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Questions for Cognitive Load Experiment - B

Finish the function has22 below 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 has22([1, 2, 2]) since there are two adjacent items equal to 2 (at index 1 and 2) and False for has22([2, 1, 2]) since the 2’s are not adjacent.

def has22(nums): ==== from unittest.gui import TestCaseGui class myTests(TestCaseGui): def testOne(self): self.assertEqual(has22([1, 2, 2]), True, 'has22([1, 2, 2])') self.assertEqual(has22([1, 2, 1, 2]), False, 'has22([1, 2, 1, 2])') self.assertEqual(has22([2, 1, 2]), False, 'has22([2, 1, 2])') self.assertEqual(has22([2, 2, 1]), True, 'has22([2, 2, 1])') self.assertEqual(has22([3, 4, 2]), False, 'has22([3, 4, 2])') self.assertEqual(has22([2]), False, 'has22([2])') self.assertEqual(has22([]), False, 'has22([])') self.assertEqual(has22([3, 3, 1]), False, 'has22([3, 3, 1])') self.assertEqual(has22([1, 4, 4]), False, 'has22([1, 4, 4])') 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 code in order 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):
---
    count = 0
---
    count = 1 #paired
---
    for index in range(start, end+1):
---
    for index in range(start, end): #paired
---
        current = numList[index]
---
        current = numList[start] #paired
---
        if current == target:
---
        if index == target: #paired
---
            count = count + 1
---
            count++ #paired
---
    return count
        

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 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): ==== from unittest.gui import TestCaseGui class myTests(TestCaseGui): def testOne(self): self.assertEqual(diffMaxMin([1, 2, 3]), 2, 'diffMaxMin([1, 2, 3])') self.assertEqual(diffMaxMin([10, 20, 30]), 20, 'diffMaxMin([10, 20, 30])') self.assertEqual(diffMaxMin([1, 10]), 9, 'diffMaxMin([1 , 10])') self.assertEqual(diffMaxMin([10, 1]), 9, 'diffMaxMin([-10 , 1])') self.assertEqual(diffMaxMin([4, 1, 10]), 9, 'diffMaxMin([4, 1 , 10])') self.assertEqual(diffMaxMin([4, -10, 1]), 14, 'diffMaxMin([4, -10 , 1])') 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 code in order to define total_values that takes a dictionary (dict) and returns the total of the values in the dictionary. For example, total_values({'red': 3, 'blue': 2, 'green': 20}) would return 25.

        def total_values(dict):
---
def total_values(): #paired
---
    total = 0
---
    for key in dict:
---
    for key in dict #paired
---
        total += dict[key]
---
        total += key #paired
---
    return total
        

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 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(d_list): ==== from unittest.gui import TestCaseGui class myTests(TestCaseGui): def testOne(self): d1 = {'first': 'Tsi', 'last': 'Choi'} d2 = {'first': 'Joy', 'last': 'Ali'} d3 = {'first': 'Jose', 'last': 'Capo'} d4 = {'first': 'Malik'} d5 = {'last': 'Owens'} d_l1 = [d1, d2] d_l2 = [d3] d_l3 = [d4] d_l4 = [d5] self.assertEqual(get_names(d_l1), ['Tsi Choi', 'Joy Ali'], "get_names([{'first': 'Tsi', 'last': 'Choi'}, {'first': 'Joy', 'last': 'Ali'}])") self.assertEqual(get_names(d_l2), ['Jose Capo'], "get_names([{'first': 'Jose', 'last': 'Capo'}])") self.assertEqual(get_names(d_l3), ['Malik Unknown'], "get_names([{'first': 'Malik'}])") self.assertEqual(get_names(d_l4), ['Unknown Owens'], "get_names([{'last': 'Owens'}])") 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

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