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Каков наиболее эффективный способ создания каждой возможной комбинации букв алфавита?Python: любая возможная комбинация букв
E.g. a, b, c, d, e ... z, aa, ab, ac, ad, ae ... ect
Я написал код, который успешно производит этот результат, но я чувствую, что он слишком неэффективен при создавая желаемый результат.
Любые идеи?
Я также попытался объяснить, что я сделал
#import module time
import time
#set starting variable "gen", make gen list.
gen =''
gen = list(gen)
#expermintal alphabet, not actually used
alpha ='abcdefghijklmnopqrstuvwxyz'
alpha = list(alpha)
#Approx. on time taken to complete x letters
useless_value = raw_input("Press Enter to continue...")
print "How many letters would you like to start at?"
useless_value = input("> ")
gen = gen*useless_value
#start time for function
start_time= time.time()
try:
gen[-1]
except IndexError:
print "INVALID LETTER RANGE"
print "DEFAULT LETTERS USED (1)"
gen = 'a'
gen = list(gen)
#function loop (will never break)
x=1
while True:
#print raw string of gen
print "".join(gen)
#since infinite loop within same function, variables have to be reset
#thus oh = 0 is reseting for later use
oh = 0
#change a to b, b to c... ect. Will only make one change
if gen[-1] =='a':
gen[-1] ='b'
elif gen[-1] =='b':
gen[-1] ='c'
elif gen[-1] =='c':
gen[-1] ='d'
elif gen[-1] =='d':
gen[-1] ='e'
elif gen[-1] =='e':
gen[-1] ='f'
elif gen[-1] =='f':
gen[-1] ='g'
elif gen[-1] =='g':
gen[-1] ='h'
elif gen[-1] =='h':
gen[-1] ='i'
elif gen[-1] =='i':
gen[-1] ='j'
elif gen[-1] =='j':
gen[-1] ='k'
elif gen[-1] =='k':
gen[-1] ='l'
elif gen[-1] =='l':
gen[-1] ='m'
elif gen[-1] =='m':
gen[-1] ='n'
elif gen[-1] =='n':
gen[-1] ='o'
elif gen[-1] =='o':
gen[-1] ='p'
elif gen[-1] =='p':
gen[-1] ='q'
elif gen[-1] =='q':
gen[-1] ='r'
elif gen[-1] =='r':
gen[-1] ='s'
elif gen[-1] =='s':
gen[-1] ='t'
elif gen[-1] =='t':
gen[-1] ='u'
elif gen[-1] =='u':
gen[-1] ='v'
elif gen[-1] =='v':
gen[-1] ='w'
elif gen[-1] =='w':
gen[-1] ='x'
elif gen[-1] =='x':
gen[-1] ='y'
#if 'y' changes to 'z'. variable 'oh' is set to '1'.
# This is so it doesn't enter unwanted equations
elif gen[-1] =='y':
gen[-1] ='z'
oh = 1
#if last string = 'z'
#Checks max length of gen through [-1,-2].. ect
if gen[-1] =='z':
#reset s
s = 0
x=1
while True:
try:
s
except NameError:
s = -1
s = s-1
try:
gen[s]
except IndexError:
s = s+1
break
#s = the amount that gen cannot be
#Therefore s+1 == max
#resets values because of loop
d= 0
q= 0
#this infinite loop checks the string to see if all string values = 'z'
#if it does it will change all values to 'a' then add 'a'
x=1
while True:
#useless piece of code #1
try:
d
except NameError:
d = 0
#useful
d = d-1
try:
gen[d]
except IndexError:
break
#if d value == 'z' it will continue, otherwise break from loop
if gen[d] =='z':
#if the max == the d value that means all values have been checked
#and = 'z'. Otherwise it would've already broken the loop
if s == d:
x=1
while True:
#if oh == 1 which was set from y changing to z
# it will not continue
#this is so that if 'zy' changes to 'zz' in the first
#loop it will not change to 'aaa' while still in the loop
# this is so it prints 'zz' and doesn't skip this
#This is important to assure all possible combinations are printed
if oh == 1:
break
else:
pass
#useless q already set to 0
try:
q
except NameError:
q = 0
#sets individually all values of 'z' to 'a'
q= q-1
try:
gen[q] ='a'
#then when q cannot exist, adds an 'a'
except IndexError:
gen = gen + ['a']
#sets oh = 1, just in case. most likely useless again
oh = 1
#prints time taken to complete amount of letters
print "Completed", ((len(gen))-1), "letters in", ((time.time() - start_time) /60), "minutes."
break
else:
continue
else:
break
#if the last value = 'z' it will find the next non 'z' value and make all
#values after it = 'a'. e.g. you have abczezzz
# it will check find the 'e' and change that 'e' to a 'f'
# then it will change all the z's after it to a's
# so final would be - abczfaaa
m = -1
if gen[-1] =='z':
x=1
while True:
if oh == 1:
break
else:
pass
m = m -1
if gen[m] =='a':
gen[m] ='b'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='b':
gen[m] ='c'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='c':
gen[m] ='d'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='d':
gen[m] ='e'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='e':
gen[m] ='f'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='f':
gen[m] ='g'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='g':
gen[m] ='h'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='h':
gen[m] ='i'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='i':
gen[m] ='j'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='j':
gen[m] ='k'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='k':
gen[m] ='l'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='l':
gen[m] ='m'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='m':
gen[m] ='n'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='n':
gen[m] ='o'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='o':
gen[m] ='p'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='p':
gen[m] ='q'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='q':
gen[m] ='r'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='r':
gen[m] ='s'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='s':
gen[m] ='t'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='t':
gen[m] ='u'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='u':
gen[m] ='v'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='v':
gen[m] ='w'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='w':
gen[m] ='x'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='x':
gen[m] ='y'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='y':
gen[m] ='z'
gen[(m+1):] = ('a')*(-(m+1))
break
elif gen[m] =='z':
continue
Ваш вопрос, вероятно, лучше подходит для [CodeReview] (http://codereview.stackexchange.com/) – Sayse
Спасибо за это, я не знал, что сайт существует –
@Sayse: это достаточно сфокусировано на конкретной задаче , который может быть разрешен в одной строке. –