Python remove element from array

Question

I have to create a python Brute-force for a school project. I have a shadow file and i have to brute force passwords.

On the list I have few passwords I don't need, so I want to remove them from my array.
So I have an array with all my passwords. I've wrote a if condition to check if password is different than "*" or "!".

for mdp in passwd:
    if mdp != '*' or mdp != '!':
        str(mdp)
        list.append(mdp)
        print(list)

But when I print my array password with "!" and "*" are still inside... I don't know why.

or should have been an and. Think about it and see if you can work out why. — Paritosh Singh
– Paritosh Singh, Commented Jan 20, 2019 at 11:41
thanx for answer @ParitoshSingh But i don"t understand why to use and... I'll check it online. THanx again — Newbiedev
– Newbiedev, Commented Jan 20, 2019 at 11:46
"or" condition is correct. I guess you should be checking for containment as mentioned in the answer below — Kannappan Sirchabesan
– Kannappan Sirchabesan, Commented Jan 20, 2019 at 11:51
@KannappanSirchabesan No it isnt. To the OP: mdp != '*' or mdp != '!'. Assume you encounter a * symbol. what happens to the conditional? It evalutes to False or True and becomes True, and the if block still executes. You need to ensure mdp is neither * nor !. That is, the condition should fail if "any one" of the 2 negation matches give a False. (The Not changes conditionals in strange ways, and it helps to think of it using pen and paper and actually working it out.) — Paritosh Singh
– Paritosh Singh, Commented Jan 20, 2019 at 11:53

Austin · Accepted Answer · 2019-01-20 11:51:27Z

1

You need a containment check, not != check. Do a list-comprehension:

password = ['ahdjd', 'xj*', '!jr']

new_password_list = [x for x in password if '*' not in x and '!' not in x]
print(new_password_list)
# ['ahdjd']

edited Jan 20, 2019 at 11:51

answered Jan 20, 2019 at 11:46

Austin

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Comments

Nullman · Accepted Answer · 2019-01-20 11:53:32Z

the major issue with your code is the or in your if statement.
think of the following, assume your password is '!' not you get into your if, it check if the password is not '' see that that is correct and goes inside, this is bad. the same issue works if your password was '', it will check if its '*' see that it is and will then will engage the or and check if its '!' see that it isnt and goes inside the if, this is still wrong. what you want to do here is have an and so your password must be both not '*' and not '!' also, you can use fancier coding styles like comprehensions and member testing also, do not use the word list as a variable, because it will shadow the type list and together you will get:
filtered = [item for item in passwd if item not in ['*', '!']]
this assumes that the passwords you want to filter out are a single character long

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