea is is that you convert the number to a list, check to first digit, if it's six, four or nine, recursively call the procedure on the cdr of the number list converted back to a string + 1...
If you are not using lists, you can work with modulo % of 10 and dividing whole numbers / with 10. Below is the recursive solution : (define (digits n) (if(not (< n 1)) (+ 1 (digits (/ n 10))) 0))
First, we must understand what the question is asking: It is asking you to write a procedure that counts the number of times the numbers 4, 6, or 9 show up in another inputted number. For instance, inputting 10345 should return 1. Let's see why: The digits of 10345 are 1, 0, 3, 4, and 5. We have to ask, "How many times do 4, 6, or 9 show up?" Well, there are no 6's or 9's in 10345. However, there is one 4. Therefore, the procedure should return 1. Another example: (digit-count 14289) Let's break it up like we did before. The digits of 14289 are 1, 4, 2, 8, and 9. There are no 6's. There are, however, 1's and 9's. How many? There is one 1 and one 9. Since there are two (total) of the desired digits present (desired digits are 4, 6, and 9), (digit-count 14289) should return 2. Some more examples: (digit-count 144) --> 2 (there are two 4's) (digit-count 1) --> 0 (there are no 4's, 6's, or 9's) (digit-count 1262) --> 1 (there is one 6) Now, let's start defining. We can take advantage of the appearances function, which takes two inputs and returns how many times the first input appears in the second. For example: (appearances 'a 'amsterdam) returns 2 because there are two a's in amsterdam. Using appearances, here is our definition (finally!): (define (count469 num) (+ (appearances 4 num) (appearances 6 num) (appearances 9 num))) This function returns the sum of the appearances of 4, the appearances of 6, and the appearances of 9. Please feel free to reply with any feedback or questions!