问题
I am trying to create a function where:
- The output list is generated from random numbers from the input list
- The output list is a specified length and adds to a specified sum
ex. I specify that I want a list that is 4 in length and adds up to 10. random numbers are pulled from the input list until the criteria is satisfied.
I feel like I am approaching this problem all wrong trying to use recursion. Any help will be greatly appreciated!!!
EDIT: for more context on this problem.... Its going to be a random enemy generator.
The end goal input list will be coming from a column in a CSV called XP. (I plan to use pandas module). But this CSV will have a list of enemy names in the one column, XP in another, Health in another, etc. So the end goal is to be able to specify the total number of enemies and what the sum XP should be between those enemies and have the list generate with the appropriate information. For ex. 5 enemies with a total of 200 XP between them. The result is maybe -> Apprentice Wizard(50 xp), Apprentice Wizard(50 xp), Grung(50), Xvart(25 xp), Xvart(25 xp). The output list will actually need to include all of the row information for the selected items. And it is totally fine to have duplicated in the output as seen in this example. That will actually make more sense in the narrative of the game that this is for.
The csv --> https://docs.google.com/spreadsheets/d/1PjnN00bikJfY7mO3xt4nV5Ua1yOIsh8DycGqed6hWD8/edit?usp=sharing
import random
from random import *
lis = [1,2,3,4,5,6,7,8,9,10]
output = []
def query (total, numReturns, myList, counter):
random_index = randrange(len(myList)-1)
i = myList[random_index]
h = myList[i]
# if the problem hasn't been solved yet...
if len(output) != numReturns and sum(output) != total:
print(output)
# if the length of the list is 0 (if we just started), then go ahead and add h to the output
if len(output) == 0 and sum(output) + h != total:
output.append(h)
query (total, numReturns, myList, counter)
#if the length of the output is greater than 0
if len(output) > 0:
# if the length plus 1 is less than or equal to the number numReturns
if len(output) +1 <= numReturns:
print(output)
#if the sum of list plus h is greater than the total..then h is too big. We need to try another number
if sum(output) + h > total:
# start counter
for i in myList:# try all numbers in myList...
print(output)
print ("counter is ", counter, " and i is", i)
counter += 1
print(counter)
if sum(output) + i == total:
output.append(i)
counter = 0
break
if sum(output) + i != total:
pass
if counter == len(myList):
del(output[-1]) #delete last item in list
print(output)
counter = 0 # reset the counter
else:
pass
#if the sum of list plus h is less than the total
if sum(output) + h < total:
output.append(h) # add h to the list
print(output)
query (total, numReturns, myList, counter)
if len(output) == numReturns and sum(output) == total:
print(output, 'It worked')
else:
print ("it did not work")
query(10, 4, lis, 0)
回答1:
I guess that it would be better to get first all n-size combinations of given array which adds to specified number, and then randomly select one of them. Random selecting and checking if sum is equal to specified value, in pessimistic scenario, can last indefinitely.
from itertools import combinations as comb
from random import randint
x = [1,1,2,4,3,1,5,2,6]
def query(arr, total, size):
combs = [c for c in list(comb(arr, size)) if sum(c)==total]
return combs[randint(0, len(combs))]
#example 4-item array with items from x, which adds to 10
print(query(x, 10, 4))
回答2:
If the numbers in your input list are consecutive numbers, then this is equivalent to the problem of choosing a uniform random output list of N integers in the range [min, max], where the output list is ordered randomly and min and max are the smallest and largest number in the input list. The Python code below shows how this can be solved. It has the following advantages:
- It does not use rejection sampling.
- It chooses uniformly at random from among all combinations that meet the requirements.
It's based on an algorithm by John McClane, which he posted as an answer to another question. I describe the algorithm in another answer.
import random # Or secrets
def _getSolTable(n, mn, mx, sum):
t = [[0 for i in range(sum + 1)] for j in range(n + 1)]
t[0][0] = 1
for i in range(1, n + 1):
for j in range(0, sum + 1):
jm = max(j - (mx - mn), 0)
v = 0
for k in range(jm, j + 1):
v += t[i - 1][k]
t[i][j] = v
return t
def intsInRangeWithSum(numSamples, numPerSample, mn, mx, sum):
""" Generates one or more combinations of
'numPerSample' numbers each, where each
combination's numbers sum to 'sum' and are listed
in any order, and each
number is in the interval '[mn, mx]'.
The combinations are chosen uniformly at random.
'mn', 'mx', and
'sum' may not be negative. Returns an empty
list if 'numSamples' is zero.
The algorithm is thanks to a _Stack Overflow_
answer (`questions/61393463`) by John McClane.
Raises an error if there is no solution for the given
parameters. """
adjsum = sum - numPerSample * mn
# Min, max, sum negative
if mn < 0 or mx < 0 or sum < 0:
raise ValueError
# No solution
if numPerSample * mx < sum:
raise ValueError
if numPerSample * mn > sum:
raise ValueError
if numSamples == 0:
return []
# One solution
if numPerSample * mx == sum:
return [[mx for i in range(numPerSample)] for i in range(numSamples)]
if numPerSample * mn == sum:
return [[mn for i in range(numPerSample)] for i in range(numSamples)]
samples = [None for i in range(numSamples)]
table = _getSolTable(numPerSample, mn, mx, adjsum)
for sample in range(numSamples):
s = adjsum
ret = [0 for i in range(numPerSample)]
for ib in range(numPerSample):
i = numPerSample - 1 - ib
# Or secrets.randbelow(table[i + 1][s])
v = random.randint(0, table[i + 1][s] - 1)
r = mn
v -= table[i][s]
while v >= 0:
s -= 1
r += 1
v -= table[i][s]
ret[i] = r
samples[sample] = ret
return samples
Example:
weights=intsInRangeWithSum(
# One sample
1,
# Count of numbers per sample
4,
# Range of the random numbers
1, 5,
# Sum of the numbers
10)
# Divide by 100 to get weights that sum to 1
weights=[x/20.0 for x in weights[0]]
来源：https://stackoverflow.com/questions/60679090/python-pull-random-numbers-from-a-list-populate-a-new-list-with-a-specified-l