Python Programming - to modify a list item in a dictionary

Question

Question: # Write a program to modify the email addresses in the records dictionary to reflect this change

records = {57394: ['Suresh Datta', '[email protected]'], 48539: ['ColetteBrowning', '[email protected]'], 58302: ['Skye Homsi','[email protected]'], 48502: ['Hiroto Yamaguchi', '[email protected]'], 48291: ['Tobias Ledford', '[email protected]'], 48293: ['Jin Xu', '[email protected]'], 23945: ['Joana Dias', '[email protected]'], 85823: ['Alton Derosa', '[email protected]']}

I have iterated through the dictionary and created a new list with the values and split the email at @ and was able to change the the email from .com to .org.

My approach was to join the changed email and change the values of the dictionary. However, I keep on getting a TypeError: sequence item 0: expected str instance, list found

my code :

lst2 = []

for value in records.values():
    lst2.append(value[1].split('@'))


for items in lst2:
    items[1] = 'examples.org'

for items in lst2:
    ','.join(lst2)

Quick note: I have not included the first part of the question. it is : The company's domain has changed from (example.com) to (example.org) — KKH
– KKH, Commented Aug 12, 2018 at 3:26
problem in last line, you cannot use ','.join(lst2) , lst2 is two-dimensional array — pwxcoo
– pwxcoo, Commented Aug 12, 2018 at 3:29
Appreciate your help. I understand the mistake I've made. Your answer worked as well. — KKH
– KKH, Commented Aug 12, 2018 at 3:37

Aeolus · Accepted Answer · 2018-08-12 03:28:57Z

1

The issue is in your final for loop:

for items in lst2:
    ','.join(lst2)

You are joining should be joining items not lst2. However if you fix that it still won't work. You need to create a third list and add the values to it like this:

lst3 = []
for items in lst2:
    lst3.append('@'.join(items))

Then, lst3 will have the properly formatted emails.

answered Aug 12, 2018 at 3:28

Aeolus

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Aeolus Over a year ago

@KKH Glad I could help! If you've solved your problem, click the green check on my answer to accept it.

U13-Forward · Accepted Answer · 2018-08-12 03:48:54Z

You can do a one-liner list-comprehension for, then iterate and do join the split of i with ',', so try this:

print([','.join(i[1].replace('.com','.org').split('@')) for i in records.values()])

Output:

['suresh,example.org', 'colette,example.org', 'skye,example.org', 'hiroto,example.org', 'tobias,example.org', 'jin,example.org', 'joana,example.org', 'alton,example.org']

Or:

print(['@'.join(i[1].replace('.com','.org').split('@')) for i in records.values()])

Output:

['[email protected]', '[email protected]', '[email protected]', '[email protected]', '[email protected]', '[email protected]', '[email protected]', '[email protected]']

Or if want to edit dict:

print({k:[i.replace('.com','.org') for i in v] for k,v in records.items()})

Output:

{57394: ['Suresh Datta', '[email protected]'], 48539: ['ColetteBrowning', '[email protected]'], 58302: ['Skye Homsi', '[email protected]'], 48502: ['Hiroto Yamaguchi', '[email protected]'], 48291: ['Tobias Ledford', '[email protected]'], 48293: ['Jin Xu', '[email protected]'], 23945: ['Joana Dias', '[email protected]'], 85823: ['Alton Derosa', '[email protected]']}

Arghya Saha · Accepted Answer · 2018-08-12 03:32:43Z

0

Try the following code

records = {57394: ['Suresh Datta', '[email protected]'], 48539: ['ColetteBrowning', '[email protected]'], 58302: ['Skye Homsi','[email protected]'], 48502: ['Hiroto Yamaguchi', '[email protected]'], 48291: ['Tobias Ledford', '[email protected]'], 48293: ['Jin Xu', '[email protected]'], 23945: ['Joana Dias', '[email protected]'], 85823: ['Alton Derosa', '[email protected]']}
for key, value in records.items():
    new_data = []
    for data in value:
        new_data.append(data.replace('.com', '.org'))
    records[key] = new_data

print(records)

{57394: ['Suresh Datta', '[email protected]'], 48539: ['ColetteBrowning', '[email protected]'], 58302: ['Skye Homsi', '[email protected]'], 48502: ['Hiroto Yamaguchi', '[email protected]'], 48291: ['Tobias Ledford', '[email protected]'], 48293: ['Jin Xu', '[email protected]'], 23945: ['Joana Dias', '[email protected]'], 85823: ['Alton Derosa', '[email protected]']}

answered Aug 12, 2018 at 3:32

Arghya Saha

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2 Comments

KKH Over a year ago

I wasn't aware of the .replace method, it makes things much easier. Thanks for the response.

Arghya Saha Over a year ago

Glad I could help. I feel you should go for the nested loop, this will simplify you code. rather than going for multiple single loops.

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