I'm trying to figure out the best way of doing the following:
I have a list of values: L
I'd like to pick a subset of this list, of size N, and get a different subset (if the list has enough members) every X minutes.
I'd like the values to be picked sequentially, or randomly, as long as all the values get used.
For example, I have a list: [google.com, yahoo.com, gmail.com]
I'd like to pick X (2 for this example) values and rotate those values every Y(60 for now) minutes:
minute 0-59: [google.com, yahoo.com]
minute 60-119: [gmail.com, google.com
minute 120-179: [google.com, yahoo.com]
etc.
Random picking is also fine, i.e:
minute 0-59: [google.com, gmail.com]
minute 60-119: [yahoo.com, google.com]
Note: The time epoch should be 0 when the user sets the rotation up, i.e, the 0 point can be at any point in time.
Finally: I'd prefer not to store a set of "used" values or anything like that, if possible. i.e, I'd like this to be as simple as possible.
Random picking is actually preferred to sequential, but either is fine.
What's the best way to go about this? Python/Pseudo-code or C/C++ is fine.
Thank you!
You can use the itertools standard module to help:
import itertools
import random
import time
a = ["google.com", "yahoo.com", "gmail.com"]
combs = list(itertools.combinations(a, 2))
random.shuffle(combs)
for c in combs:
print(c)
time.sleep(3600)
EDIT: Based on your clarification in the comments, the following suggestion might help.
What you're looking for is a maximal-length sequence of integers within the range [0, N). You can generate this in Python using something like:
def modseq(n, p):
r = 0
for i in range(n):
r = (r + p) % n
yield r
Given an integer n and a prime number p (which is not a factor of n, making p greater than n guarantees this), you will get a sequence of all the integers from 0 to n-1:
>>> list(modseq(10, 13))
[3, 6, 9, 2, 5, 8, 1, 4, 7, 0]
From there, you can filter this list to include only the integers that contain the desired number of 1 bits set (see Best algorithm to count the number of set bits in a 32-bit integer? for suggestions). Then choose the elements from your set based on which bits are set to 1. In your case, you would use pass n as 2N if N is the number of elements in your set.
This sequence is deterministic given a time T (from which you can find the position in the sequence), a number N of elements, and a prime P.
Related
I am in need of finding prime numbers in arithmetic progression
80218110*n+8021749, n=1 to 100,000
I was told that using Sage would be a good option, since my computer is old. I happen to be new to Sage and I haven't found it to solve my problem, I guess it shouldn't be difficult, does anyone have a good reference for printing primes in arithmetic progressions?
SageMath is based on Python, and Python provides a syntax which should be comfortable for mathematicians:
[80218110*n + 8021749 for n in range(100)]
range(100) is the ordered set 0, 1, 2, ..., 99, and so the previous line evaluates 80218110*n + 8021749 for these values of n. We can also test whether the entries are prime:
INPUT: [80218110*n+8021749 for n in range(100) if (80218110*n+8021749).is_prime()]
OUTPUT:
[8021749,
489330409,
569548519,
970639069,
1050857179,
1131075289,
1772820169,
2093692609,
2173910719,
3136528039,
3617836699,
4660672129,
4740890239,
5382635119,
6425470549,
7067215429,
7227651649,
7548524089]
You can of course make the argument to range larger, but maybe it's not a good idea to print the whole list.
INPUT: len([80218110*n+8021749 for n in range(100000) if (80218110*n+8021749).is_prime()])
OUTPUT: 15273
(len returns the length of the list.) Producing this list is pretty quick, at least on my computer:
INPUT: %time L = [80218110*n+8021749 for n in range(100000) if (80218110*n+8021749).is_prime()]
OUTPUT CPU times: user 94.6 ms, sys: 1.13 ms, total: 95.8 ms
Wall time: 95.5 ms
(ms is milliseconds.)
Problem =====>
Basically there are three .rrd which are generated for three departments.
From that we fetch three values (MIN, MAX, CURRENT) and print ins 3x3 format. There is a python script which does that.
eg -
Dept1: Min=10 Max=20 Cur=15
Dept2: Min=0 Max=10 Cur=5
Dept3: Min=10 Max=30 Cur=25
Now I want to add the values together (Min, Max, Cur) and print in one line.
eg -
Dept: Min=20 Max=60 Cur=45
Issue I am facing =====>
No matter what CDEF i write, I am breaking the graph. :(
This is the part I hate as i do not get any error message.
As far as I understand(please correct me if i am wrong) I definitely cannot store the value anywhere in my program as a graph is returned.
What would be a proper way to add the values in this condition.
Please let me know if my describing the problem is lacking more detail.
You can do this with a VDEF over a CDEF'd sum.
DEF:a=dept1.rrd:ds0:AVERAGE
DEF:b=dept2.rrd:ds0:AVERAGE
DEF:maxa=dept1.rrd:ds0:MAXIMUM
DEF:maxb=dept2.rrd:ds0:MAXIMUM
CDEF:maxall=maxa,maxb,+
CDEF:all=a,b,+
VDEF:maxalltime=maxall,MAXIMUM
VDEF:alltimeavg=all,AVERAGE
PRINT:maxalltime:Max=%f
PRINT:alltimeavg:Avg=%f
LINE:all#ff0000:AllDepartments
However, you should note that, apart form at the highest granularity, the Min and Max totals will be wrong! This is because max(a+b) != max(a) + max(b). If you dont calculate the min/max aggregate at time of storage, the granularity will be gone at time of display.
For example, if a = (1, 2, 3) and b = (3, 2, 1), then max(a) + max(b) = 6; however the maximum at any point in time is in fact 4. The same issue applies to using min(a) + min(b).
This question already has answers here:
How to get the n next values of a generator into a list
(5 answers)
Fetch first 10 results from a list in Python
(4 answers)
Closed 9 days ago.
With linq I would
var top5 = array.Take(5);
How to do this with Python?
Slicing a list
top5 = array[:5]
To slice a list, there's a simple syntax: array[start:stop:step]
You can omit any parameter. These are all valid: array[start:], array[:stop], array[::step]
Slicing a generator
import itertools
top5 = itertools.islice(my_list, 5) # grab the first five elements
You can't slice a generator directly in Python. itertools.islice() will wrap an object in a new slicing generator using the syntax itertools.islice(generator, start, stop, step)
Remember, slicing a generator will exhaust it partially. If you want to keep the entire generator intact, perhaps turn it into a tuple or list first, like: result = tuple(generator)
import itertools
top5 = itertools.islice(array, 5)
#Shaikovsky's answer is excellent, but I wanted to clarify a couple of points.
[next(generator) for _ in range(n)]
This is the most simple approach, but throws StopIteration if the generator is prematurely exhausted.
On the other hand, the following approaches return up to n items which is preferable in many circumstances:
List:
[x for _, x in zip(range(n), records)]
Generator:
(x for _, x in zip(range(n), records))
In my taste, it's also very concise to combine zip() with xrange(n) (or range(n) in Python3), which works nice on generators as well and seems to be more flexible for changes in general.
# Option #1: taking the first n elements as a list
[x for _, x in zip(xrange(n), generator)]
# Option #2, using 'next()' and taking care for 'StopIteration'
[next(generator) for _ in xrange(n)]
# Option #3: taking the first n elements as a new generator
(x for _, x in zip(xrange(n), generator))
# Option #4: yielding them by simply preparing a function
# (but take care for 'StopIteration')
def top_n(n, generator):
for _ in xrange(n):
yield next(generator)
The answer for how to do this can be found here
>>> generator = (i for i in xrange(10))
>>> list(next(generator) for _ in range(4))
[0, 1, 2, 3]
>>> list(next(generator) for _ in range(4))
[4, 5, 6, 7]
>>> list(next(generator) for _ in range(4))
[8, 9]
Notice that the last call asks for the next 4 when only 2 are remaining. The use of the list() instead of [] is what gets the comprehension to terminate on the StopIteration exception that is thrown by next().
Do you mean the first N items, or the N largest items?
If you want the first:
top5 = sequence[:5]
This also works for the largest N items, assuming that your sequence is sorted in descending order. (Your LINQ example seems to assume this as well.)
If you want the largest, and it isn't sorted, the most obvious solution is to sort it first:
l = list(sequence)
l.sort(reverse=True)
top5 = l[:5]
For a more performant solution, use a min-heap (thanks Thijs):
import heapq
top5 = heapq.nlargest(5, sequence)
With itertools you will obtain another generator object so in most of the cases you will need another step the take the first n elements. There are at least two simpler solutions (a little bit less efficient in terms of performance but very handy) to get the elements ready to use from a generator:
Using list comprehension:
first_n_elements = [generator.next() for i in range(n)]
Otherwise:
first_n_elements = list(generator)[:n]
Where n is the number of elements you want to take (e.g. n=5 for the first five elements).
This should work
top5 = array[:5]
I am having problems with list of prolog. I want to make this:
[1,2,3,4,5]
[5,6,9,12,10]
You take a number for example 3, and you do a plus operation with the neighbours so the operation is 2+3+4 = 9. For the first and the last element you pretend there is an imaginary 1 there.
I have this now:
sum_list([A,X,B|T], [Xs|Ts]):-
add(A,X,B,Xs),
sum_list([X,B|T], Ts).
I haven't consider the first and the last element. My problem is I don't know how to get the element before and the next and then how to move on.
Note: I not allow to use meta-predicates.
Thanks.
I'm not sure how you calculated the first 5. The last 10 would be 4 + 5 + implicit 1. But following that calculation, the first element of your result should be 4 instead of 5?
Anyways, that doesn't really matter in terms of writing this code. You are actually close to your desired result. There are of course multiple ways of tackling this problem, but I think the simplest one would be to write a small 'initial' case in which you already calculate the first sum and afterwards recursively calculate all of the other sums. We can then write a case in which only 2 elements are left to calculate the last 'special' sum:
% Initial case for easily distinguishing the first sum
initial([X,Y|T],[Sum|R]) :-
Sum is X+Y+1,
others([X,Y|T],R).
% Match on 2 last elements left
others([X,Y],[Sum|[]]) :-
Sum is X+Y+1.
% Recursively keep adding neighbours
others([X,Y,Z|T],[Sum|R]) :-
Sum is X+Y+Z,
others([Y,Z|T],R).
Execution:
?- initial([1,2],Result)
Result = [4,4]
?- initial([1,2,3,4,5],Result)
Result = [4, 6, 9, 12, 10]
Note that we now don't have any cases (yet) for an empty list or a list with just one element in it. This still needs to be covered if necessary.
We are given an integer N and we need to count the total number of permutations of numbers less than N. We are also given N-1 constraints. e.g.:
if N=4 then count permutations of 0,1,2,3 given:
0>1
0>2
0>3
I thought about making a graph and then counting total no of permutation of numbers at same level and multiply it with permutations at other level.e.g.:
For above example:
0
/ | \
/ | \
1 2 3 ------> 3!=6 So total no of permutations are 6.
But I have difficulty in implementing it in C++. Also, this question was asked in Facebook hacker cup, the competition is over now. I have seen code of other people and found that they did it using DFS. Any help?
The simplest way to do this is to use a standard permutation generator and filter out each permutation that violates the conditions. This is obviously very inefficient and for larger values of N is not computable. Doing this is sort of the "booby" option that these contests have which allows the less smart contestants to complete the problem.
The skilled approach requires insight into the ways of counting combinations and permutations. To illustrate the method I will use an example. Inputs:
N = 7
2 < 4
0 < 3
3 < 6
We first simplify this by combining the dependent conditions into a single condition, as follows:
2 < 4
0 < 3 < 6
Start with the longest condition, and determine the combination count of the gaps (this is the key insight). For example, some of the combinations are as follows:
XXXX036
XXX0X36
XXX03X6
XXX036X
XX0XX36
etc.
Now, you can see there are 4 gaps: ? 0 ? 3 ? 6 ?. We need to count the possible partitions of X's in these four gaps. The number of such partitions is (7 choose 3) = 35 (do you see why?). Now, we next multiply by the combinations of the next condition, which is 2 < 4 over the remaining blank spots (the Xs). We can multiply because this condition is fully independent of the 0<3<6 condition. This combination count is (4 choose 2) = 6. The final condition has 2 values in 2 spots = 2! = 2. Thus, the answer is 35 x 6 x 2 = 420.
Now, let's make it a little more complicated. Add the condition:
1 < 6
The way this changes the calculation is that before 036 had to appear in that order. But, now, we have three possible arrangements:
1036
0136
0316
Thus, the total count is now (7 choose 4) x 3 x (3 choose 2) = 35 x 3 x 3 = 315.
So, to recap, the procedure is you isolate the problem into independent conditions. For each independent condition you calculate the combinations of partitions, then you multiply them together.
I have walked through this example manually, but you can program the same procedure.