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P: 4
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I was attempting to change a value within a dictionary list via a list variable. However it does not change. Is this a feature of a Dictionary list or am I doing something incorrect? Thank you in advance for any assistance. -
user_id = ['username', 'password', 'first_name', 'last_name', 'email', 'active']
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account = {'username':user_id[0], 'password':user_id[1], 'first_name':user_id[2],
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'last_name':user_id[3], 'email':user_id[4], 'active':user_id[5]}
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run = 1
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while run == 1:
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user_id[1] = raw_input('Please enter a new password: ')
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if user_id[1] is account['password']:
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print "Password Successfully Changed!"
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run = 0
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else:
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print "Password Change FAILED, please try again"
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Output:
Password Change FAILED, please try again
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| Expert Mod 2.5K+
P: 2,851
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When the assignment to account is made, user_id[1] is evaluated as a string and is not a reference to user_id. Why do you need the list user_id anyway?
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P: 4
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I really do not have to use it. Really just curious more then anything. I noticed when I added a print syntax it showed me the correct value. I figured if anything it had to be how the value was stored. Thank you for your assistance.
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| Expert 100+
P: 626
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Strings are immutable so you would have to use a container that is mutable, like a list, to have the dictionary point to the same memory location (otherwise the dictionary stays with the original entry). Also "is" is different from "==", but would work in my code (not in yours) because they both point to the same object. - user_id = ['username', ['password'], 'first_name', 'last_name', 'email', 'active']
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account = {'username':user_id[0], 'password':user_id[1], 'first_name':user_id[2],
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'last_name':user_id[3], 'email':user_id[4], 'active':user_id[5]}
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run = 1
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while run == 1:
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user_id[1][0] = raw_input('Please enter a new password: ')
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if user_id[1] == account['password']:
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print "Password Successfully Changed!"
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run = 0
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else:
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print "Password Change FAILED, please try again"
- a="abc"
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b="abc"
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print a, b, a is b
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c="ab"
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print a, c, a is c
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c += "c"
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print a, c, a is c, a == c
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P: 4
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So in order to get the desired results I needed to create a 2D array/list which is considered a mutable container?
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| Expert 100+
P: 626
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The mutable container has to be whatever you want to link, so both reflect the change. So if you want to change a certain element of a list, then that element has to be a sub-list and not a string, which is different from a 2D array as the rest of the elements are still strings. - user_id = ['username', 'password', 'first_name', 'last_name', 'email', 'active']
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account = {'username':user_id[0], 'password':user_id[1], 'first_name':user_id[2],
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'last_name':user_id[3], 'email':user_id[4], 'active':user_id[5]}
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print id(account['password']), id(user_id[1]) ## they are the same
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user_id[1]="new"
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print id(account['password']), id(user_id[1]) ## different
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user_id = ['username', ['password'], 'first_name', 'last_name', 'email', 'active']
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account = {'username':user_id[0], 'password':user_id[1], 'first_name':user_id[2],
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'last_name':user_id[3], 'email':user_id[4], 'active':user_id[5]}
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print "-"*30
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print id(account['password']), id(user_id[1]) ## they are the same
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user_id[1][0]="new"
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print id(account['password']), id(user_id[1]) ## still the same
| | | |  Question stats - viewed: 1598
- replies: 5
- date asked: Dec 12 '12
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