How do I generate a string of size N, made of numbers and uppercase English letters such as:
6U1S75
4Z4UKK
U911K4
Answer in one line:
''.join(random.choice(string.ascii_uppercase + string.digits) for _ in range(N))
or even shorter starting with Python 3.6 using random.choices():
''.join(random.choices(string.ascii_uppercase + string.digits, k=N))
A cryptographically more secure version: see this post
''.join(random.SystemRandom().choice(string.ascii_uppercase + string.digits) for _ in range(N))
In details, with a clean function for further reuse:
>>> import string
>>> import random
>>> def id_generator(size=6, chars=string.ascii_uppercase + string.digits):
... return ''.join(random.choice(chars) for _ in range(size))
...
>>> id_generator()
'G5G74W'
>>> id_generator(3, "6793YUIO")
'Y3U'
How does it work ?
We import string, a module that contains sequences of common ASCII characters, and random, a module that deals with random generation.
string.ascii_uppercase + string.digits just concatenates the list of characters representing uppercase ASCII chars and digits:
>>> string.ascii_uppercase
'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
>>> string.digits
'0123456789'
>>> string.ascii_uppercase + string.digits
'ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'
Then we use a list comprehension to create a list of 'n' elements:
>>> range(4) # range create a list of 'n' numbers
[0, 1, 2, 3]
>>> ['elem' for _ in range(4)] # we use range to create 4 times 'elem'
['elem', 'elem', 'elem', 'elem']
In the example above, we use [ to create the list, but we don't in the id_generator function so Python doesn't create the list in memory, but generates the elements on the fly, one by one (more about this here).
Instead of asking to create 'n' times the string elem, we will ask Python to create 'n' times a random character, picked from a sequence of characters:
>>> random.choice("abcde")
'a'
>>> random.choice("abcde")
'd'
>>> random.choice("abcde")
'b'
Therefore random.choice(chars) for _ in range(size) really is creating a sequence of size characters. Characters that are randomly picked from chars:
>>> [random.choice('abcde') for _ in range(3)]
['a', 'b', 'b']
>>> [random.choice('abcde') for _ in range(3)]
['e', 'b', 'e']
>>> [random.choice('abcde') for _ in range(3)]
['d', 'a', 'c']
Then we just join them with an empty string so the sequence becomes a string:
>>> ''.join(['a', 'b', 'b'])
'abb'
>>> [random.choice('abcde') for _ in range(3)]
['d', 'c', 'b']
>>> ''.join(random.choice('abcde') for _ in range(3))
'dac'
This Stack Overflow quesion is the current top Google result for "random string Python". The current top answer is:
''.join(random.choice(string.ascii_uppercase + string.digits) for _ in range(N))
This is an excellent method, but the PRNG in random is not cryptographically secure. I assume many people researching this question will want to generate random strings for encryption or passwords. You can do this securely by making a small change in the above code:
''.join(random.SystemRandom().choice(string.ascii_uppercase + string.digits) for _ in range(N))
Using random.SystemRandom() instead of just random uses /dev/urandom on *nix machines and CryptGenRandom() in Windows. These are cryptographically secure PRNGs. Using random.choice instead of random.SystemRandom().choice in an application that requires a secure PRNG could be potentially devastating, and given the popularity of this question, I bet that mistake has been made many times already.
If you're using python3.6 or above, you can use the new secrets module as mentioned in MSeifert's answer:
''.join(secrets.choice(string.ascii_uppercase + string.digits) for _ in range(N))
The module docs also discuss convenient ways to generate secure tokens and best practices.
Simply use Python's builtin uuid:
If UUIDs are okay for your purposes, use the built-in uuid package.
One Line Solution:
import uuid; uuid.uuid4().hex.upper()[0:6]
In Depth Version:
Example:
import uuid
uuid.uuid4() #uuid4 => full random uuid
# Outputs something like: UUID('0172fc9a-1dac-4414-b88d-6b9a6feb91ea')
If you need exactly your format (for example, "6U1S75"), you can do it like this:
import uuid
def my_random_string(string_length=10):
"""Returns a random string of length string_length."""
random = str(uuid.uuid4()) # Convert UUID format to a Python string.
random = random.upper() # Make all characters uppercase.
random = random.replace("-","") # Remove the UUID '-'.
return random[0:string_length] # Return the random string.
print(my_random_string(6)) # For example, D9E50C
A simpler, faster but slightly less random way is to use random.sample instead of choosing each letter separately, If n-repetitions are allowed, enlarge your random basis by n times e.g.
import random
import string
char_set = string.ascii_uppercase + string.digits
print ''.join(random.sample(char_set*6, 6))
Note:
random.sample prevents character reuse, multiplying the size of the character set makes multiple repetitions possible, but they are still less likely then they are in a pure random choice. If we go for a string of length 6, and we pick 'X' as the first character, in the choice example, the odds of getting 'X' for the second character are the same as the odds of getting 'X' as the first character. In the random.sample implementation, the odds of getting 'X' as any subsequent character are only 6/7 the chance of getting it as the first character
import uuid
lowercase_str = uuid.uuid4().hex
lowercase_str is a random value like 'cea8b32e00934aaea8c005a35d85a5c0'
uppercase_str = lowercase_str.upper()
uppercase_str is 'CEA8B32E00934AAEA8C005A35D85A5C0'
From Python 3.6 on you should use the secrets module if you need it to be cryptographically secure instead of the random module (otherwise this answer is identical to the one of #Ignacio Vazquez-Abrams):
from secrets import choice
import string
''.join([choice(string.ascii_uppercase + string.digits) for _ in range(N)])
One additional note: a list-comprehension is faster in the case of str.join than using a generator expression!
A faster, easier and more flexible way to do this is to use the strgen module (pip install StringGenerator).
Generate a 6-character random string with upper case letters and digits:
>>> from strgen import StringGenerator as SG
>>> SG("[\u\d]{6}").render()
u'YZI2CI'
Get a unique list:
>>> SG("[\l\d]{10}").render_list(5,unique=True)
[u'xqqtmi1pOk', u'zmkWdUr63O', u'PGaGcPHrX2', u'6RZiUbkk2i', u'j9eIeeWgEF']
Guarantee one "special" character in the string:
>>> SG("[\l\d]{10}&[\p]").render()
u'jaYI0bcPG*0'
A random HTML color:
>>> SG("#[\h]{6}").render()
u'#CEdFCa'
etc.
We need to be aware that this:
''.join(random.choice(string.ascii_uppercase + string.digits) for _ in range(N))
might not have a digit (or uppercase character) in it.
strgen is faster in developer-time than any of the above solutions. The solution from Ignacio is the fastest run-time performing and is the right answer using the Python Standard Library. But you will hardly ever use it in that form. You will want to use SystemRandom (or fallback if not available), make sure required character sets are represented, use unicode (or not), make sure successive invocations produce a unique string, use a subset of one of the string module character classes, etc. This all requires lots more code than in the answers provided. The various attempts to generalize a solution all have limitations that strgen solves with greater brevity and expressive power using a simple template language.
It's on PyPI:
pip install StringGenerator
Disclosure: I'm the author of the strgen module.
Based on another Stack Overflow answer, Most lightweight way to create a random string and a random hexadecimal number, a better version than the accepted answer would be:
('%06x' % random.randrange(16**6)).upper()
much faster.
I thought no one had answered this yet lol! But hey, here's my own go at it:
import random
def random_alphanumeric(limit):
#ascii alphabet of all alphanumerals
r = (range(48, 58) + range(65, 91) + range(97, 123))
random.shuffle(r)
return reduce(lambda i, s: i + chr(s), r[:random.randint(0, len(r))], "")
If you need a random string rather than a pseudo random one, you should use os.urandom as the source
from os import urandom
from itertools import islice, imap, repeat
import string
def rand_string(length=5):
chars = set(string.ascii_uppercase + string.digits)
char_gen = (c for c in imap(urandom, repeat(1)) if c in chars)
return ''.join(islice(char_gen, None, length))
This method is slightly faster, and slightly more annoying, than the random.choice() method Ignacio posted.
It takes advantage of the nature of pseudo-random algorithms, and banks on bitwise and and shift being faster than generating a new random number for each character.
# must be length 32 -- 5 bits -- the question didn't specify using the full set
# of uppercase letters ;)
_ALPHABET = 'ABCDEFGHJKLMNPQRSTUVWXYZ23456789'
def generate_with_randbits(size=32):
def chop(x):
while x:
yield x & 31
x = x >> 5
return ''.join(_ALPHABET[x] for x in chop(random.getrandbits(size * 5))).ljust(size, 'A')
...create a generator that takes out 5 bit numbers at a time 0..31 until none left
...join() the results of the generator on a random number with the right bits
With Timeit, for 32-character strings, the timing was:
[('generate_with_random_choice', 28.92901611328125),
('generate_with_randbits', 20.0293550491333)]
...but for 64 character strings, randbits loses out ;)
I would probably never use this approach in production code unless I really disliked my co-workers.
edit: updated to suit the question (uppercase and digits only), and use bitwise operators & and >> instead of % and //
Use Numpy's random.choice() function
import numpy as np
import string
if __name__ == '__main__':
length = 16
a = np.random.choice(list(string.ascii_uppercase + string.digits), length)
print(''.join(a))
Documentation is here http://docs.scipy.org/doc/numpy-1.10.0/reference/generated/numpy.random.choice.html
I'd do it this way:
import random
from string import digits, ascii_uppercase
legals = digits + ascii_uppercase
def rand_string(length, char_set=legals):
output = ''
for _ in range(length): output += random.choice(char_set)
return output
Or just:
def rand_string(length, char_set=legals):
return ''.join( random.choice(char_set) for _ in range(length) )
Sometimes 0 (zero) & O (letter O) can be confusing. So I use
import uuid
uuid.uuid4().hex[:6].upper().replace('0','X').replace('O','Y')
>>> import string
>>> import random
the following logic still generates 6 character random sample
>>> print ''.join(random.sample((string.ascii_uppercase+string.digits),6))
JT7K3Q
No need to multiply by 6
>>> print ''.join(random.sample((string.ascii_uppercase+string.digits)*6,6))
TK82HK
I used this method to generate random string of length n from a -> z
import random
s = ''.join(random.choice([chr(i) for i in range(ord('a'),ord('z'))]) for _ in range(10))
Security Oriented Approach
Our recommendation for anything security related is to avoid "rolling you own" and to use the secrets module which is specifically vetted for security.
This is from the best practices section of the docs:
import string
import secrets
alphabet = string.ascii_letters + string.digits
password = ''.join(secrets.choice(alphabet) for i in range(8))
Since you specifically asked for uppercase letters, you can either substitute ascii_uppercase for ascii_letters, or just uppercase the password with:
password = password.upper()
Standard Approach Not Aiming for Security
The canonical approach to this problem (as specified) uses the choices() function in the random module:
>>> from random import choices
>>> from string import ascii_uppercase, digits
>>> population = ascii_uppercase + digits
>>> str.join('', choices(population, k=6))
'6JWF1H'
>>> import random
>>> str = []
>>> chars = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890'
>>> num = int(raw_input('How long do you want the string to be? '))
How long do you want the string to be? 10
>>> for k in range(1, num+1):
... str.append(random.choice(chars))
...
>>> str = "".join(str)
>>> str
'tm2JUQ04CK'
The random.choice function picks a random entry in a list. You also create a list so that you can append the character in the for statement. At the end str is ['t', 'm', '2', 'J', 'U', 'Q', '0', '4', 'C', 'K'], but the str = "".join(str) takes care of that, leaving you with 'tm2JUQ04CK'.
Hope this helps!
For those of you who enjoy functional python:
from itertools import imap, starmap, islice, repeat
from functools import partial
from string import letters, digits, join
from random import choice
join_chars = partial(join, sep='')
identity = lambda o: o
def irand_seqs(symbols=join_chars((letters, digits)), length=6, join=join_chars, select=choice, breakup=islice):
""" Generates an indefinite sequence of joined random symbols each of a specific length
:param symbols: symbols to select,
[defaults to string.letters + string.digits, digits 0 - 9, lower and upper case English letters.]
:param length: the length of each sequence,
[defaults to 6]
:param join: method used to join selected symbol,
[defaults to ''.join generating a string.]
:param select: method used to select a random element from the giving population.
[defaults to random.choice, which selects a single element randomly]
:return: indefinite iterator generating random sequences of giving [:param length]
>>> from tools import irand_seqs
>>> strings = irand_seqs()
>>> a = next(strings)
>>> assert isinstance(a, (str, unicode))
>>> assert len(a) == 6
>>> assert next(strings) != next(strings)
"""
return imap(join, starmap(breakup, repeat((imap(select, repeat(symbols)), None, length))))
It generates an indefinite [infinite] iterator, of joined random sequences, by first generating an indefinite sequence of randomly selected symbol from the giving pool, then breaking this sequence into length parts which is then joined, it should work with any sequence that supports getitem, by default it simply generates a random sequence of alpha numeric letters, though you can easily modify to generate other things:
for example to generate random tuples of digits:
>>> irand_tuples = irand_seqs(xrange(10), join=tuple)
>>> next(irand_tuples)
(0, 5, 5, 7, 2, 8)
>>> next(irand_tuples)
(3, 2, 2, 0, 3, 1)
if you don't want to use next for generation you can simply make it callable:
>>> irand_tuples = irand_seqs(xrange(10), join=tuple)
>>> make_rand_tuples = partial(next, irand_tuples)
>>> make_rand_tuples()
(1, 6, 2, 8, 1, 9)
if you want to generate the sequence on the fly simply set join to identity.
>>> irand_tuples = irand_seqs(xrange(10), join=identity)
>>> selections = next(irand_tuples)
>>> next(selections)
8
>>> list(selections)
[6, 3, 8, 2, 2]
As others have mentioned if you need more security then set the appropriate select function:
>>> from random import SystemRandom
>>> rand_strs = irand_seqs(select=SystemRandom().choice)
'QsaDxQ'
the default selector is choice which may select the same symbol multiple times for each chunk, if instead you'd want the same member selected at most once for each chunk then, one possible usage:
>>> from random import sample
>>> irand_samples = irand_seqs(xrange(10), length=1, join=next, select=lambda pool: sample(pool, 6))
>>> next(irand_samples)
[0, 9, 2, 3, 1, 6]
we use sample as our selector, to do the complete selection, so the chunks are actually length 1, and to join we simply call next which fetches the next completely generated chunk, granted this example seems a bit cumbersome and it is ...
(1) This will give you all caps and numbers:
import string, random
passkey=''
for x in range(8):
if random.choice([1,2]) == 1:
passkey += passkey.join(random.choice(string.ascii_uppercase))
else:
passkey += passkey.join(random.choice(string.digits))
print passkey
(2) If you later want to include lowercase letters in your key, then this will also work:
import string, random
passkey=''
for x in range(8):
if random.choice([1,2]) == 1:
passkey += passkey.join(random.choice(string.ascii_letters))
else:
passkey += passkey.join(random.choice(string.digits))
print passkey
this is a take on Anurag Uniyal 's response and something that i was working on myself.
import random
import string
oneFile = open('Numbers.txt', 'w')
userInput = 0
key_count = 0
value_count = 0
chars = string.ascii_uppercase + string.digits + string.punctuation
for userInput in range(int(input('How many 12 digit keys do you want?'))):
while key_count <= userInput:
key_count += 1
number = random.randint(1, 999)
key = number
text = str(key) + ": " + str(''.join(random.sample(chars*6, 12)))
oneFile.write(text + "\n")
oneFile.close()
import string
from random import *
characters = string.ascii_letters + string.punctuation + string.digits
password = "".join(choice(characters) for x in range(randint(8, 16)))
print password
import random
q=2
o=1
list =[r'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','s','0','1','2','3','4','5','6','7','8','9','0']
while(q>o):
print("")
for i in range(1,128):
x=random.choice(list)
print(x,end="")
Here length of string can be changed in for loop i.e for i in range(1,length)
It is simple algorithm which is easy to understand. it uses list so you can discard characters that you do not need.
I was looking at the different answers and took time to read the documentation of secrets
The secrets module is used for generating cryptographically strong random numbers suitable for managing data such as passwords, account authentication, security tokens, and related secrets.
In particularly, secrets should be used in preference to the default pseudo-random number generator in the random module, which is designed for modelling and simulation, not security or cryptography.
Looking more into what it has to offer I found a very handy function if you want to mimic an ID like Google Drive IDs:
secrets.token_urlsafe([nbytes=None])
Return a random URL-safe text string, containing nbytes random bytes. The text is Base64 encoded, so on average each byte results in approximately 1.3 characters. If nbytes is None or not supplied, a reasonable default is used.
Use it the following way:
import secrets
import math
def id_generator():
id = secrets.token_urlsafe(math.floor(32 / 1.3))
return id
print(id_generator())
Output a 32 characters length id:
joXR8dYbBDAHpVs5ci6iD-oIgPhkeQFk
I know this is slightly different from the OP's question but I expect that it would still be helpful to many who were looking for the same use-case that I was looking for.
A simple one:
import string
import random
character = string.lowercase + string.uppercase + string.digits + string.punctuation
char_len = len(character)
# you can specify your password length here
pass_len = random.randint(10,20)
password = ''
for x in range(pass_len):
password = password + character[random.randint(0,char_len-1)]
print password
I would like to suggest you next option:
import crypt
n = 10
crypt.crypt("any sring").replace('/', '').replace('.', '').upper()[-n:-1]
Paranoic mode:
import uuid
import crypt
n = 10
crypt.crypt(str(uuid.uuid4())).replace('/', '').replace('.', '').upper()[-n:-1]
Two methods :
import random, math
def randStr_1(chars:str, length:int) -> str:
chars *= math.ceil(length / len(chars))
chars = letters[0:length]
chars = list(chars)
random.shuffle(characters)
return ''.join(chars)
def randStr_2(chars:str, length:int) -> str:
return ''.join(random.choice(chars) for i in range(chars))
Benchmark :
from timeit import timeit
setup = """
import os, subprocess, time, string, random, math
def randStr_1(letters:str, length:int) -> str:
letters *= math.ceil(length / len(letters))
letters = letters[0:length]
letters = list(letters)
random.shuffle(letters)
return ''.join(letters)
def randStr_2(letters:str, length:int) -> str:
return ''.join(random.choice(letters) for i in range(length))
"""
print('Method 1 vs Method 2', ', run 10 times each.')
for length in [100,1000,10000,50000,100000,500000,1000000]:
print(length, 'characters:')
eff1 = timeit("randStr_1(string.ascii_letters, {})".format(length), setup=setup, number=10)
eff2 = timeit("randStr_2(string.ascii_letters, {})".format(length), setup=setup, number=10)
print('\t{}s : {}s'.format(round(eff1, 6), round(eff2, 6)))
print('\tratio = {} : {}\n'.format(eff1/eff1, round(eff2/eff1, 2)))
Output :
Method 1 vs Method 2 , run 10 times each.
100 characters:
0.001411s : 0.00179s
ratio = 1.0 : 1.27
1000 characters:
0.013857s : 0.017603s
ratio = 1.0 : 1.27
10000 characters:
0.13426s : 0.151169s
ratio = 1.0 : 1.13
50000 characters:
0.709403s : 0.855136s
ratio = 1.0 : 1.21
100000 characters:
1.360735s : 1.674584s
ratio = 1.0 : 1.23
500000 characters:
6.754923s : 7.160508s
ratio = 1.0 : 1.06
1000000 characters:
11.232965s : 14.223914s
ratio = 1.0 : 1.27
The performance of first method is better.
Generate random 16-byte ID containig letters, digits, '_' and '-'
os.urandom(16).translate((f'{string.ascii_letters}{string.digits}-_'*4).encode('ascii'))
import string, random
lower = string.ascii_lowercase
upper = string.ascii_uppercase
digits = string.digits
special = '!"£$%^&*.,##/?'
def rand_pass(l=4, u=4, d=4, s=4):
p = []
[p.append(random.choice(lower)) for x in range(l)]
[p.append(random.choice(upper)) for x in range(u)]
[p.append(random.choice(digits)) for x in range(d)]
[p.append(random.choice(special)) for x in range(s)]
random.shuffle(p)
return "".join(p)
print(rand_pass())
# #5U,#A4yIZvnp%51
If you want an easy-to-use but highly customisable key generator, use key-generator pypi package.
Here is the GitHub repo where you can find the complete documentation.
You can customise it to give a string jist like you want with many more options. Here's an example:
from key_generator.key_generator import generate
custom_key = generate(2, ['-', ':'], 3, 10, type_of_value = 'char', capital = 'mix', seed = 17).get_key()
print(custom_key) # ZLFdHXIUe-ekwJCu
Hope this helps :)
Disclaimer: This uses the key-generator library which I made.
Related
I am making a small program that guesses a password.
I am making this program just for the purpose of learning, I want to improve my python skills by making a program that have a real meaning.
for example:
using_characts = "abcdefghijklmnopqrstuvwxyz" # I have other characters in my alphabetic system
What I want to do is something like this:
for char in myCharacters:
print(char)
for char_1 in myCharacters:
print(char + char_1)
for char_2 in myCharacters:
print(char + char_1 + char_2)
...etc
which makes this method non dynamic, and hard in the same time.
the output should be something like this:
a
b
c
d
e
f
..etc
aa
ab
ac
..etc
ba
bb
bc
..etc
You can use itertools.product but you should really limit yourself with a small number. Generating cartesian product for higher numbers can take really long time:
from itertools import chain, product
chars = "abcdefghijklmnopqrstuvwxyz"
limit = 2
for perm in chain.from_iterable(product(chars, repeat=i) for i in range(1, limit+1)):
print("".join(perm))
a
b
c
.
.
.
aa
ab
ac
.
.
.
zy
zz
Here you go, this will work. Let me know if you want me to explain any part.
import itertools
using_characts = "abc"
for str_length in range(1,len(using_characts)+1):
for q in itertools.product(using_characts,repeat=str_length):
print("".join(q))
So, the other answers have given you code that will probably work, but I wanted to explain a general approach. This algorithm uses a stack to keep track of the next things that need to be generated, and continues generating until it reaches the maximum length that you've specified.
from collections import deque
from typing import Deque, Iterator, Optional
def generate_next_strings(chars: str, base: str = "") -> Iterator[str]:
# This function appends each letter of a given alphabet to the given base.
# At its first run, it will generate all the single-length letters of the
# alphabet, since the default base is the empty string.
for c in chars:
yield f"{base}{c}"
def generate_all_strings(chars: str, maxlen: Optional[int] = None) -> Iterator[str]:
# We "seed" the stack with a generator. This generator will produce all the
# single-length letters of the alphabet, as noted above.
stack: Deque[Iterator[str]] = deque([generate_next_strings(chars)])
# While there are still items (generators) in the stack...
while stack:
# ...pop the next one off for processing.
next_strings: Iterator[str] = stack.popleft()
# Take each item from the generator that we popped off,
for string in next_strings:
# and send it back to the caller. This is a single "result."
yield string
# If we're still generating strings -- that is, we haven't reached
# our maximum length -- we add more generators to the stack for the
# next length of strings.
if maxlen is None or len(string) < maxlen:
stack.append(generate_next_strings(chars, string))
You can try it using print("\n".join(generate_all_strings("abc", maxlen=5))).
The following code will give you all combinations with lengths between 1 and max_length - 1:
import itertools
combs = []
for i in range(1, max_length):
c = [list(x) for x in itertools.combinations(using_characts, i)]
combs.extend(c)
I need to make sequence of random strings, which increase(decrease) for alphabetic oder. For example: "ajikfk45kJDk", "bFJIPH7CDd", "c".
The simplest thing to do is to create N random strings and then sort them.
So, how do you create a random string? Well, you haven't specified what your rule is, but your three examples are strings of 1 to 12 characters taken from the set of ASCII lowercase, uppercase, and digits, so let's do that.
length = random.randrange(1, 13)
letters = random.choices(string.ascii_letters + string.digits, k=length)
string = ''.join(letters)
So, just do this N times, then sort it.
Putting it together:
chars = string.ascii_letters + string.digits
def make_string():
return ''.join(random.choices(chars, k=random.randrange(1, 13)))
def make_n_strings(n):
return sorted(make_string() for _ in range(n))
This should be simple enough that you can customize it however you want. Want case-insensitive sorting? Just add key=str.upper to the sorted. Want some other distribution of lengths? Just replace the randrange. And so on.
You can use the chr() Python 3 function in a loop while generating random number in the ASCII category range you want.
You can find all the ASCII categories here or on Wikipedia.
For exemple :
chr(99)
-> equals c
More information about the chr() function on Python 3 official documentation.
The simplest way I can think of is
from random import randint
a = ''.join(sorted([chr(randint(33,127)) for i in range(randint(1,20))], reverse = False))
print(a)
reverse = True makes it descending
There's a lot of ways to do that, and this an easy and simple example to do that in Python 3 using Ascii char codes:-
from random import randint
def generateString(minLength, maxLength):
result = "";
resultLength = randint(minLength, maxLength)
for i in range(resultLength):
charType = randint(1,3)
if(charType == 1):
#number
result += chr(randint(48, 57))
elif(charType == 2):
#upper letter
result += chr(randint(65, 90))
elif(charType == 3):
#lower letter
result += chr(randint(97, 122))
return result;
#Example
print(generateString(1,20))
Given a string which contains only lowercase letters, remove duplicate letters so that every letter appear once and only once. I must make sure your result is the smallest in lexicographical order among all possible results.
def removeDuplicates(str):
dict = {}
word = []
for i in xrange(len(str)):
if str[i] not in word:
word.append(str[i])
dict[str[i]] = i
else:
word.remove(str[i])
word.append(str[i])
dict[str[i]] = i
ind = dict.values()
# Second scan
for i in xrange(len(str)):
if str.index(str[i]) in ind:
continue
temp = dict[str[i]]
dict[str[i]] = i
lst = sorted(dict.keys(),key = lambda d:dict[d])
if ''.join(lst) < ''.join(word):
word = lst
else:
dict[str[i]] = temp
return ''.join(word)
I am not getting the desired result
print removeDuplicateLetters("cbacdcbc")
Input:
"cbacdcbc"
Output:
"abcd"
Expected:
"acdb"
Use a set. A set is a data structure similar to a list, but it removes all duplicates. You can instantiate a set by doing set(), or setting a variable to a set by using curly brackets. However, this isn't very good for instantiating empty sets, because then Python will think that it's a dictionary. So to achieve what you're doing, you could make the following function:
def removeDuplicates(string):
return ''.join(sorted(set(string)))
Dorian's answer IS the way to go for any practical application, so my addition is mostly toying around.
If a word is really long, it's more efficient to just search whether each letter in the alphabet is in the string and keep only those that are present. Explicitly,
from string import ascii_lowercase
def removeDuplicates(string):
return ''.join(letter for letter in ascii_lowercase if letter in string)
Code to test timings
import random
import timeit
def compare(string, n):
s1 = "''.join(sorted(set('{}')))".format(string)
print timeit.timeit(s1, number=n)
s2 = "from string import ascii_lowercase; ''.join(letter for letter in ascii_lowercase if letter in '{}')".format(string)
print timeit.timeit(s2, number=n)
Tests:
>>> word = 'cbacdcbc'
>>> compare(word, 1000)
0.00385931823843
0.013727678263
>>> word = ''.join(random.choice(ascii_lowercase) for _ in xrange(100000))
>>> compare(word, 1000)
2.21139290323
0.0071371927042
>>> word = 'a'*100000 + ascii_lowercase
>>> compare(word, 1000)
2.20644530225
1.63490857359
This shows that Dorian's answer should perform equally well or even better for small words, even though the speed isn't noticeable by humans. However, for very large strings, this method is much faster. Even for an edge case, where every letter is the same and the rest of the letters can only be found by transversing the whole string it performs better.
Still, Dorian's answer is more elegant and practical.
This is what makes the test succeed.
def removeDuplicates(my_string):
for char in sorted(set(my_string)):
suffix = my_string[my_string.index(char):]
if set(suffix) == set(my_string):
return char + removeDuplicates(suffix.replace(char, ''))
return ''
print removeDuplicates('cbacdcbc')
acdb
This question already has answers here:
Generate a random letter in Python
(23 answers)
Closed 10 years ago.
I want to generate a 10 alphanumeric character long string in python . So here is one part of selecting random index from a list of alphanumeric chars.
My plan :
set_list = ['a','b','c' ........] # all the way till I finish [a-zA-Z0-9]
index = random() # will use python's random generator
some_char = setlist[index]
Is there a better way of choosing a character randomly ?
The usual way is random.choice()
>>> import string
>>> import random
>>> random.choice(string.ascii_letters + string.digits)
'v'
try this:
def getCode(length = 10, char = string.ascii_uppercase +
string.digits +
string.ascii_lowercase ):
return ''.join(random.choice( char) for x in range(length))
run:
>>> import random
>>> import string
>>> getCode()
'1RZLCRBBm5'
>>> getCode(5, "mychars")
'ahssh'
if you have a list then you can do like this:
>>> set_list = ['a','b','c','d']
>>> getCode(2, ''.join(set_list))
'da'
if you want to use special symbols , you can use string's punctuation:
>>> print string.punctuation
!"#$%&'()*+,-./:;<=>?#[\]^_`{|}~
random() isn't a function on its own. The traditional way in Python 3 would be:
import random
import string
random.choice(string.ascii_letters + string.digits)
string.letters is contingent on the locale, and was removed in Python 3.
I have never used python in my life. I need to make a little fix to a given code.
I need to replace this
new_q = q[:q.index('?')] + str(random.randint(1,rand_max)) + q[q.index('?')+1:]
with something that replace all of the occurrence of ? with a random, different number.
how can I do that?
import re
import random
a = 'abc?def?ghi?jkl'
rand_max = 9
re.sub(r'\?', lambda x:str(random.randint(1,rand_max)), a)
# returns 'abc3def4ghi6jkl'
or without regexp:
import random
a = 'abc?def?ghi?jkl'
rand_max = 9
while '?' in a:
a = a[:a.index('?')] + str(random.randint(1,rand_max)) + a[a.index('?')+1:]
If you need all the numbers to be different, just using a new random number for each occurrence of ? won't be enough -- a random number might occur twice. You could use the following code in this case:
random_numbers = iter(random.sample(range(1, rand_max + 1), q.count("?")))
new_q = "".join(c if c != "?" else str(next(random_numbers)) for c in q)