Users report that the database on the main server cannot be accessed. A database administrator verifies the issue and notices th
1 answer:
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Answer:
Explanation:
The following Python code asks the user for inputs for the name to remove and add in the set. It then tries to remove the given name from the list. If it fails it prints out to the user saying that the name does not exist. If it does exist then it removes the name and adds the new name. Finally it prints out the current list of names.
male_names = {'oliver', 'Declan', 'Henry'}
name_to_remove = input("Enter name to remove: ")
name_to_add = input("Enter name to add: ")
try:
male_names.remove(name_to_remove)
male_names.add(name_to_add)
except:
print("Name that you are trying to remove does not exist")
print("List of Names: ")
print(male_names)
Answer:
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.
