Write a function that takes in a sequence s and a function fn and returns a dictionary. The values of the dictionary are lists of elements from s. Each element e in a list should be constructed such that fn(e) is the same for all elements in that list. Finally, the key for each value should be fn(e) def group_by(s, fn):
""" >>> group_by([12, 23, 14, 45], lambda p: p // 10)
{1: [12, 14], 2: [23], 4: [45]}
>>> group_by(range(-3, 4), lambda x: x * x)
{0: [0], 1: [-1, 1], 4: [-2, 2], 9: [-3, 3]}
"""
Use Python

Here is the Python function:

def group_by(s, fn):  #method definition

    group = {}  #dictionary

    for e in s:  #for each element from s

         key = fn(e)  #set key to fn(e)

         if key in group:  #if key is in dictionary group

              group[key].append(e)  #append e to the group[key]

         else:  #if key is not in group

              group[key] = [e]  #set group[key] to [e]

    return group #returns dictionary group

Explanation:

To check the working of the above function call the function by passing a sequence and a function and use print function to display the results produced by the function:

print(group_by([12, 23, 14, 45], lambda p: p // 10))

print(group_by(range(-3, 4), lambda x: x * x))

function group_by takes in a sequence s and a function fn as parameters. It then creates an empty dictionary group. The values of the dictionary are lists of elements from s. Each element e in a list is constructed such that fn(e) is the same for all elements in that list. Finally, the key for each value is fn(e). The function returns a dictionary. The screenshot of the program along with its output is attached.

when listing items that have an order of priority.

answer; ///400 ibs;