Question

Problem 1. Write a Python function times_i_at_odd(L) that takes as arguments a
list L and returns a list consisting of the elements of L multiplied by the index number of the element at odd positions. (Use list comprehensions)

>>> times_i_at_odd([1,2,3,4,5,6,7,8,9,10]) [2, 12, 30, 56, 90]

Problem 2. Write a recursive function sum_cols(grid, n) that takes a list of lists of integers grid and integer n and returns the sum of column n in grid. For example, the call

sum_cols([[1,2,3,4], [10,20,30,40], [100,200,300,400]], 2) should return 333.

Problem 3. Write a recursive function replace(s, a, b) that takes as arguments the strings s, a, and b and replaces all the occurrences of string a with string b in string s.

a) For your solution, assume that string a is only one character long.

b) For your solution, assume that string a can be of any size. Use s.startswith(a), which returns True if string s starts with a and False otherwise.

Answer 1

Thanks for the question.

Here is the completed code for this problem.  Let me know if you have any doubts or if you need anything to change.

Thank You !!

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def times_i_at_odd(num_list):
    return [num_list[i]*num_list[i+1] for i in range(0,len(num_list),2)]
print(times_i_at_odd([1,2,3,4,5,6,7,8,9,10]))

def sum_cols(grid,n):

    if len(grid)==1:
        return grid[0][n]
    else:
        return grid[0][n]+sum_cols(grid[1:],n)

print(sum_cols([[1,2,3,4], [10,20,30,40], [100,200,300,400]], 2) )

def replace(s,a,b):
    if len(s)==1:
        if s==a:return b
        else:return s
    else:
        if s[0]==a:return b+replace(s[1:],a,b)
        else:return s[0]+replace(s[1:],a,b)

print(replace('elephant','e','<e>'))