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Python: get random number from range (or two ranges) and single number

I have that little script that generates me a random number from sets of ranges like:
(1-6).
Now I would like to read not only one range, but add another range or single numbers like:
('1-6', '10', '12-17')
My Script looks like that so far, "choose_frame_set" is the parameter for the ranges:
def retime_frames():
# choose frame set from presets
fr_range = hou.evalParm("choose_frame_set")
# extract frame range from list into readable string
fr_sequence = []
for frames in fr_range.split(','):
if '-' in frames:
a, b = frames.split('-')
a, b = int(a), int(b)
fr_sequence.extend(range(a, b + 1))
else:
a = int(frames)
# set random frame from frame range
random_frame = str(random.sample(fr_sequence, 1))
# replace special characters
for char in '[]':
random_frame = random_frame.replace(char,'')
hou.parm('retime_frame').set(random_frame)
print random_frame
return random_frame
How can I add extra ranges or single numbers, to have more control about what numbers I randomize?

You can use the choice method:
from random import choice
def ran(*r):
num = []
for n in r:
num += [m for m in range(n[0],n[1]+1)]
return choice(num)
print(ran((1,4),(5,7),(12,13)))
All possible output: 1,2,3,4,5,6,7,12,13

Referring to splash58's comment under my question, the solution is:
Seems, your code works if change else: a = int(frames) to else:
fr_sequence.append(int(frames))

Trying to optimize quicksort for larger files

Does anyone know how I can optimize this code better to run larger files. It works with smaller inputs, but I need it to run a file with over 200,000 words. Any suggestions?
Thank you.
import random
import re
def quick_sort(a,i,n):
if n <= 1:
return
mid = (len(a)) // 2
x = a[random.randint(0,len(a)-1)]
p = i - 1
j = i
q = i + n
while j < q:
if a[j] < x:
p = p + 1
a[j],a[p] = a[p],a[j]
j = j + 1
elif a[j] > x:
q = q - 1
a[j],a[q] = a[q],a[j]
else:
j = j + 1
quick_sort(a,i,p-i+1)
quick_sort(a,q,n-(q-i))
file_name = input("Enter file name: ")
my_list = []
with open(file_name,'r') as f:
for line in f:
line = re.sub('[!#?,.:";\']', '', line).lower()
token = line.split()
for t in token:
my_list.append(t)
a = my_list
quick_sort(a,0,len(my_list))
print("List After Calling Quick Sort: ",a)

Your random selection of an index to use for your pivot x is using the whole size of the input list a, not just the part you're supposed to be sorting on the current call. This means that very often your pivot won't be in the current section at all, and so you won't be able to usefully reduce your problem (because all of the values will be on the same side of the pivot). This leads to lots and lots of recursion, and for larger inputs you'll almost always hit the recursion cap.
The fix is simple, just change how you get x:
x = a[random.randrange(i, i+n)]
I like randrange a lot better than randint, but you could use randint(i, i+n-1) if you feel the other way.

Must you use a quicksort? If you can use a heapq or PriorityQueue, the .get/(.pop()) methods automatically implement the sort:
import sys
from queue import PriorityQueue
pq = PriorityQueue()
inp = open(sys.stdin.fileno(), newline='\n')
#inp = ['dag', 'Rug', 'gob', 'kex', 'mog', 'Wes', 'pox', 'sec', 'ego', 'wah'] # for testing
for word in inp:
word = word.rstrip('\n')
pq.put(word)
while not pq.empty():
print(pq.get())
Then test with some large random word input or file e.g.:
shuf /usr/share/dict/words | ./word_pq.py
where shuf is Gnu /usr/local/bin/shuf.

How can I optimize this function which is related to reversal of string?

I have a string: "String"
The first thing you do is reverse it: "gnirtS"
Then you will take the string from the 1st position and reverse it again: "gStrin"
Then you will take the string from the 2nd position and reverse it again: "gSnirt"
Then you will take the string from the 3rd position and reverse it again: "gSntri"
Continue this pattern until you have done every single position, and then you will return the string you have created. For this particular string, you would return: "gSntir"
And I have to repeat this entire procedure for x times where the string and x can be very big . (million or billion)
My code is working fine for small strings but it's giving timeout error for very long strings.
def string_func(s,x):
def reversal(st):
n1=len(st)
for i in range(0,n1):
st=st[0:i]+st[i:n1][::-1]
return st
for i in range(0,x):
s=reversal(s)
return s

This linear implementation could point you in the right direction:
from collections import deque
from itertools import cycle
def special_reverse(s):
d, res = deque(s), []
ops = cycle((d.pop, d.popleft))
while d:
res.append(next(ops)())
return ''.join(res)
You can recognize the slice patterns in the following examples:
>>> special_reverse('123456')
'615243'
>>> special_reverse('1234567')
'7162534'

This works too:
my_string = "String"
my_string_len = len(my_string)
result = ""
for i in range(my_string_len):
my_string = my_string[::-1]
result += my_string[0]
my_string = my_string[1:]
print(result)
And this, though it looks spaghetti :D
s = "String"
lenn = len(s)
resultStringList = []
first_half = list(s[0:int(len(s) / 2)])
second_half = None
middle = None
if lenn % 2 == 0:
second_half = list(s[int(len(s) / 2) : len(s)][::-1])
else:
second_half = list(s[int(len(s) / 2) + 1 : len(s)][::-1])
middle = s[int(len(s) / 2)]
lenn -= 1
for k in range(int(lenn / 2)):
print(k)
resultStringList.append(second_half.pop(0))
resultStringList.append(first_half.pop(0))
if middle != None:
resultStringList.append(middle)
print(''.join(resultStringList))

From the pattern of the original string and the result I constructed this algorithm. It has minimal number of operations.
str = 'Strings'
lens = len(str)
lensh = int(lens/2)
nstr = ''
for i in range(lensh):
nstr = nstr + str[lens - i - 1] + str[i]
if ((lens % 2) == 1):
nstr = nstr + str[lensh]
print(nstr)

or a short version using iterator magic:
def string_func(s):
ops = (iter(reversed(s)), iter(s))
return ''.join(next(ops[i % 2]) for i in range(len(s)))
which does the right thing for me, while if you're happy using some library code, you can golf it down to:
from itertools import cycle, islice
def string_func(s):
ops = (iter(reversed(s)), iter(s))
return ''.join(map(next, islice(cycle(ops), len(s))))
my original version takes 80microseconds for a 512 character string, this updated version takes 32µs, while your version took 290µs and schwobaseggl's solution is about 75µs.
I've had a play in Cython and I can get runtime down to ~0.5µs. Measuring this under perf_event_open I can see my CPU is retiring ~8 instructions per character, which seems pretty good, while a hard-coded loop in C gets this down to ~4.5 instructions per ASCII char. These don't seem to be very "Pythonic" solutions so I'll leave them out of this answer. But included this paragraph to show that the OP has options to make things faster, and that running this a billion times on a string consisting of ~500 characters will still take hundreds of seconds even with relatively careful C code.

python intelligent hexadecimal numbers generator

I want to be able to generate 12 character long chain, of hexadecimal, BUT with no more than 2 identical numbers duplicate in the chain: 00 and not 000
Because, I know how to generate ALL possibilites, including 00000000000 to FFFFFFFFFFF, but I know that I won't use all those values, and because the size of the file generated with ALL possibilities is many GB long, I want to reduce the size by avoiding the not useful generated chains.
So my goal is to have results like 00A300BF8911 and not like 000300BF8911
Could you please help me to do so?
Many thanks in advance!

if you picked the same one twice, remove it from the choices for a round:
import random
hex_digits = set('0123456789ABCDEF')
result = ""
pick_from = hex_digits
for digit in range(12):
cur_digit = random.sample(hex_digits, 1)[0]
result += cur_digit
if result[-1] == cur_digit:
pick_from = hex_digits - set(cur_digit)
else:
pick_from = hex_digits
print(result)
Since the title mentions generators. Here's the above as a generator:
import random
hex_digits = set('0123456789ABCDEF')
def hexGen():
while True:
result = ""
pick_from = hex_digits
for digit in range(12):
cur_digit = random.sample(hex_digits, 1)[0]
result += cur_digit
if result[-1] == cur_digit:
pick_from = hex_digits - set(cur_digit)
else:
pick_from = hex_digits
yield result
my_hex_gen = hexGen()
counter = 0
for result in my_hex_gen:
print(result)
counter += 1
if counter > 10:
break
Results:
1ECC6A83EB14
D0897DE15E81
9C3E9028B0DE
CE74A2674AF0
9ECBD32C003D
0DF2E5DAC0FB
31C48E691C96
F33AAC2C2052
CD4CEDADD54D
40A329FF6E25
5F5D71F823A4
You could also change the while true loop to only produce a certain number of these based on a number passed into the function.

I interpret this question as, "I want to construct a rainbow table by iterating through all strings that have the following qualities. The string has a length of 12, contains only the characters 0-9 and A-F, and it never has the same character appearing three times in a row."
def iter_all_strings_without_triplicates(size, last_two_digits = (None, None)):
a,b = last_two_digits
if size == 0:
yield ""
else:
for c in "0123456789ABCDEF":
if a == b == c:
continue
else:
for rest in iter_all_strings_without_triplicates(size-1, (b,c)):
yield c + rest
for s in iter_all_strings_without_triplicates(12):
print(s)
Result:
001001001001
001001001002
001001001003
001001001004
001001001005
001001001006
001001001007
001001001008
001001001009
00100100100A
00100100100B
00100100100C
00100100100D
00100100100E
00100100100F
001001001010
001001001011
...
Note that there will be several hundred terabytes' worth of values outputted, so you aren't saving much room compared to just saving every single string, triplicates or not.

import string, random
source = string.hexdigits[:16]
result = ''
while len(result) < 12 :
idx = random.randint(0,len(source))
if len(result) < 3 or result[-1] != result[-2] or result[-1] != source[idx] :
result += source[idx]

You could extract a random sequence from a list of twice each hexadecimal digits:
digits = list('1234567890ABCDEF') * 2
random.shuffle(digits)
hex_number = ''.join(digits[:12])
If you wanted to allow shorter sequences, you could randomize that too, and left fill the blanks with zeros.
import random
digits = list('1234567890ABCDEF') * 2
random.shuffle(digits)
num_digits = random.randrange(3, 13)
hex_number = ''.join(['0'] * (12-num_digits)) + ''.join(digits[:num_digits])
print(hex_number)

You could use a generator iterating a window over the strings your current implementation yields. Sth. like (hex_str[i:i + 3] for i in range(len(hex_str) - window_size + 1)) Using len and set you could count the number of different characters in the slice. Although in your example it might be easier to just compare all 3 characters.

You can create an array from 0 to 255, and use random.sample with your list to get your list

Another IndexError with python

This is supposed to become a random name generator in the end, all the random part is working. Only problem is that it is REALLY random, getting weird stuff like aaaaaaaa etc.
So I'm trying to add a rule to not allow 2 vowels after each other (same goes with consonants).
So yeah, guys please help me out here. I've been looking throu' this code for 2 hours now and I cant find the problem.
Just pasting my entire code here.
import random
import string
import numpy as np
from sys import argv
import csv
# abcdefghijklmnopqrstuvwxyz
# Example output: floke fl0ke flok3 fl0k3
#
class facts:
kons = list('bcdfghjklmnpqrstvwxz') #20
voks = list('aeiouy') #6
abc = list('abcdefghijklmnopqrstuvwxyz')
def r_trfa(): #True Or False (1/0)
x = random.randrange(0, 2)
return x;
def r_kons(): #Konsonant
y = random.randrange(0, 20)
x = facts.kons[y]
return x;
def r_vok(): #Vokal
y = random.randrange(0, 6)
x = facts.voks[y]
return x;
def r_len(): #Langd
x = random.randrange(4, 8)
return x;
def r_type():
x = random.randrange(1, 4)
return x;
def r_structure(length): #Skapar strukturen
y = r_type()
if y == 0:
no1 = 1
else:
no1 = 2
i = 0
x = [no1]
y = r_type()
if not no1 == y:
x.append(y)
while i < length:
y = r_type()
if not x[i] == y:
x.append(y)
i = i + 1
x2 = list(x)
return x2;
def name(): #Final product
struct = r_structure(r_len())
name = struct

You've got several bugs. For example, you're checking the value y against 0 even though it is always in the range 1-4, probably unintended behavior. Furthermore, you never actually call a function that gets you a character, and you never create a string. Thus it's not clear what you're trying to do.
Here's how I'd rewrite things based on my guess of what you want to do.
import random, itertools
voks = frozenset('aeiouy')
abc = 'abcdefghijklmnopqrstuvwxyz'
def r_gen():
last=None #both classes ok
while 1:
new = random.choice(abc)
if (new in voks) != last:
yield new
last = (new in voks)
def name(): #Final product
length = random.randrange(4, 8)
return ''.join(itertools.islice(r_gen(), length))

The problem you're having is that your loop increments i always, but only adds an additional value to your x list if the random value doesn't match x[i]. This means that if you get several matches in a row, i may become larger than the largest index into x and so you'll get an IndexError exception.
I'm not entirely sure I understand what you're trying to do, but I think this will do something similar to your current r_structure function:
def r_structure(length):
"""Returns a list of random "types", avoiding any immediate repeats"""
x = [r_type()]
while len(x) < length:
y = r_type()
if y != x[-1]: # check against the last item in the list
x.append(y)
return x
If your goal is simply to randomly generate a sequence of alternating vowels and consonants, there's an easier way than what you seem to be doing. First off, you can use random.choice to pick your characters. Further, rather than picking many letters and rejecting ones that are of the wrong type, you can simply pick from one string, then pick from the other, for as long as you need:
import random
def alternating_characters(length):
characters = ["aeiouy", "bcdfghjklmnpqrstvwxz"]
char_type = random.randrange(2) # pick a random letter type to start with
results = []
while len(char_list) < length:
results.append(random.choice(characters[char_type])) # pick random char
char_type = 1-char_type # pick from the other list next time
return "".join(char_list)

Well it's unclear what you want to do.. As the conditions on vowels and consonants is the same, so why do you need to differentiate between them?
So all you need to do is take a random letter and check that it doesn't match with the last letter.
Here's some code:
import random
abc = 'abcdefghijklmnopqrstuvwxyz'
def gen_word(length):
last = ''
while length > 0:
l = random.choice(abc)
if l != last:
length -= 1
yield l
if __name__ == '__main__':
word = ''.join(gen_word(10))
print word