The best way to write this "double-barreled list" in Python
The for loop on line 12 and one nested inside of it, I mean. I have met such situations more than once. I would use a list, but it doesn't seem to work here.
import random
import string
def password_generator():
key = zip(string.digits, string.ascii_uppercase)
cruft, x = str(random.random()).split('.')
pw = ''
for item in x:
for element in key:
if item in element:
Q = random.random()
if Q > 0.7:
pw += element[1].lower()
else:
pw += element[1]
print pw
Thank.
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A list comprehension can be used here:
def pw_gen():
key = zip(string.digits, string.ascii_uppercase)
cruft, x = str(random.random()).split('.')
def f(i,e):
Q = random.random()
if Q > 0.7:
return e[1].lower()
else:
return e[1]
return [ f(item,element) for item in x for element in key if item in element ]
Returns a list of characters. Use "".join( pw_gen() )
to convert to string.
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It seems to me that the difficult thing for you to do is that in the middle of the loop you get a random part. () <0.7. Make it a separate function, and it's easier to turn it into a list comprehension.
def f(str):
if random.random()>0.7:
return str.lower()
else:
return str
''.join([f(element[1]) for element in key for item in x if item in element])
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You can use itertools.product to combine the two, but I feel like this task can be simplified by using a higher level method from random.
symbols=string.ascii_uppercase[:10]
pw = ''
for i in range(15):
letter = random.choice(symbols)
if random.random() > 0.7:
letter = letter.lower()
pw += letter
Or even shorter:
symbols = (7*string.ascii_uppercase[:10] +
3*string.ascii_lowercase[:10])
pw = ''.join(random.choice(symbols) for i in range(15))
The way it is constructed key
makes it a very inefficient lookup table. You can replace multiple lines to use a more efficient method:
key = dict(zip(string.digits, string.ascii_uppercase))
#...
for item in x:
letter = key[item]
They all admittedly produce a fixed length password, and string conversions have a small chance of getting a shorter number. However, for the same reasons, the last figure was less random than the others.
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Your logic can be much simpler than using a list comprehension to replace the inner loop. This will be the equivalent of your code:
import random
import string
key = dict(zip(string.digits, string.ascii_uppercase))
cruft, x = str(random.random()).split('.', 1)
pw = ''
for item in x:
if item in key: # This if is not required as all the single digits are present
Q = random.random()
if Q>0.7:
pw += key[item].lower()
else:
pw += key[item]
print pw
However, I would advise you to go with a UUID if you really want the password to be unique and not required for easy conversion.If you want a better password generator (with alphanumeric and symbols) you can do something like this .
Please note that this is not an official python site and could be improved in many ways. As usual, this will be your challenge to shoot , I mean to use it in some way.
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