Question at position 7 If the missing lines are filled in correctly, the program should go through a list of emails and for each email remove the at sign (@) and all characters after it. [table]
1 emails = ['user321@gmail.com',2 'mywork@proton.me',3 'info1110@sydney.edu.au']4 5 i = 06 while i

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Question at position 7 If the missing lines are filled in correctly, the program should go through a list of emails and for each email remove the at sign (@) and all characters after it. [table]
1  emails = ['user321@gmail.com',2 'mywork@proton.me',3 'info1110@sydney.edu.au']4 5  i = 06 while i < len(emails):7 j = 08 while # ( code WILL go here ) :9 if # ( code WILL go here ) == '@':10 # ( code WILL go inside IF )11 j = j + 112 i = i + 113 14 print(emails)
[/table] After execution of the code fragment, the value of emails will be:['user321', 'mywork', 'info1110']Which of the following options is the correct missing code? (Code filled in has been bolded.) Note: string[:n] returns the first n characters in string.8 while j < len(emails):9 if emails[i][j] == '@':10 emails[i] = emails[i][:j]8 while j < len(emails[i]):9 if emails[j][i] == '@': 10 emails[j] = emails[j][:i]8 while j < len(emails[i]):9 if emails[i][j] == '@':10 emails[i] = emails[i][:j]8 while i <= j:9 if emails[i][j] == '@':10 emails[i] = emails[i][:j]Clear my selection

BlackTom AI · Accepted Answer

Question recap: We’re given a Python code fragment that processes a list of emails and, for each email, removes everything from the at-sign '@' onward. The task is to pick the missing code snippet that, when filled in, results in emails being ['user321', 'mywork', 'info1110'] after execution.

Option 1: 8    while j < len(emails):
9          if emails[i][j] == '@':
10             emails[i] = emails[i][:j]
- What’s wrong here is the loop condition uses len(emails), which is the number of emails in the list, not the length of the current email string. Iterating j up to len(emails) would access invalid indices for many emails and doesn’t focus on the characters of a single email. The inner check emails[i][j] could still work if j stayed within the email length, but the outer bound is incorrect for this purpose.
- The intended approach is to scan the characters of the current email string emails[i], not the list length, so this option won’t reliably strip the substring after '@' for every item.

Option 2: 8     while j < len(emails[i]):
9          if emails[j][i] == '@': 
10             emails[j] = emails[j][:i]
- Here, the indexing inside the condition uses emails[j][i], which flips the indices and attempts to access the i-th character of the j-th email. That changes the dimension of the data being inspected and would corrupt the intended per-email processing. It won’t correctly remove characters after '@' in each individual email string.
- Additionally, the slice emails[j] = emails[j][:i] truncates at i, which is inconsistent with removing from the position of '@' in the current email. Overall, this misplaces the logic from per-email processing to a mixed per-column approach, which is incorrect for the goal.

Option 3: 8      while j < len(emails[i]):
9          if emails[i][j] == '@':
10             emails[i] = emails[i][:j]
- This option keeps the loop bounded by the length of the current email (len(emails[i])) and examines each character emails[i][j]. When it encounters '@', it slices the string up to index j, effectively removing the '@' and everything after it for that email. This precisely achieves the required transformation for each email in the list, leaving characters before '@' intact.
- The logic is straightforward: scan until the '@' is found, then cut off the string at that point. This aligns with the described behavior and the target output ['user321', 'mywork', 'info1110'] when applied to the given list.

Option 4: 8     while i <= j:
9          if emails[i][j] == '@':
10             emails[i] = emails[i][:j]
- The loop condition while i <= j uses two indices, i and j, in a way that mixes two looping variables that are intended to traverse different dimensions (the outer loop over emails and the inner loop over characters). This form would likely cause out-of-bounds errors or incorrect traversal order. It does not represent a correct per-email character scan.
- Moreover, the condition emails[i][j] accesses the j-th character of the i-th email, but the controlling relation i <= j doesn’t guarantee a proper progression through the string, so the slicing would be inconsistent and unreliable for removing the portion after '@'.

Synthesis: To achieve the described outcome, you need to examine each character of the current email string emails[i] and, upon finding '@', truncate the string at that position. The only option that correctly mirrors this approach is the one that iterates with while j < len(emails[i]) and slices emails[i] at the found index j where emails[i][j] == '@'.

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