I have a random int in the range of 30-60 which I get using randint(30,60). Let's say it's 40. I want to split this number in exactly 7 random whole ints. So for instance [5,5,5,5,5,5,10] is a valid result. But there are many possible solutions, like this one as well [6,6,6,6,6,6,4] or [4,2,9,13,8,1,3] ...
I know there are many solutions but I am searching for a fast way to go through them. I am not trying to get every single solution but rather looking for a fast way to iterate over a lot of them in short time. One way to achieve it is to randomly pick a number (let's say in the range from 1-15) and save it to a list, then do a while loop until the sum is exactly 40. I tried that and it is not efficient at all. I think choosing a start value like [5,5,5,5,5,5,10] and altering the numbers in a precise way like "1st digit -2" and 3rd +2 to yield [3,5,7,5,5,5,10] would be a much faster solution. Does anyone know how to do that or has a good suggestion? Thanks. I prefer python 3.
A set of whole numbers that sum to a number n is called a partition of n; if order matters then it's called a composition.
Here's a reasonably fast way to produce random compositions.
import random
def random_partition(n, size):
seq = []
while size > 1:
x = random.randint(1, 1 + n - size)
seq.append(x)
n -= x
size -= 1
seq.append(n)
return seq
n = 40
for _ in range(20):
print(random_partition(n, 7))
typical output
[26, 2, 8, 1, 1, 1, 1]
[30, 2, 1, 3, 1, 1, 2]
[26, 5, 3, 1, 2, 2, 1]
[2, 25, 9, 1, 1, 1, 1]
[28, 2, 2, 2, 1, 2, 3]
[23, 1, 9, 3, 2, 1, 1]
[3, 26, 1, 7, 1, 1, 1]
[25, 1, 7, 1, 2, 1, 3]
[10, 8, 11, 5, 3, 1, 2]
[19, 16, 1, 1, 1, 1, 1]
[12, 23, 1, 1, 1, 1, 1]
[1, 14, 15, 7, 1, 1, 1]
[29, 5, 1, 1, 2, 1, 1]
[25, 1, 3, 3, 1, 2, 5]
[10, 12, 10, 4, 1, 2, 1]
[13, 4, 6, 14, 1, 1, 1]
[31, 3, 1, 1, 1, 1, 2]
[16, 11, 9, 1, 1, 1, 1]
[3, 26, 5, 3, 1, 1, 1]
[31, 2, 1, 2, 2, 1, 1]
We use 1 + n - size as the upper limit because the other size - 1 numbers are at least 1.
Here's a fairly efficient way to generate all partitions of a given integer. Note that these are ordered; you could use random.shuffle if you want to produce random compositions from these partitions.
We first print all partitions of 16 of size 5, and then we count the number of partitions of 40 of size 7 (= 2738).
This code was derived from an algorithm by Jerome Kelleher.
def partitionR(num, size):
a = [0, num] + [0] * (num - 1)
size -= 1
k = 1
while k > 0:
x = a[k - 1] + 1
y = a[k] - 1
k -= 1
while x <= y and k < size:
a[k] = x
y -= x
k += 1
a[k] = x + y
if k == size:
yield a[:k + 1]
for u in partitionR(16, 5):
print(u)
print('- ' * 32)
print(sum(1 for _ in partitionR(40, 7)))
output
[1, 1, 1, 1, 12]
[1, 1, 1, 2, 11]
[1, 1, 1, 3, 10]
[1, 1, 1, 4, 9]
[1, 1, 1, 5, 8]
[1, 1, 1, 6, 7]
[1, 1, 2, 2, 10]
[1, 1, 2, 3, 9]
[1, 1, 2, 4, 8]
[1, 1, 2, 5, 7]
[1, 1, 2, 6, 6]
[1, 1, 3, 3, 8]
[1, 1, 3, 4, 7]
[1, 1, 3, 5, 6]
[1, 1, 4, 4, 6]
[1, 1, 4, 5, 5]
[1, 2, 2, 2, 9]
[1, 2, 2, 3, 8]
[1, 2, 2, 4, 7]
[1, 2, 2, 5, 6]
[1, 2, 3, 3, 7]
[1, 2, 3, 4, 6]
[1, 2, 3, 5, 5]
[1, 2, 4, 4, 5]
[1, 3, 3, 3, 6]
[1, 3, 3, 4, 5]
[1, 3, 4, 4, 4]
[2, 2, 2, 2, 8]
[2, 2, 2, 3, 7]
[2, 2, 2, 4, 6]
[2, 2, 2, 5, 5]
[2, 2, 3, 3, 6]
[2, 2, 3, 4, 5]
[2, 2, 4, 4, 4]
[2, 3, 3, 3, 5]
[2, 3, 3, 4, 4]
[3, 3, 3, 3, 4]
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
2738
If you only care about getting an arbitrary set of numbers that add up to your total rather than an exhaustive iteration over all combinations, the following should get you what you need.
def get_parts(total, num_parts=7, max_part=15):
running_total = 0
for i in range(num_parts - 1):
remaining_total = total - running_total
upper_limit = min(max_part, remaining_total - num_parts + 1 + i)
# need to make sure there will be enough left
lower_limit = max(1, remaining_total - max_part*(num_parts - i - 1))
part = randint(lower_limit, upper_limit)
running_total += part
yield part
yield total - running_total
>>> list(get_parts(40))
[2, 7, 10, 11, 1, 4, 5]
>>> list(get_parts(40))
[7, 13, 11, 6, 1, 1, 1]
>>> list(get_parts(50, 4))
[6, 14, 15, 15]
Of course, the items in each list above is not truly random and will favor larger numbers earlier in the list and smaller numbers later. You can feed these lists through random.shuffle() if you want more of an element of pseudorandomness.
From Python Integer Partitioning with given k partitions
def partitionfunc(n,k,l=1):
'''n is the integer to partition, k is the length of partitions, l is the min partition element size'''
if k < 1:
raise StopIteration
if k == 1:
if n >= l:
yield (n,)
raise StopIteration
for i in range(l,n//k+1):
for result in partitionfunc(n-i,k-1,i):
yield (i,)+result
list(partitionfunc(40,7))
You can do a simple iteration over all possible combinations of the first 6 values (where the sum does not exceed 40), and calculate the 7th value.
for a in range(41):
for b in range(41-a):
for c in range(41-(a+b)):
for d in range(41-(a+b+c)):
for e in range(41-(a+b+c+d)):
for f in range(41-(a+b+c+d+e)):
g = 40 - (a+b+c+d+e+f)
# Do what you need to do here
You can cut the amount of time required by the loop almost in half (according to tests using timeit) by precomputing the sums:
for a in range(41):
for b in range(41-a):
ab = a + b
for c in range(41-ab):
abc = ab + c
for d in range(41-abc):
abcd = abc + d
for e in range(41-abcd):
abcde = abcd + e
for f in range(41-abcde):
g = 40 - (abcde + f)
# Do what you need to do here
Related
I want to graph the execution time of my algorithm in python, I've been trying with graph, but it doesn't work, any recommendation?
this is my algorithm:
def countingSort(lista, maximo):
maximo += 1
conteo = [0] * maximo
for n in lista:
conteo[n] += 1
aux = 0
for j in range(maximo):
for k in range(conteo[j]):
lista[aux] = j
aux += 1
return lista
lista = [3, 7, 4, 1, 1, 1, 3, 1, 2, 7, 2, 4, 2, 6, 6, 5]
print("input: ",lista)
lista = countingSort(lista, max(lista))
print("output: ",lista)
input: [3, 7, 4, 1, 1, 1, 3, 1, 2, 7, 2, 4, 2, 6, 6, 5]
output: [1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 5, 6, 6, 7, 7]
I got this code
A = [1, 1, 1, 2, 2, 2, 3, 4, 5, 6, 6, 6, 7, 8, 8, 9]
B = [0 for b in range(16)]
skipped = 0
for i in range(16):
if A[i] == A[i-1]:
skipped += 1
else:
B[i-skipped] = A[i]
print(B)
The output:
[1, 2, 3, 4, 5, 2, 7, 8, 9, 0, 0, 0, 0, 0, 0, 0]
it eliminates the doubles. But if i got an array where doubles are at more random index it fails, like:
The Array#2:
A = [1, 1, 1, 2, 2, 2, 3, 4, 5, 2, 2, 2, 7, 8, 8, 9]
The output#2
[1, 2, 3, 4, 5, 2, 7, 8, 9, 0, 0, 0, 0, 0, 0, 0]
In the output#2 there is the value 2 at index 1 and index 5, but i just want to eliminate all the doubles.
Sum:
So basically my algorithm should copy the values from Array A to Array B and eliminate all doubles independent from their index.
EDIT: i have to put it in pseudocode so i cant use convert methods or functions like SET
You can use set to do it:
A = [1, 1, 1, 2, 2, 2, 3, 4, 5, 6, 6, 6, 7, 8, 8, 9]
B = set(A)
print(B)
This code returns a set. To convert set to list you can write some_list = list(B).
Another way to do what you need:
A = [1, 1, 1, 2, 2, 2, 3, 4, 5, 6, 6, 6, 7, 8, 8, 9]
B = []
for x in A:
if x not in B:
B.append(x)
print(B)
Hey everyone how can I sort array index to index.
So I have code here
a = [0, 1, 2, 3, 4, 4, 3, 2, 1, 0, 4, 3, 2, 1, 0, 0, 1, 2, 3, 4]
how can i sort to?
[0, 4, 1, 3, 2, 2, 3, 1, 4, 0, 4, 0, 3, 1, 2, 2, 1, 3, 0, 4]
this is my idea
I could be wrong, but it sounds like you would like to return a list that is sorted like this:
[first_item, last_item, second_item, second_to_last_item, third_item, third_to_last_item,...]
I don't know of a one-line way to do that, but here's one way you could do it:
import numpy as np
a = [0, 1, 2, 3, 7] # length of list is an odd number
# create indexes that are all positive
index_values = np.repeat(np.arange(0, len(a)//2 + 1), 2) # [0,0,1,1,.....]
# make every other one negative
index_values[::2] *= -1 #[-0, 0, -1, 1, ....]
# return a[i]
[a[i] for i in index_values[1:(len(a)+1)]]
### Output: [0, 7, 1, 3, 2]
It also works for lists with even length:
a = [0, 1, 2, 3, 7, 5] # list length is an even number
index_values = np.repeat(np.arange(0, len(a)//2 + 1), 2) # [0,0,1,1,.....]
index_values[::2] *= -1 #[-0, 0, -1, 1, ....]
[a[i] for i in index_values[1:(len(a)+1)]]
### Output: [0, 5, 1, 7, 2, 3]
Here’s an almost one liner (based on #Callin’s sort method) for those that want one and that can’t/don’t want to use pandas:
from itertools import zip_longest
def custom_sort(a):
half = len(a)//2
return [n for fl in zip_longest(a[:half], a[:half-1:-1]) for n in fl if n is not None])
Examples:
custom_sort([0, 1, 2, 3, 7])
#[0, 7, 1, 3, 2]
custom_sort([0, 1, 2, 3, 7, 5])
#[0, 5, 1, 7, 2, 3]
This can be done in one line, although you’d be repeating the math to find the halfway point
[n for x in zip_longest(a[:len(a)//2], a[:(len(a)//2)-1:-1]) for n in x if n is not None]
Sometimes we want to sort in place, that is without creating a new list. Here is what I came up with
l=[1,2,3,4,5,6,7]
for i in range(1, len(l), 2):
l.insert(i, l.pop())
I know how to get ALL combinations of a list in Python with itertools, but what if I want to limit the amount of repeats?
So, if I have [1, 2, 3, 4, 5]
But I want to limit combinations to only 3 repeats of each item (with a fixed length of the final list, say 10):
[1, 1, 1, 2, 3, 3, 5, 5, 5, 4]
[1, 2, 3, 3, 3, 4, 5, 5, 4, 4]
[4, 4, 1, 1, 1, 5, 2, 2, 2, 3]
and so on. How do I do this?
This would work:
import random
L = [1, 2, 3, 4, 5]
L3 = L * 3
random.shuffle(L3)
L3[:10]
I don't know if there is a more elegant way but that seems to work:
from itertools import combinations_with_replacement
a = [1, 2, 3, 4, 5] # can contain letters as well
all_combinations = combinations_with_replacement(a, 10)
filtered_results = []
for current in all_combinations:
throw_away = False
for item in current:
if current.count(item) > 3:
throw_away = True
break
if not throw_away:
filtered_results.append( current )
for j in filtered_results:
print(j)
You could do it the following way: Use [1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5] and use itertools.combination(the list from above, 10). That would mean that each element occurs at most three times in a resulting list.
Example with smaller list:
Edited to ensure every combination occurs exactly once, the combinations occur in random order and, finally, every combination has a random order.
Also incorporated the idea of Mike Müller to just multiply the list by 3.
import itertools
import random
combinations = set()
for c in itertools.combinations(sorted([1, 2, 3] * 3), 5):
combinations.add(c)
shuffledCombinations = list(combinations)
random.shuffle(shuffledCombinations)
for combination in shuffledCombinations:
copiedCombination = list(combination)
random.shuffle(copiedCombination)
print copiedCombination
which will give:
[2, 1, 3, 2, 2]
[2, 2, 1, 1, 2]
[3, 3, 3, 1, 2]
[2, 2, 3, 2, 3]
[1, 2, 3, 1, 2]
[1, 1, 1, 2, 3]
[2, 1, 1, 1, 2]
[3, 3, 1, 1, 1]
[1, 3, 3, 3, 1]
[3, 2, 3, 2, 3]
[3, 2, 1, 2, 3]
[1, 1, 2, 3, 3]
I want to create new list according cumulative sums of numbers in a list. Input is ideal - can be splitting to subset, sum of each subset is equal. Length of subset is not equal. Number of subset is input.
Each subset of output represents increment integers [0,1,2,3,...], which replace original input. Quantity of integers is number of subsets.
Example:
number of subsets = 2
input = [1, 4, 5]
#cumsum = [1, 5, 10]
subsets = [1,5], [10]
output-subsets = [0,0], [1]
output = [0, 0, 1]
Example1:
number of subsets = 4
input = [1, 2, 3, 4, 2, 5, 1, 6]
#cumsum = [1, 3, 6, 10, 12, 17, 18, 24]
subsets = [1,3,6], [10, 12],[17, 18], [24]
output-subsets = [0, 0, 0], [1, 1], [2, 2], [3]
output = [0, 0, 0, 1, 1, 2, 2, 3]
number of subsets = 2
input = [1, 2, 3, 4, 2, 5, 1, 6]
#cumsum = [1, 3, 6, 10, 12, 17, 18, 24]
subsets = [1, 3, 6, 10, 12],[17, 18, 24]
output-subsets = [0, 0, 0, 0, 0], [1, 1, 1]
output = [0, 0, 0, 0, 0, 1, 1, 1]
I try modified SO question:
def changelist(lis, t):
total = 0
s = sum(lis)
subset = s/t
for x in lis:
total += x
i= 1
if(total <= subset):
i = 0
yield i
#changelist([input array], number of subset)
print list(changelist([1, 2, 3, 4, 2, 5, 1, 6], 4))
but only first subset is correct:
output = [0, 0, 0, 1, 1, 1, 1, 1]
I think numpy.array_split is problematic strange behaviour of numpy array_split.
I would really love any kind of explanation or help.
This should solve your problem:
def changelist (l, t):
subset = sum(l) / t
current, total = 0, 0
for x in l:
total += x
if total > subset:
current, total = current + 1, x
yield current
Examples:
>>> list(changelist([1, 4, 5], 2))
[0, 0, 1]
>>> list(changelist([1, 2, 3, 4, 2, 5, 1, 6], 4))
[0, 0, 0, 1, 1, 2, 2, 3]
>>> list(changelist([1, 2, 3, 4, 2, 5, 1, 6], 2))
[0, 0, 0, 0, 0, 1, 1, 1]
How does it work?
current stores the "id" of the current subset, total the sum of the current subset.
For each element x in your initial list l, you add its value to the current total, if this total is greater than the expected sum of each subset (subset in my code), then you know that you are in the next subset (current = current + 1) and you "reset" the total of the current subset to the actuel element (total = x).
You can use NumPy here after converting the input to an array for a vectorized solution, assuming N as the number of subsets, as listed here -
def modified_cumsum(input,N):
A = np.asarray(input).cumsum()
return np.append(False,np.in1d(A,(1+np.arange(N))*A[-1]/N))[:-1].cumsum()
Sample runs -
In [31]: N = 2 #number of subsets
...: input = [1, 4, 5]
...:
In [32]: modified_cumsum(input,N)
Out[32]: array([0, 0, 1])
In [33]: N = 4 #number of subsets
...: input = [1, 2, 3, 4, 2, 5, 1, 6]
...:
In [34]: modified_cumsum(input,N)
Out[34]: array([0, 0, 0, 1, 1, 2, 2, 3])
In [35]: N = 2 #number of subsets
...: input = [1, 2, 3, 4, 2, 5, 1, 6]
...:
In [36]: modified_cumsum(input,N)
Out[36]: array([0, 0, 0, 0, 0, 1, 1, 1])