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Questions for Cognitive Load Experiment - A - Original

Try to solve the following mixed up code problems. You can use the “Help Me” button to make the problem easier if you have made at least three attempts to solve the problem. After you solve each problem, please answer the poll as well.

Put the blocks in order to define the function has22 to return True if there are at least two items in the list nums that are adjacent and both equal to 2, otherwise return False. For example, return True for [1, 2, 2] since there are two adjacent items equal to 2 (at index 1 and 2) and False for [2, 1, 2] since the 2’s are not adjacent.

        def has22(nums):
---
    i = 1
---
    i = 0 #distractor
---
    while i < len(nums):
---
        if nums[i] == 2 and nums[i-1] == 2:
---
        if nums[i] == 2 and nums[i+1] == 2: #distractor
---
            return True
---
            return true #distractor
---
        i += 1
---
    return False
        

In solving the preceding problem I invested:
• 1. Very, very low mental effort
• 2. Very low mental effort
• 3. Low mental effort
• 4. Rather low mental effort
• 5. Neither low nor high mental effort
• 6. Rather high mental effort
• 7. High mental effort
• 8. Very high mental effort
• 9. Very, very high mental effort

Finish the function to define countInRange that returns a count of the number of times that a target value appears in a list between the start and end indices (inclusive). For example, countInRange(1,2,4,[1, 2, 1, 1, 1, 1]) should return 3 since there are three 1’s between index 2 and 4 inclusive.

def countInRange(target, start, end, numList): ==== from unittest.gui import TestCaseGui class myTests(TestCaseGui): def testOne(self): self.assertEqual(countInRange(2, 0, 2, [1, 2, 2]), 2, 'countInRange(2, 0, 2, [1, 2, 2])') self.assertEqual(countInRange(1, 2, 4, [1, 2, 1, 1, 1, 1]), 3, 'countInRange(1, 2, 4, [1, 2, 1, 1, 1, 1])') self.assertEqual(countInRange(1, 0, 4, [1, 2, 1, 1, 1, 1]), 4, 'countInRange(1, 0, 4, [1, 2, 1, 1, 1, 1])') self.assertEqual(countInRange(2, 1, 2, [1, 2, 2]), 2, 'countInRange(2, 1, 2, [1, 2, 2])') self.assertEqual(countInRange(3, 1, 2, [1, 2, 2]), 0, 'countInRange(3, 1, 2, [1, 2, 2])') self.assertEqual(countInRange(3, 1, 2, [3, 3, 3, 3]), 2, 'countInRange(3, 1, 2, [3, 3, 3, 3])') myTests().main()

In solving the preceding problem I invested:
• 1. Very, very low mental effort
• 2. Very low mental effort
• 3. Low mental effort
• 4. Rather low mental effort
• 5. Neither low nor high mental effort
• 6. Rather high mental effort
• 7. High mental effort
• 8. Very high mental effort
• 9. Very, very high mental effort

Put the blocks in order to define the function diffMaxMin to return the difference between the largest and smallest value in the passed list of numbers (nums). For example, diffMaxMin([1,2,3]) should return 2 since the difference between 3 and 1 is 2.

        def diffMaxMin(nums):
---
def diffMaxMin(nums) #paired
---
    large = max(nums)
    small = min(nums)
---
    return large - small
---
    return small - large #paired
        

In solving the preceding problem I invested:
• 1. Very, very low mental effort
• 2. Very low mental effort
• 3. Low mental effort
• 4. Rather low mental effort
• 5. Neither low nor high mental effort
• 6. Rather high mental effort
• 7. High mental effort
• 8. Very high mental effort
• 9. Very, very high mental effort

Finish the function total_values that takes a dictionary (dict) and returns the total of the values in the dictionary. For example, total_dict_values({'red': 3, 'blue': 2, 'green’: 20}) would return 25.

def total_values(dict): ==== from unittest.gui import TestCaseGui class myTests(TestCaseGui): def testOne(self): self.assertEqual(total_values({'red':2, 'blue':3, 'green':20}),25, "total_values({'red':2, 'blue':3, 'green':20})") self.assertEqual(total_values({'a': 3}), 3, "total_values({'a': 3})") self.assertEqual(total_values({'a': 3, 'b': -5}), -2, "total_values({'a': 3, 'b': -5})") self.assertEqual(total_values({'a': 3, 'b': -3}), 0, "total_values({'a': 3, 'b': -3})") self.assertEqual(total_values({}), 0, "total_values({})") myTests().main()

In solving the preceding problem I invested:
• 1. Very, very low mental effort
• 2. Very low mental effort
• 3. Low mental effort
• 4. Rather low mental effort
• 5. Neither low nor high mental effort
• 6. Rather high mental effort
• 7. High mental effort
• 8. Very high mental effort
• 9. Very, very high mental effort

Put the blocks in order to define the function get_names that takes a list of dictionaries and returns a list of strings with the names from the dictionaries. The key for the first name is first and the key for the last name is last. Return a list of the full names (first last) as a string. If the first or last key is missing in the dictionary use Unknown. For example, [{'first': 'Ann', 'last': 'Brown'}, {'first': 'Darius'}] should return ['Ann Brown', 'Darius Unknown'].

        def get_names(list_of_dict):
---
 name_list = []
---
 for p_dict in list_of_dict:
---
     first = p_dict.get('first', 'Unknown')
     last = p_dict.get('last', 'Unknown')
---
     first = p_dict.get('first', None)
     last = p_dict.get('last',  None) #paired
---
     name = first + " " + last
---
     name = first + last #paired
---
     name_list.append(name)
---
 return name_list
        

In solving the preceding problem I invested:
• 1. Very, very low mental effort
• 2. Very low mental effort
• 3. Low mental effort
• 4. Rather low mental effort
• 5. Neither low nor high mental effort
• 6. Rather high mental effort
• 7. High mental effort
• 8. Very high mental effort
• 9. Very, very high mental effort

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