List all possible words with n letters - python

I want to list all possible words with n letters where the first letter can be a1 or a2, the second can be b1, b2 or b3, the third can be c1 or c2, ... Here's a simple example input-output for n=2 with each letter having 2 alternatives:
input = [["a","b"],["c","d"]]
output = ["ac", "ad", "bc", "bd"]
I tried doing this recursively by creating all possible words with the first 2 letters first, so something like this:
def go(l):
if len(l) > 2:
head = go(l[0:2])
tail = l[2:]
tail.insert(0, head)
go(tail)
elif len(l) == 2:
res = []
for i in l[0]:
for j in l[1]:
res.append(i+j)
return res
elif len(l) == 1:
return l
else:
return None
However, this becomes incredibly slow for large n or many alternatives per letter. What would be a more efficient way to solve this?
Thanks


I think you just want itertools.product here:
>>> from itertools import product
>>> lst = ['ab', 'c', 'de']
>>> words = product(*lst)
>>> list(words)
[('a', 'c', 'd'), ('a', 'c', 'e'), ('b', 'c', 'd'), ('b', 'c', 'e')]`
Or, if you wanted them joined into words:
>>> [''.join(word) for word in product(*lst)]
['acd', 'ace', 'bcd', 'bce']
Or, with your example:
>>> lst = [["a","b"],["c","d"]]
>>> [''.join(word) for word in product(*lst)]
['ac', 'ad', 'bc', 'bd']
Of course for very large n or very large sets of letters (size m), this will be slow. If you want to generate an exponentially large set of outputs (O(m**n)), that will take exponential time. But at least it has constant rather than exponential space (it generates one product at a time, instead of a giant list of all of them), and will be faster than what you were on your way to by a decent constant factor, and it's a whole lot simpler and harder to get wrong.


You can use the permutations from the built-in itertools module to achieve this, like so
>>> from itertools import permutations
>>> [''.join(word) for word in permutations('abc', 2)]
['ab', 'ac', 'ba', 'bc', 'ca', 'cb']


Generating all strings of some length with given alphabet :
test.py :
def generate_random_list(alphabet, length):
if length == 0: return []
c = [[a] for a in alphabet[:]]
if length == 1: return c
c = [[x,y] for x in alphabet for y in alphabet]
if length == 2: return c
for l in range(2, length):
c = [[x]+y for x in alphabet for y in c]
return c
if __name__ == "__main__":
for p in generate_random_list(['h','i'],2):
print p
$ python2 test.py
['h', 'h']
['h', 'i']
['i', 'h']
['i', 'i']
Next Way :
def generate_random_list(alphabet, length):
c = []
for i in range(length):
c = [[x]+y for x in alphabet for y in c or [[]]]
return c
if __name__ == "__main__":
for p in generate_random_list(['h','i'],2):
print p
Next Way :
import itertools
if __name__ == "__main__":
chars = "hi"
count = 2
for item in itertools.product(chars, repeat=count):
print("".join(item))
import itertools
print([''.join(x) for x in itertools.product('hi',repeat=2)])
Next Way :
from itertools import product
#from string import ascii_letters, digits
#for i in product(ascii_letters + digits, repeat=2):
for i in product("hi",repeat=2):
print(''.join(i))

Related

How to count sequentially using letters instead of numbers?

Is there a simple way to count using letters in Python? Meaning, 'A' will be used as 1, 'B' as 2 and so on, and after 'Z' will be 'AA', 'AB' and so on. So below code would generate:
def get_next_letter(last_letter):
return last_letter += 1 # pseudo
>>> get_next_letter('a')
'b'
>>> get_next_letter('b')
'c'
>>> get_next_letter('c')
'd'
...
>>> get_next_letter('z')
'aa'
>>> get_next_letter('aa')
'ab'
>>> get_next_letter('ab')
'ac'
...
>>> get_next_letter('az')
'ba'
>>> get_next_letter('ba')
'bb'
...
>>> get_next_letter('zz')
'aaa'

Based on #Charlie Clark's implementation of the openpyxl util get_column_letter, we can have:
def get_number_letter(n):
letters = []
while n > 0:
n, remainder = divmod(n, 26)
# check for exact division and borrow if needed
if remainder == 0:
remainder = 26
n-= 1
letters.append(chr(remainder+64))
return ''.join(reversed(letters))
This gives the letter representation of a number. Now, to increment, we need the reverse. Based on that logic (and the general number base logic), I wrote:
def number_from_string(letters):
n = 0
for i, c in enumerate(reversed(letters)):
n += (ord(c)-64)*26**i
return n
And now we can combine them to:
def get_next_letter(letters):
return get_number_letter(number_from_string(letters)+1)
Original answer:
This kind of "counting" is very similar to how Excel indexes its columns. Therefore it is possible to take advantage of the openpyxl package, which has two utility functions: get_column_letter and column_index_from_string:
from openpyxl.utils import get_column_letter, column_index_from_string
def get_next_letter(letters):
return get_column_letter(column_index_from_string(letters)+1)
NOTE: as this is based on Excel, it is limited to count up-to 'ZZZ'. i.e. calling the function with 'ZZZ' will raise an exception.
Output example for both implementations:
>>> get_next_letter('A')
'B'
>>> get_next_letter('Z')
'AA'
>>> get_next_letter('BD')
'BE'

Let's start with the simple special case of getting just the single-character strings.
from string import ascii_lowercase
def population():
yield from ascii_lowercase
Then
>>> x = population()
>>> list(x)
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
>>> x = population()
>>> next(x)
'a'
>>> next(x)
'b'
So we'd like to add the two-character sequences next:
from string import ascii_lowercase
from itertools import product
def population():
yield from ascii_lowercase
yield from map(''.join, product(ascii_lowercase, repeat=2)
Note that the single-character strings are just a special case of the product with repeat=1, so we could have written
from string import ascii_lowercase
from itertools import product
def population():
yield from map(''.join, product(ascii_lowercase, repeat=1)
yield from map(''.join, product(ascii_lowercase, repeat=2)
We can write this with a loop:
def population():
for k in range(1, 3):
yield from map(''.join, product(ascii_lowercase, repeat=k)
but we don't necessarily want an artificial upper limit on what strings we can produce; we want, in theory, to produce all of them. For that, we replace range with itertools.count.
from string import ascii_lowercase
from itertools import product, count
def population():
for k in count(1):
yield from map(''.join, product(ascii_lowercase, repeat=k)

all proposed are just way too complicated
I came up with below, using a recursive call,
this is it!
def getNextLetter(previous_letter):
"""
'increments' the provide string to the next letter recursively
raises TypeError if previous_letter is not a string
returns "a" if provided previous_letter was emtpy string
"""
if not isinstance(previous_letter, str):
raise TypeError("the previous letter should be a letter, doh")
if previous_letter == '':
return "a"
for letter_location in range(len(previous_letter) - 1, -1, -1):
if previous_letter[letter_location] == "z":
return getNextLetter(previous_letter[:-1])+"a"
else:
return (previous_letter[:-1])+chr(ord(previous_letter[letter_location])+1)
# EOF

How to generate combination of characters in a string at a particular position?

I have a string list :
li = ['a', 'b', 'c', 'd']
Using the following code in Python, I generated all the possible combination of characters for list li and got a result of 256 strings.
from itertools import product
li = ['a', 'b', 'c', 'd']
for comb in product(li, repeat=4):
print(''.join(comb))
Say for example, I know the character of the second and fourth position of the string in the list li which is 'b' and 'c'.
So the result will be a set of only 16 strings which is :
abac
abbc
abcc
abdc
bbac
bbbc
bbcc
bbdc
cbac
cbbc
cbcc
cbdc
dbac
dbbc
dbcc
dbdc
How to get this result? Is there a Pythonic way to achieve this?
Thanks.
Edit : My desired size of list li is a to z and the value for repeat is 13. When I tried the above code, compiler throwed memory error!

Use list comprehension:
from itertools import product
li = ['a', 'b', 'c', 'd']
combs = [list(x) for x in product(li, repeat=4)]
selected_combs = [comb for comb in combs if (comb[1] == 'b' and comb[3] == 'c')]
print(["".join(comb) for comb in selected_combs])
# ['abac', 'abbc', 'abcc', 'abdc', 'bbac', 'bbbc', 'bbcc', 'bbdc', 'cbac', 'cbbc', 'cbcc', 'cbdc', 'dbac', 'dbbc', 'dbcc', 'dbdc']
To save memory in case you do not need all the combinations combs, you can simply do:
li = ['a', 'b', 'c', 'd']
selected_combs = [comb for comb in product(li, repeat=4) if (comb[1] == 'b' and comb[3] == 'c')]
print(["".join(comb) for comb in selected_combs])

def permute(s):
out = []
if len(s) == 1:
return s
else:
for i,let in enumerate(s):
for perm in permute(s[:i] + s[i+1:]):
out += [let + perm]
return out
per=permute(['a', 'b', 'c', 'd'])
print(per)
Do you want this?

How to find all possible permutation of two strings within constant length

I want to find all possible permutation of two list of strings within a constant length (5). Assume list_1 = ["A"] and list_2 = ["BB"].
All possible combinations are:
A A A A A
A A A BB
A A BB A
A BB A A
BB A A A
A BB BB
BB A BB
BB BB A
I was trying to implement it with the code below, but I am not sure how to define the length 5 for it.
import itertools
from itertools import permutations
list_1 = ["A"]
list_2 = ["BB"]
unique_combinations = []
permut = itertools.permutations(list_1, 5)
for comb in permut:
zipped = zip(comb, list_2)
unique_combinations.append(list(zipped))
print(unique_combinations)

Use recursion:
list_1 = ["A"]
list_2 = ["BB"]
size = 5
strs = list_1 + list_2
res = []
def helper(strs, size, cur, res):
if size == 0:
res.append(cur)
return
if size < 0:
return
for s in strs:
helper(strs, size-len(s), cur+[s], res)
helper(strs, size, [], res)
print(res)
No recursion:
list_1 = ["A"]
list_2 = ["BB"]
size = 5
strs = list_1 + list_2
res = []
q = [[]]
while q:
t = q.pop()
for s in strs:
cur = t + [s]
cursize = len(''.join(cur))
if cursize == size:
res.append(cur)
elif cursize < size:
q.append(cur)
print(res)

You could do the following:
import itertools
unique_combinations = []
permut = itertools.product(["A","B"], repeat=5)
for comb in permut:
l = "".join(comb)
c_bb = l.count("BB")
c_a = l.count("A")
if 2*c_bb + c_a == 5:
unique_combinations.append(l)
print(unique_combinations)
This will give:
['AAAAA', 'AAABB', 'AABBA', 'ABBAA', 'ABBBB', 'BBAAA', 'BBABB', 'BBBBA']
First find all the string-like of length 5 consists of 5 elements, either "A" or "B". Then use string.count to count the occurrences of each substring you are interested in and if it is equal 5 save it.

You can use itertools.product to find all the possible combinations of 'A' and 'BB' (of repeat from 3 to 5, as these are the number of elements in the acceptable answers), and then filter than based on their total length being 5 characters:
import itertools
all_options = []
for i in range(3,6):
all_options += list(itertools.product(['A', 'BB'], repeat=i))
all_options = [i for i in all_options if len(''.join(i)) == 5]
print(all_options)
Output:
[('A', 'BB', 'BB'), ('BB', 'A', 'BB'), ('BB', 'BB', 'A'), ('A', 'A', 'A', 'BB'), ('A', 'A', 'BB', 'A'), ('A', 'BB', 'A', 'A'), ('BB', 'A', 'A', 'A'), ('A', 'A', 'A', 'A', 'A')]

You need a recursive function like this:
def f(words, N, current = ""):
if(len(current)<N):
for i in words:
f(words, N, current+i)
elif(len(current)==N):
print(current)
f(["A", "BB"], 5)
Edit: Unfortunately, this function return duplicates, if two or more words in your list share the same letter. So the correct approach should be to fill a list with all return and then eliminate the duplicates.

All possible substring in Python

Can anyone help me with finding all the possible substring in a string using python?
E.g:
string = 'abc'
output
a, b, c, ab, bc, abc
P.s : I am a beginner and would appreciate if the solution is simple to understand.

You could do something like:
for length in range(len(string)):
for index in range(len(string) - length):
print(string[index:index+length+1])
Output:
a
b
c
ab
bc
abc

else one way is using the combinations
from itertools import combinations
s = 'abc'
[
''.join(x)
for size in range(1, len(s) + 1)
for x in (combinations(s, size))
]
Out
['a', 'b', 'c', 'ab', 'ac', 'bc', 'abc']

Every substring contains a unique start index and a unique end index (which is greater than the start index). You can use two for loops to get all unique combinations of indices.
def all_substrings(s):
all_subs = []
for end in range(1, len(s) + 1):
for start in range(end):
all_subs.append(s[start:end])
return all_subs
s = 'abc'
print(all_substrings(s)) # prints ['a', 'ab', 'b', 'abc', 'bc', 'c']

You can do like:
def subString(s):
for i in range(len(s)):
for j in range(i+1,len(s)+1):
print(s[i:j])
subString("aashu")
a
aa
aas
aash
aashu
a
as
ash
ashu
s
sh
shu
h
hu
u

Python Brute Force algorithm [closed]

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I need to generate every possible combination from a given charset to a given range.
Like,
charset=list(map(str,"abcdefghijklmnopqrstuvwxyz"))
range=10
And the out put should be,
[a,b,c,d..................,zzzzzzzzzy,zzzzzzzzzz]
I know I can do this using already in use libraries.But I need to know how they really works.If anyone can give me a commented code of this kind of algorithm in Python or any programming language readable,I would be very grateful.

Use itertools.product, combined with itertools.chain to put the various lengths together:
from itertools import chain, product
def bruteforce(charset, maxlength):
return (''.join(candidate)
for candidate in chain.from_iterable(product(charset, repeat=i)
for i in range(1, maxlength + 1)))
Demonstration:
>>> list(bruteforce('abcde', 2))
['a', 'b', 'c', 'd', 'e', 'aa', 'ab', 'ac', 'ad', 'ae', 'ba', 'bb', 'bc', 'bd', 'be', 'ca', 'cb', 'cc', 'cd', 'ce', 'da', 'db', 'dc', 'dd', 'de', 'ea', 'eb', 'ec', 'ed', 'ee']
This will efficiently produce progressively larger words with the input sets, up to length maxlength.
Do not attempt to produce an in-memory list of 26 characters up to length 10; instead, iterate over the results produced:
for attempt in bruteforce(string.ascii_lowercase, 10):
# match it against your password, or whatever
if matched:
break

If you REALLY want to brute force it, try this, but it will take you a ridiculous amount of time:
your_list = 'abcdefghijklmnopqrstuvwxyz'
complete_list = []
for current in xrange(10):
a = [i for i in your_list]
for y in xrange(current):
a = [x+i for i in your_list for x in a]
complete_list = complete_list+a
On a smaller example, where list = 'ab' and we only go up to 5, this prints the following:
['a', 'b', 'aa', 'ba', 'ab', 'bb', 'aaa', 'baa', 'aba', 'bba', 'aab', 'bab', 'abb', 'bbb', 'aaaa', 'baaa', 'abaa', 'bbaa', 'aaba', 'baba', 'abba', 'bbba', 'aaab', 'baab', 'abab', 'bbab', 'aabb', 'babb', 'abbb', 'bbbb', 'aaaaa', 'baaaa', 'abaaa', 'bbaaa', 'aabaa', 'babaa', 'abbaa', 'bbbaa', 'aaaba','baaba', 'ababa', 'bbaba', 'aabba', 'babba', 'abbba', 'bbbba', 'aaaab', 'baaab', 'abaab', 'bbaab', 'aabab', 'babab', 'abbab', 'bbbab', 'aaabb', 'baabb', 'ababb', 'bbabb', 'aabbb', 'babbb', 'abbbb', 'bbbbb']

I found another very easy way to create dictionaries using itertools.
generator=itertools.combinations_with_replacement('abcd', 4 )
This will iterate through all combinations of 'a','b','c' and 'd' and create combinations with a total length of 1 to 4. ie. a,b,c,d,aa,ab.........,dddc,dddd. generator is an itertool object and you can loop through normally like this,
for password in generator:
''.join(password)
Each password is infact of type tuple and you can work on them as you normally do.

If you really want a bruteforce algorithm, don't save any big list in the memory of your computer, unless you want a slow algorithm that crashes with a MemoryError.
You could try to use itertools.product like this :
from string import ascii_lowercase
from itertools import product
charset = ascii_lowercase # abcdefghijklmnopqrstuvwxyz
maxrange = 10
def solve_password(password, maxrange):
for i in range(maxrange+1):
for attempt in product(charset, repeat=i):
if ''.join(attempt) == password:
return ''.join(attempt)
solved = solve_password('solve', maxrange) # This worked for me in 2.51 sec
itertools.product(*iterables) returns the cartesian products of the iterables you entered.
[i for i in product('bar', (42,))] returns e.g. [('b', 42), ('a', 42), ('r', 42)]
The repeat parameter allows you to make exactly what you asked :
[i for i in product('abc', repeat=2)]
Returns
[('a', 'a'),
('a', 'b'),
('a', 'c'),
('b', 'a'),
('b', 'b'),
('b', 'c'),
('c', 'a'),
('c', 'b'),
('c', 'c')]
Note:
You wanted a brute-force algorithm so I gave it to you. Now, it is a very long method when the password starts to get bigger because it grows exponentially (it took 62 sec to find the word 'solved').

itertools is ideally suited for this:
itertools.chain.from_iterable((''.join(l)
for l in itertools.product(charset, repeat=i))
for i in range(1, maxlen + 1))

A solution using recursion:
def brute(string, length, charset):
if len(string) == length:
return
for char in charset:
temp = string + char
print(temp)
brute(temp, length, charset)
Usage:
brute("", 4, "rce")

import string, itertools
#password = input("Enter password: ")
password = "abc"
characters = string.printable
def iter_all_strings():
length = 1
while True:
for s in itertools.product(characters, repeat=length):
yield "".join(s)
length +=1
for s in iter_all_strings():
print(s)
if s == password:
print('Password is {}'.format(s))
break

Simple solution using the itertools and string modules
# modules to easily set characters and iterate over them
import itertools, string
# character limit so you don't run out of ram
maxChar = int(input('Character limit for password: '))
# file to save output to, so you can look over the output without using so much ram
output_file = open('insert filepath here', 'a+')
# this is the part that actually iterates over the valid characters, and stops at the
# character limit.
x = list(map(''.join, itertools.permutations(string.ascii_lowercase, maxChar)))
# writes the output of the above line to a file
output_file.write(str(x))
# saves the output to the file and closes it to preserve ram
output_file.close()
I piped the output to a file to save ram, and used the input function so you can set the character limit to something like "hiiworld". Below is the same script but with a more fluid character set using letters, numbers, symbols, and spaces.
import itertools, string
maxChar = int(input('Character limit for password: '))
output_file = open('insert filepath here', 'a+')
x = list(map(''.join, itertools.permutations(string.printable, maxChar)))
x.write(str(x))
x.close()

from random import choice
sl = 4 #start length
ml = 8 #max length
ls = '9876543210qwertyuiopasdfghjklzxcvbnm' # list
g = 0
tries = 0
file = open("file.txt",'w') #your file
for j in range(0,len(ls)**4):
while sl <= ml:
i = 0
while i < sl:
file.write(choice(ls))
i += 1
sl += 1
file.write('\n')
g += 1
sl -= g
g = 0
print(tries)
tries += 1
file.close()

Try this:
import os
import sys
Zeichen=["a","b","c","d","e","f","g","h"­,"i","j","k","l","m","n","o","p","q­","r","s","­;t","u","v","w","x","y","z"]
def start(): input("Enter to start")
def Gen(stellen): if stellen==1: for i in Zeichen: print(i) elif stellen==2: for i in Zeichen: for r in Zeichen: print(i+r) elif stellen==3: for i in Zeichen: for r in Zeichen: for t in Zeichen: print(i+r+t) elif stellen==4: for i in Zeichen: for r in Zeichen: for t in Zeichen: for u in Zeichen: print(i+r+t+u) elif stellen==5: for i in Zeichen: for r in Zeichen: for t in Zeichen: for u in Zeichen: for o in Zeichen: print(i+r+t+u+o) else: print("done")
#*********************
start()
Gen(1)
Gen(2)
Gen(3)
Gen(4)
Gen(5)

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