I am looking for to take one particular number or range of numbers from a set of number?
Example
A = [-10,-2,-3,-8, 0 ,1, 2, 3, 4 ,5,7, 8, 9, 10, -100];
How can I just take number 5 from the set of above number and
How can I take a range of number for example from -3 to 4 from A.
Please help.
Thanks
I don't know what you are trying to accomplish by this. But you could check each entry of the set and test it it's in the specified range of numbers. The test for a single number could be accomplished by testing each number explicitly or as a special case of range check where the lower and the upper bound are the same number.
looping and testing, no matter what the programming language is, although most programming languages have builtin methods for accomplishing this type of task (so you may want to specify what language are you supposed to use for your homework):
procfun get_element:
index=0
for element in set:
if element is 5 then return (element,index)
increment index
your "5" is in element and at set[index]
getting a range:
procfun getrange:
subset = []
index = 0
for element in set:
if element is -3:
push element in subset
while index < length(set)-1:
push set[index] in subset
if set[index] is 4:
return subset
increment index
#if we met "-3" but we didn't met "4" then there's no such range
return None
#keep searching for a "-3"
increment index
return None
if ran against A, subset would be [-3,-8, 0 ,1, 2, 3, 4]; this is a "first matched, first grabbed" poorman's algorithm. on sorted sets the algorithms can get smarter and faster.
Related
In the following Swift code:
self.numbers[2] = Int.random(in: 0...self.symbols.count-1)
Why do we have to write -1?
I don't understand the code.
When you are using A...B, that range includes B in it. See ...(_:_:) docs. So if you say -
Int.random(in: 0...self.symbols.count-1)
It means range starts at 0 and ends at symbols.count-1 including both.
Say an array has 2 elements, it's count is 2, but valid indices are 0, 1 (2 is not a valid index), so you are just making sure that it is restricted to valid index values.
Other way to write the same would be - A..<B, in this range, B is not included. See ..<(_:_:) docs. Following is same as above.
Int.random(in: 0..<self.symbols.count)
You are using ClosedRange here. It is a range which includes both start and end value of the range. You can also used simple Range, which is open so to say. It only includes start value and not the final value.
Lets take an example.
let a = 1 ... 10 // includes all values from 1 to 10
let b = 1 ..< 10 // includes values from 1 to 9
So, basically what you are writing here,
self.numbers[2] = Int.random(in: 0 ... self.symbols.count - 1)
is equivalent to open range using,
self.numbers[2] = Int.random(in: 0 ..< self.symbols.count)
such that you dont need to add -1 to it.
Int.random can take both ClosedRange and Range, here and here.
Here are your code:
self.numbers[2] = Int.random(in: 0...self.symbols.count-1)
It says: your third item in numbers array will be assigned by the random item of symbols array
But how to get the random item of symbols array? You'll get it by its index.
The indexes of an array starting from 0, so, for the last element of array, its index will be the array length - 1, for example, with [1,2,3], the last element's index will be 2.
Here, you used Closed Range, it will include the last element, so if you don't minus count by one, the last index will be 3 (because array.count will return the length of the array), which is produce out of index error
I have an array of 1 x 400, where all element values are above 1500. However, I have some elements that have values<50 which are wrong measures and I would like to have the mean of the elements before and after the wrong measured data points and replace it in the main array.
For instance, element number 17 is below 50 so I want to take the mean of elements 16 and 18 and replace element 17 with the new mean.
Can someone help me, please? many thanks in advance.
No language is specified in the question, but for Python you could work with List Comprehension:
# array with 400 values, some of which are incorrect
arr = [...]
arr = [arr[i] if arr[i] >= 50 else (arr[i-1]+arr[i+1])/2 for i in range(len(arr))]
That is, if arr[i] is less than 50, it'll be replaced by the average value of the element before and after it. There are two issues with this approach.
If i is the first or last element, then one of the two values will be undefined, and no mean can be obtained. This can be fixed by just using the value of the available neighbour, as specified below
If two values in a row are very low, the leftmost one will use the rightmost one to calculate its value, which will result in a very low value. This is a problem that may not occur for you in practice, but it is an inherent result of the way you wish to recalculate values, and you might want to keep it in mind.
Improved version, keeping in mind the edge cases:
# don't alter the first and last item, even if they're low
arr = [arr[i] if arr[i] >= 50 or i == 0 or i+1 == len(arr) else (arr[i-1]+arr[i+1])/2 for i in range(len(arr))]
# replace the first and last element if needed
if arr[0] < 50:
arr[0] = arr[1]
if arr[len(arr)-1] < 50:
arr[len(arr)-1] = arr[len(arr)-2]
I hope this answer was useful for you, even if you intend to use another language or framework than python.
I understand that the command compares and can subtract values, but I don't see exactly how that works. I've used a TI BASIC programming tutorial site (http://tibasicdev.wikidot.com/movement-explanation) and I need clarification on 🔺List as a whole.
This portion of the code with 🔺List is as follows,:
:min(8,max(1,A+sum(ΔList(Ans={25,34→A
:min(16,max(1,B+sum(ΔList(K={24,26→B
and the website explains the code like this.:
"This is how this code works. When you press a key, its value is stored to K. We check to see if K equals one of the keys we pressed by comparing it to the lists {24,26 and {25,34. This results in a list {0,1}, {1,0}, or {0,0}. We then take the fancy command Δlist( to see whether to move up, down, left or right. What Δlist( does is quite simple. Δlist( subtracts the first element from the second in the previous list, and stores that as a new one element list, {1}, {-1}, or {0}. We then turn the list into a real number by taking the sum of the one byte list. This 1, -1, or 0 is added to A."
The ΔList( command subtracts every element in a list from its previous element. This code uses some trickery with it to compactly return 1 if a key is pressed and -1
ΔList( calculates the differences between consecutive terms of a list, and returns them in a new list.
ΔList({0,1,4,9,16,25,36})
{1 3 5 7 9 11}
That is, ΔList({0,1,4,9,16,25,36}) = {1-0, 4-1, 9-4, 16-9, 25-16, 36-25} = {1 3 5 7 9 11}.
When there are only two elements in a list, ΔList({a,b}) is therefore equal to {b-a}. Then sum(ΔList({a,b})) is equal to b-a, since that's the only term in the list. Let's say that K is 26 in your example; that is, the > key is pressed.
B+sum(ΔList(K={24,26→B Result of expression:
K 26
K={24,26 {0,1}
ΔList(K={24,26 {1} = {0 - 1}
sum(ΔList(K={24,26 -1
B [current x-position of player]
B+sum(ΔList(K={24,26→B [add 1 to current x-pos. of player]
Similarly, B will be decreased if key 24, the left key, is pressed.
Let's say I want to generate a random number between 1 and 100, but I don't want to include 42. How would I do this without repeating the random method until it is not 42.
Updated for Swift 5.1
Excluding 1 value
var nums = [Int](1...100)
nums.remove(at: 42)
let random = Int(arc4random_uniform(UInt32(nums.count)))
print(nums[random])
Excluding multiple values
This extension of Range does provide a solution when you want to exclude more than 1 value.
extension ClosedRange where Element: Hashable {
func random(without excluded:[Element]) -> Element {
let valid = Set(self).subtracting(Set(excluded))
let random = Int(arc4random_uniform(UInt32(valid.count)))
return Array(valid)[random]
}
}
Example
(1...100).random(without: [40,50,60])
I believe the computation complexity of this second solution is O(n) where n is the number of elements included in the range.
The assumption here is the no more than n excluded values are provided by the caller.
appzYourLife has some great general purpose solutions, but I want to tackle the specific problem in a lightweight way.
Both of these approaches work roughly the same way: Narrow the range to the random number generator to remove the impossible answer (99 answers instead of 100), then map the result so it isn't the illegal value.
Neither approach increases the probability of an outcome relative to another outcome. That is, assuming your random number function is perfectly random the result will still be random (and no 2x chance of 43 relative to 5, for instance).
Approach 1: Addition.
Get a random number from 1 to 99. If it's greater than or equal to the number you want to avoid, add one to it.
func approach1()->Int {
var number = Int(arc4random_uniform(99)+1)
if number >= 42 {
number = number + 1
}
return number
}
As an example, trying to generate a random number from 1-5 that's not 3, take a random number from 1 to 4 and add one if it's greater than or equal to 3.
rand(1..4) produces 1, +0, = 1
rand(1..4) produces 2, +0, = 2
rand(1..4) produces 3, +1, = 4
rand(1..4) produces 4, +1, = 5
Approach 2: Avoidance.
Another simple way would be to get a number from 1 to 99. If it's exactly equal to the number you're trying to avoid, make it 100 instead.
func approach2()->Int {
var number = Int(arc4random_uniform(99)+1)
if number == 42 {
number = 100
}
return number
}
Using this algorithm and narrowing the range to 1-5 (while avoiding 3) again, we get these possible outcomes:
rand(1..4) produces 1; allowed, so Result = 1
rand(1..4) produces 2, allowed, so Result = 2
rand(1..4) produces 3; not allowed, so Result = 5
rand(1..4) produces 4, allowed, so Result = 4
I want to count the number of set bits in a uint in Specman:
var x: uint;
gen x;
var x_set_bits: uint;
x_set_bits = ?;
What's the best way to do this?
One way I've seen is:
x_set_bits = pack(NULL, x).count(it == 1);
pack(NULL, x) converts x to a list of bits.
count acts on the list and counts all the elements for which the condition holds. In this case the condition is that the element equals 1, which comes out to the number of set bits.
I don't know Specman, but another way I've seen this done looks a bit cheesy, but tends to be efficient: Keep a 256-element array; each element of the array consists of the number of bits corresponding to that value. For example (pseudocode):
bit_count = [0, 1, 1, 2, 1, ...]
Thus, bit_count2 == 1, because the value 2, in binary, has a single "1" bit. Simiarly, bit_count[255] == 8.
Then, break the uint into bytes, use the byte values to index into the bit_count array, and add the results. Pseudocode:
total = 0
for byte in list_of_bytes
total = total + bit_count[byte]
EDIT
This issue shows up in the book Beautiful Code, in the chapter by Henry S. Warren. Also, Matt Howells shows a C-language implementation that efficiently calculates a bit count. See this answer.