Object range with conditions - range

In groovy I can write
def n = 10
print 1..<n
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9]
Are there other language that allow to specify range with conditions?
examples
def n = 10
print 1<=..n
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
def n = -2
print 1<=..n
Output: [1]
def n = -2
print 1..n
Output: [1, 0, -1, -2]

Python has the range() method which does a similar thing. While it does not use operators for the condition you can specify a start value, stop value and step value. It then creates a list containing all values starting with the start value, then start+step, ... until it reaches the end value (which is not included).

Related

Contract m-repeated numbers in list to n-repeated (n<m) in place in O(1) space

I want to write a python 3.7 function that has a sorted list of numbers as an input, and a number n which is the max number each one of the integers can be repeated and modifies the list in place, so that any numbers that are repeated more than n times, would be cut to n repeats, and it should be done in O(1) space, no additional data structures allowed (e.g. set()). Special case - remove duplicates where n = 1. Example:
dup_list = [1, 1, 1, 2, 3, 7, 7, 7, 7, 12]
dedup(dup_list, n = 1)
print(dup_list)
[1, 2, 3, 7, 12]
dup_list = [1, 1, 1, 2, 3, 7, 7, 7, 7, 12]
dedup(dup_list, n = 2)
print(dup_list)
[1, 1, 2, 3, 7, 7, 12]
dup_list = [1, 1, 1, 2, 3, 7, 7, 7, 7, 12]
dedup(dup_list, n = 3)
print(dup_list)
[1, 1, 1, 2, 3, 7, 7, 7, 12]
Case n = 1 is easy, the code is below (code is taken from Elements of Prograqmming Interviews, 2008, page 49 except the last line return dup_list[:write_index]):
def dedup(dup_list):
if not dup_list:
return 0
write_index = 1
for i in range(1, len(dup_list)):
if dup_list[write_index-1] != dup_list[i]:
dup_list[write_index] = dup_list[i]
write_index += 1
return dup_list[:write_index]
This should work:
def dedup2(dup_list, n):
count = 1
list_len = len(dup_list)
i = 1
while i < list_len:
if dup_list[i - 1] != dup_list[i]:
count = 1
else:
count += 1
if count > n:
del(dup_list[i])
i -= 1
list_len -= 1
i += 1
return dup_list
print(dedup2([1, 2, 3, 3, 4, 4, 5, 5, 5, 5, 8, 9], 1))

Can you merge two Flux, without blocking, such that the result only contains unique elements?

Is there a way to merge two Flux such that the result only contains unique elements? I can block on the output and then convert it to a set, but is there a way that does not depend on blocking?
Source (Kotlin)
val set1 = Flux.just(1, 2, 3, 4, 5)
val set2 = Flux.just(2, 4, 6, 8, 10)
val mergedSet = set1.mergeWith(set2)
println(mergedSet.collectList().block())
Output
[1, 2, 3, 4, 5, 2, 4, 6, 8, 10]
Desired Output (order is not important)
[1, 2, 3, 4, 5, 6, 8, 10]
You can use the Flux's merge method and then apply distinct() to it.
Flux.merge (Flux.just(1, 2, 3, 4, 5), Flux.just(2, 4, 6, 8, 10)).distinct();
This way you get a flux which produces only distinct values.

Adding lists by element in pyspark

I'd like to take a RDD of integer lists and reduce it down to one list. For example...
[1, 2, 3, 4]
[2, 3, 4, 5]
to
[3, 5, 7, 9]
I can do this in python using the zip function but not sure how to replicate it in spark besides doing collect on the object but I want to keep the data in the rdd.
If all elements in rdd are of the same length, you can use reduce with zip:
rdd = sc.parallelize([[1,2,3,4],[2,3,4,5]])
rdd.reduce(lambda x, y: [i+j for i, j in zip(x, y)])
# [3, 5, 7, 9]

Do you have any idea or documentation about why we have arc4random_stir() in swift?

I have written the program below for generating random unique numbers for several number of times by invoking the function, but it seems like I'm getting the same pattern with minimal changes.
func generateRandom(withinNumber: Int) {
var i:Int = 0
var elements = Set<Int>()
while i != withinNumber {
let num:Int = Int(arc4random())%withinNumber + 1
if elements.count <= withinNumber && elements.contains(num) == false {
elements.insert(num)
}
else {
i = i-1
}
i=i+1
}
print(elements)
elements.removeAll()
}
generateRandom(withinNumber: 10)
How does I make my program effectively run to generate several random unique numbers.
Please let me know it would be very helpful for me.
You are storing your numbers in a Set and sets are not ordered, so the order the elements are shown by print is unrelated to the order in which they were added to the set.
Rather the elements of a set are stored in some manner which enables fast checking for .contains(), and this is one reason you seeing similar sequences.
If you wish to preserve order of insertion use a collection which does this, i.e. an array. Changing to an array in your code produced the following results from 9 calls:
[8, 9, 7, 10, 5, 6, 2, 3, 1, 4]
[4, 9, 10, 3, 6, 2, 1, 7, 8, 5]
[8, 3, 5, 1, 6, 4, 9, 10, 7, 2]
[5, 7, 2, 9, 8, 1, 6, 10, 3, 4]
[2, 3, 7, 6, 9, 1, 8, 10, 5, 4]
[9, 10, 2, 4, 6, 8, 5, 7, 1, 3]
[9, 10, 2, 5, 4, 7, 3, 8, 1, 6]
[1, 6, 4, 5, 8, 2, 3, 9, 7, 10]
[6, 10, 5, 3, 2, 8, 1, 9, 7, 4]
You are also generating 10 random numbers in the range 1 to 10 and avoiding duplicates, so the results is always going to be the numbers 1 to 10 in some order.
To generate a random number in a given range do not use %, instead use the provided arc4random_uniform() which will give better a better distribution.
The function mention in your title arc4random_stir() is available in Swift.
BTW (somewhat opinion based): It is better to write !e (! being the boolean not operator) rather than e == false, and never ever write e == true which is the long form of e!
BTW (SO etiquette): Don't link to your code (or paste in images of it). Reduce to a small example which demonstrates the issue (not required in your case) and insert directly in the question. Keep tags minimal and appropriate. These edits were done for you this time by myself and others, you will know for next time.
HTH

Is it possible to match N elements with a coffeescript splat?

Is it possible to specify how many elements a splat should match? Something like:
foo = [1, 2, 3, 4, 5, 6]
[firstThree...(3), fourth, rest...] = foo
console.log firstThree // [1, 2, 3]
console.log forth // 4
console.log rest // [5, 6]
As far as I know there is no way of adding a limit to the amount of arguments a splat can take.
But you can use ranges(search for range in the Loops and Comprehensions Docs) to get a similar syntax in your destructuring assignment:
foo = [1, 2, 3, 4, 5, 6]
[firstThree, fourth, rest] = [foo[0..2], foo[3], foo[4..-1]]
firstThree
# => [1, 2, 3]
fourth
# => 4
rest
# => [5, 6]