Test the Strength of a Password in Python
A string is a WEAK password if:
either, it is less than 8 characters long,
or, it is an English word,which is function is_english_word( ) is True.
A string is a STRONG password if:
it contains at least 11 characters
AND it contains at least 1 lower case letter
AND it contains at least 1 capital letter
AND it contains at least 1 numerical digit.
A string is a MEDIUM password if it is NOT a WEAK password AND is NOT a STRONG password.
def is_english_word( string ):
with open("english_words.txt") as f:
word_list =
for line in f.readlines():
word_list.append(line.strip())
if string in word_list:
return True
elif string == string.upper() and string.lower() in word_list:
return True
elif string == string.title() and string.lower() in word_list:
return True
else:
return False
def password_strength( string ):
lower = ["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"]
upper = ["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"]
for item in string:
if item in lower:
string = string.replace(item, "x")
elif item in upper:
string = string.replace(item, "y")
elif item.isnumeric():
string = string.replace(item, "n")
for item in string:
if len( string ) < 8 or is_english_word( string ) :
return 'WEAK'
elif len( string ) >= 11 and string.count("x") >= 1 and string.count("y") >= 1 and string.count("n") >= 1:
return 'STRONG'
else:
return 'MEDIUM'
print( password_strength( 'Unimaginatively' ) )
This password should be "WEAK",but the output is "MEDIUM",I don't know what's the problem of my codes.Many thanks.
python python-3.x data-science data-analysis
add a comment |
A string is a WEAK password if:
either, it is less than 8 characters long,
or, it is an English word,which is function is_english_word( ) is True.
A string is a STRONG password if:
it contains at least 11 characters
AND it contains at least 1 lower case letter
AND it contains at least 1 capital letter
AND it contains at least 1 numerical digit.
A string is a MEDIUM password if it is NOT a WEAK password AND is NOT a STRONG password.
def is_english_word( string ):
with open("english_words.txt") as f:
word_list =
for line in f.readlines():
word_list.append(line.strip())
if string in word_list:
return True
elif string == string.upper() and string.lower() in word_list:
return True
elif string == string.title() and string.lower() in word_list:
return True
else:
return False
def password_strength( string ):
lower = ["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"]
upper = ["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"]
for item in string:
if item in lower:
string = string.replace(item, "x")
elif item in upper:
string = string.replace(item, "y")
elif item.isnumeric():
string = string.replace(item, "n")
for item in string:
if len( string ) < 8 or is_english_word( string ) :
return 'WEAK'
elif len( string ) >= 11 and string.count("x") >= 1 and string.count("y") >= 1 and string.count("n") >= 1:
return 'STRONG'
else:
return 'MEDIUM'
print( password_strength( 'Unimaginatively' ) )
This password should be "WEAK",but the output is "MEDIUM",I don't know what's the problem of my codes.Many thanks.
python python-3.x data-science data-analysis
add a comment |
A string is a WEAK password if:
either, it is less than 8 characters long,
or, it is an English word,which is function is_english_word( ) is True.
A string is a STRONG password if:
it contains at least 11 characters
AND it contains at least 1 lower case letter
AND it contains at least 1 capital letter
AND it contains at least 1 numerical digit.
A string is a MEDIUM password if it is NOT a WEAK password AND is NOT a STRONG password.
def is_english_word( string ):
with open("english_words.txt") as f:
word_list =
for line in f.readlines():
word_list.append(line.strip())
if string in word_list:
return True
elif string == string.upper() and string.lower() in word_list:
return True
elif string == string.title() and string.lower() in word_list:
return True
else:
return False
def password_strength( string ):
lower = ["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"]
upper = ["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"]
for item in string:
if item in lower:
string = string.replace(item, "x")
elif item in upper:
string = string.replace(item, "y")
elif item.isnumeric():
string = string.replace(item, "n")
for item in string:
if len( string ) < 8 or is_english_word( string ) :
return 'WEAK'
elif len( string ) >= 11 and string.count("x") >= 1 and string.count("y") >= 1 and string.count("n") >= 1:
return 'STRONG'
else:
return 'MEDIUM'
print( password_strength( 'Unimaginatively' ) )
This password should be "WEAK",but the output is "MEDIUM",I don't know what's the problem of my codes.Many thanks.
python python-3.x data-science data-analysis
A string is a WEAK password if:
either, it is less than 8 characters long,
or, it is an English word,which is function is_english_word( ) is True.
A string is a STRONG password if:
it contains at least 11 characters
AND it contains at least 1 lower case letter
AND it contains at least 1 capital letter
AND it contains at least 1 numerical digit.
A string is a MEDIUM password if it is NOT a WEAK password AND is NOT a STRONG password.
def is_english_word( string ):
with open("english_words.txt") as f:
word_list =
for line in f.readlines():
word_list.append(line.strip())
if string in word_list:
return True
elif string == string.upper() and string.lower() in word_list:
return True
elif string == string.title() and string.lower() in word_list:
return True
else:
return False
def password_strength( string ):
lower = ["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"]
upper = ["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"]
for item in string:
if item in lower:
string = string.replace(item, "x")
elif item in upper:
string = string.replace(item, "y")
elif item.isnumeric():
string = string.replace(item, "n")
for item in string:
if len( string ) < 8 or is_english_word( string ) :
return 'WEAK'
elif len( string ) >= 11 and string.count("x") >= 1 and string.count("y") >= 1 and string.count("n") >= 1:
return 'STRONG'
else:
return 'MEDIUM'
print( password_strength( 'Unimaginatively' ) )
This password should be "WEAK",but the output is "MEDIUM",I don't know what's the problem of my codes.Many thanks.
python python-3.x data-science data-analysis
python python-3.x data-science data-analysis
asked Nov 12 '18 at 21:26
CeciliaCecilia
587
587
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
There are a number of problems with your code; Notably, you are replacing the lowercase chars with x
, uppercase with y
and digits with n
, before calling is_english_word
- that means is_english_word()
will be called with 'Xyyyyyyyyyyyyyy'
which is not an english word. That is making your password not 'WEAK'
.
Since it is also not 'STRONG'
, it ends up being 'MEDIUM'
.
For the record, here is an example of a correct code to do what you want:
import string
def password_strength(string):
if len(string) < 8 or is_english_word(string):
return 'WEAK'
elif (len(string) > 11 and
any(ch in string.ascii_lowercase for ch in string) and
any(ch in string.ascii_uppercase for ch in string) and
any(ch.isdigit() for ch in string)):
return 'STRONG'
else:
return 'MEDIUM'
add a comment |
If you think of those 3 restrictions on the types of letters in terms of the same operation (checking it is true for at least one character) performed using string methods, you can simplify your code significantly:
def is_weak(word):
return len(word < 8 or is_english_word(word)
def is_strong(word):
return len(word) >= 11 and all(any(method(c) for c in word)
for method in (str.islower, str.isupper, str.isdigit))
def password_strength(password):
if is_weak(password):
return 'WEAK'
elif is_strong(password):
return 'STRONG'
else:
return 'MEDIUM'
Thanks for the answer,could you please tell me the format of this code'all(any(method(c) for c in word) for method in (str.islower, str.isupper, str.isdigit))',is it in one line?
– Cecilia
Nov 12 '18 at 21:52
@Cecilia Yes, I just split it across lines for readability. If you have an expression inside brackets ({}
,, or
()
), you can have line breaks in the middle, and it will still be interpreted as a single expression.
– Patrick Haugh
Nov 12 '18 at 21:56
Got it!Thank you so much
– Cecilia
Nov 12 '18 at 22:07
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
There are a number of problems with your code; Notably, you are replacing the lowercase chars with x
, uppercase with y
and digits with n
, before calling is_english_word
- that means is_english_word()
will be called with 'Xyyyyyyyyyyyyyy'
which is not an english word. That is making your password not 'WEAK'
.
Since it is also not 'STRONG'
, it ends up being 'MEDIUM'
.
For the record, here is an example of a correct code to do what you want:
import string
def password_strength(string):
if len(string) < 8 or is_english_word(string):
return 'WEAK'
elif (len(string) > 11 and
any(ch in string.ascii_lowercase for ch in string) and
any(ch in string.ascii_uppercase for ch in string) and
any(ch.isdigit() for ch in string)):
return 'STRONG'
else:
return 'MEDIUM'
add a comment |
There are a number of problems with your code; Notably, you are replacing the lowercase chars with x
, uppercase with y
and digits with n
, before calling is_english_word
- that means is_english_word()
will be called with 'Xyyyyyyyyyyyyyy'
which is not an english word. That is making your password not 'WEAK'
.
Since it is also not 'STRONG'
, it ends up being 'MEDIUM'
.
For the record, here is an example of a correct code to do what you want:
import string
def password_strength(string):
if len(string) < 8 or is_english_word(string):
return 'WEAK'
elif (len(string) > 11 and
any(ch in string.ascii_lowercase for ch in string) and
any(ch in string.ascii_uppercase for ch in string) and
any(ch.isdigit() for ch in string)):
return 'STRONG'
else:
return 'MEDIUM'
add a comment |
There are a number of problems with your code; Notably, you are replacing the lowercase chars with x
, uppercase with y
and digits with n
, before calling is_english_word
- that means is_english_word()
will be called with 'Xyyyyyyyyyyyyyy'
which is not an english word. That is making your password not 'WEAK'
.
Since it is also not 'STRONG'
, it ends up being 'MEDIUM'
.
For the record, here is an example of a correct code to do what you want:
import string
def password_strength(string):
if len(string) < 8 or is_english_word(string):
return 'WEAK'
elif (len(string) > 11 and
any(ch in string.ascii_lowercase for ch in string) and
any(ch in string.ascii_uppercase for ch in string) and
any(ch.isdigit() for ch in string)):
return 'STRONG'
else:
return 'MEDIUM'
There are a number of problems with your code; Notably, you are replacing the lowercase chars with x
, uppercase with y
and digits with n
, before calling is_english_word
- that means is_english_word()
will be called with 'Xyyyyyyyyyyyyyy'
which is not an english word. That is making your password not 'WEAK'
.
Since it is also not 'STRONG'
, it ends up being 'MEDIUM'
.
For the record, here is an example of a correct code to do what you want:
import string
def password_strength(string):
if len(string) < 8 or is_english_word(string):
return 'WEAK'
elif (len(string) > 11 and
any(ch in string.ascii_lowercase for ch in string) and
any(ch in string.ascii_uppercase for ch in string) and
any(ch.isdigit() for ch in string)):
return 'STRONG'
else:
return 'MEDIUM'
edited Nov 12 '18 at 21:39
answered Nov 12 '18 at 21:33
nosklonosklo
152k46249272
152k46249272
add a comment |
add a comment |
If you think of those 3 restrictions on the types of letters in terms of the same operation (checking it is true for at least one character) performed using string methods, you can simplify your code significantly:
def is_weak(word):
return len(word < 8 or is_english_word(word)
def is_strong(word):
return len(word) >= 11 and all(any(method(c) for c in word)
for method in (str.islower, str.isupper, str.isdigit))
def password_strength(password):
if is_weak(password):
return 'WEAK'
elif is_strong(password):
return 'STRONG'
else:
return 'MEDIUM'
Thanks for the answer,could you please tell me the format of this code'all(any(method(c) for c in word) for method in (str.islower, str.isupper, str.isdigit))',is it in one line?
– Cecilia
Nov 12 '18 at 21:52
@Cecilia Yes, I just split it across lines for readability. If you have an expression inside brackets ({}
,, or
()
), you can have line breaks in the middle, and it will still be interpreted as a single expression.
– Patrick Haugh
Nov 12 '18 at 21:56
Got it!Thank you so much
– Cecilia
Nov 12 '18 at 22:07
add a comment |
If you think of those 3 restrictions on the types of letters in terms of the same operation (checking it is true for at least one character) performed using string methods, you can simplify your code significantly:
def is_weak(word):
return len(word < 8 or is_english_word(word)
def is_strong(word):
return len(word) >= 11 and all(any(method(c) for c in word)
for method in (str.islower, str.isupper, str.isdigit))
def password_strength(password):
if is_weak(password):
return 'WEAK'
elif is_strong(password):
return 'STRONG'
else:
return 'MEDIUM'
Thanks for the answer,could you please tell me the format of this code'all(any(method(c) for c in word) for method in (str.islower, str.isupper, str.isdigit))',is it in one line?
– Cecilia
Nov 12 '18 at 21:52
@Cecilia Yes, I just split it across lines for readability. If you have an expression inside brackets ({}
,, or
()
), you can have line breaks in the middle, and it will still be interpreted as a single expression.
– Patrick Haugh
Nov 12 '18 at 21:56
Got it!Thank you so much
– Cecilia
Nov 12 '18 at 22:07
add a comment |
If you think of those 3 restrictions on the types of letters in terms of the same operation (checking it is true for at least one character) performed using string methods, you can simplify your code significantly:
def is_weak(word):
return len(word < 8 or is_english_word(word)
def is_strong(word):
return len(word) >= 11 and all(any(method(c) for c in word)
for method in (str.islower, str.isupper, str.isdigit))
def password_strength(password):
if is_weak(password):
return 'WEAK'
elif is_strong(password):
return 'STRONG'
else:
return 'MEDIUM'
If you think of those 3 restrictions on the types of letters in terms of the same operation (checking it is true for at least one character) performed using string methods, you can simplify your code significantly:
def is_weak(word):
return len(word < 8 or is_english_word(word)
def is_strong(word):
return len(word) >= 11 and all(any(method(c) for c in word)
for method in (str.islower, str.isupper, str.isdigit))
def password_strength(password):
if is_weak(password):
return 'WEAK'
elif is_strong(password):
return 'STRONG'
else:
return 'MEDIUM'
answered Nov 12 '18 at 21:44
Patrick HaughPatrick Haugh
27.7k82546
27.7k82546
Thanks for the answer,could you please tell me the format of this code'all(any(method(c) for c in word) for method in (str.islower, str.isupper, str.isdigit))',is it in one line?
– Cecilia
Nov 12 '18 at 21:52
@Cecilia Yes, I just split it across lines for readability. If you have an expression inside brackets ({}
,, or
()
), you can have line breaks in the middle, and it will still be interpreted as a single expression.
– Patrick Haugh
Nov 12 '18 at 21:56
Got it!Thank you so much
– Cecilia
Nov 12 '18 at 22:07
add a comment |
Thanks for the answer,could you please tell me the format of this code'all(any(method(c) for c in word) for method in (str.islower, str.isupper, str.isdigit))',is it in one line?
– Cecilia
Nov 12 '18 at 21:52
@Cecilia Yes, I just split it across lines for readability. If you have an expression inside brackets ({}
,, or
()
), you can have line breaks in the middle, and it will still be interpreted as a single expression.
– Patrick Haugh
Nov 12 '18 at 21:56
Got it!Thank you so much
– Cecilia
Nov 12 '18 at 22:07
Thanks for the answer,could you please tell me the format of this code'all(any(method(c) for c in word) for method in (str.islower, str.isupper, str.isdigit))',is it in one line?
– Cecilia
Nov 12 '18 at 21:52
Thanks for the answer,could you please tell me the format of this code'all(any(method(c) for c in word) for method in (str.islower, str.isupper, str.isdigit))',is it in one line?
– Cecilia
Nov 12 '18 at 21:52
@Cecilia Yes, I just split it across lines for readability. If you have an expression inside brackets (
{}
,
, or ()
), you can have line breaks in the middle, and it will still be interpreted as a single expression.– Patrick Haugh
Nov 12 '18 at 21:56
@Cecilia Yes, I just split it across lines for readability. If you have an expression inside brackets (
{}
,
, or ()
), you can have line breaks in the middle, and it will still be interpreted as a single expression.– Patrick Haugh
Nov 12 '18 at 21:56
Got it!Thank you so much
– Cecilia
Nov 12 '18 at 22:07
Got it!Thank you so much
– Cecilia
Nov 12 '18 at 22:07
add a comment |
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