在python中生成随机的weigted字符串文件

Question

i'm trying to generate a string from characters of ['A','B','C','D','E'] with length of 3900, and every character should have probability of: {'A':0.1, 'B':0.3, 'C':0.3, 'D':0.1, 'E':0.2 } in this string i wrote the following code:

from random import random
from bisect import bisect

def weighted_choice(choices):
    values, weights = zip(*choices)
    total = 0
    cum_weights = []
    for w in weights:
        total += w
        cum_weights.append(total)
    x = random() * total
    i = bisect(cum_weights, x)
    return values[i]
string_ = ''
for i in range(0,3900):
    string_ = string_ + weighted_choice([("A",10), ("B",30), ("C",30),("D",10),("E",20)])

with open("rand_file","w") as f:
        f.write(string_)

but it doesn't generate the string(file) based on the probabilities. it generates with probabilities like this:

C 0.2500264583 
B 0.2499284457 
E 0.1666428313 
D 0.0833782424 
A 0.0833758065 

probability cause the for loop runs separately every time, without considering previous results.

any help please to solve this problem ?

Answer 1

If you just use the list ['A','B','B','B','C','C','C','D','E','E'] and choose an item from it at random, you can get rid off all that weighting stuff in your code totally, and the weighting will be built in.

You can see that in the following example (yes, I don't doubt it could be written better but it's only meant to be a proof-of-concept, not production-ready, pure-as-snow-white code):

from random import random, seed

def choice(lst):
    return lst[int(random() * len(lst))];

seed()

(a, b, c, d, e, t) = (0, 0, 0, 0, 0, 0)

for i in range(1000):
    x = choice('ABBBCCCDEE')
    if (x == 'A'): a += 1
    if (x == 'B'): b += 1
    if (x == 'C'): c += 1
    if (x == 'D'): d += 1
    if (x == 'E'): e += 1
    t += 1

print ("a =", a, "which is", a * 100 / t, "%")
print ("b =", b, "which is", b * 100 / t, "%")
print ("c =", c, "which is", c * 100 / t, "%")
print ("d =", d, "which is", d * 100 / t, "%")
print ("e =", e, "which is", e * 100 / t, "%")

with the output matching (roughly) the desired distribution:

a = 101 which is 10.1 %
b = 297 which is 29.7 %
c = 299 which is 29.9 %
d = 102 which is 10.2 %
e = 201 which is 20.1 %

Now that's obviously going to be annoying if your distribution is 99.9% A and 0.1% B (it'll be a rather long string passed to choice ) but this should be adequate for the distribution you have.

Answer 2

You can generate all letters according to the weighting, then randomly shuffle them and finally join them. Something like:

from random import shuffle
N = 3900 # the string length
doc = {'A':0.1, 'B':0.3, 'C':0.3, 'D':0.1, 'E':0.2 } #weights
letters = []
for key in doc.keys():
    m = int(doc[key] * N) #generate correct number of letter
    letters.append(list(key * m))

letters = [item for sublist in letters for item in sublist] # flatten the list
shuffle(letters) # shuffle all letters randomly
result = ''.join(letters) # join all letter to make one string

print(len(result))
# 3900

Answer 3

this is actually the same as paxdiablo's solution, except a little more general (for your simple example, his solution is better. +1):

import random

choice = [("A",10), ("B",30), ("C",30),("D",10),("E",20)]
choose_from = ''.join(x * letter for letter, x in choice)

print(choose_from)
#  AAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCDDetc...

print(random.choice(choose_from))

Answer 4

This is my solution hope it helps at least a bit: import random letters = [1,2,2,2,3,3,3,4,5,5] #Each number represents a letter! for n in range(0,3900): output = letters[random.randint(0,9)] if output == 1: A += 1 import random letters = [1,2,2,2,3,3,3,4,5,5] #Each number represents a letter! for n in range(0,3900): output = letters[random.randint(0,9)] if output == 1: A += 1

And obviously you can add more if statements but i'm not sure if this is what you where asking