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Consider the following code snippet:1. list = [ [ ] ] * 52. list  # output?3. list[0].append(10)4. list  # output?5. list[1].append(20)6. list  # output?7. list.append(30)8. list  # output?Analyse and select the outputs for the lines 2, 4, 6, and 8?Select one:[[], [], [], [], []][[10], [10], [10], [10], [10]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], [10,30]][[], [], [], [], []][[], [], [], [], [],[10]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], 30][[], [], [], [], []][[10], [10], [10], [10], [10]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], 30][[], [], [], [], []][[10], [10], [10], [10], [10]][[10], [10], [10], [10], [10],[20]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], [10,30]]

Question

Consider the following code snippet:1. list = [ [ ] ] * 52. list  # output?3. list[0].append(10)4. list  # output?5. list[1].append(20)6. list  # output?7. list.append(30)8. list  # output?Analyse and select the outputs for the lines 2, 4, 6, and 8?Select one:[[], [], [], [], []][[10], [10], [10], [10], [10]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], [10,30]][[], [], [], [], []][[], [], [], [], [],[10]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], 30][[], [], [], [], []][[10], [10], [10], [10], [10]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], 30][[], [], [], [], []][[10], [10], [10], [10], [10]][[10], [10], [10], [10], [10],[20]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], [10,30]]

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Solution

The correct answer is:

[[], [], [], [], []] [[10], [10], [10], [10], [10]] [[10, 20], [10, 20], [10, 20], [10, 20], [10, 20]] [[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], 30]

Here's why:

  1. Initially, the list is created with 5 empty sublists. So, the output of line 2 is [[], [], [], [], []].

  2. When we append 10 to the first sublist at line 3, it is added to all sublists because all sublists are referencing (or pointing to) the same memory location. So, the output of line 4 is [[10], [10], [10], [10], [10]].

  3. Similarly, when we append 20 to the second sublist at line 5, it is added to all sublists for the same reason. So, the output of line 6 is [[10, 20], [10, 20], [10, 20], [10, 20], [10, 20]].

  4. Finally, when we append 30 to the list at line 7, it is added as a separate element, not a part of any sublist. So, the output of line 8 is [[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], 30].

This problem has been solved

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Consider the following code snippet:1. list = [ [ ] ] * 52. list  # output?3. list[0].append(10)4. list  # output?5. list[1].append(20)6. list  # output?7. list.append(30)8. list  # output?Analyse and select the outputs for the lines 2, 4, 6, and 8?Select one:[[], [], [], [], []][[10], [10], [10], [10], [10]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], [10,30]][[], [], [], [], []][[], [], [], [], [],[10]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], 30][[], [], [], [], []][[10], [10], [10], [10], [10]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], 30][[], [], [], [], []][[10], [10], [10], [10], [10]][[10], [10], [10], [10], [10],[20]][[10, 20], [10, 20], [10, 20], [10, 20], [10, 20], [10,30]]

. What will be the output of the following code snippet?x = [i for i in range(6)]y = [i**3 for i in x]print(y) [0, 1, 8, 27, 64, 125][1, 8, 27, 64, 125, 216][0, 1, 4, 9, 16, 25][0, 1, 8, 27, 64]

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