Task from python mid exam, list in dictionary, dictionary in list
Question:
This was the task in Python middle exam at my university(first course) and I could not figure it out how to complete it.
So, there was a dictionary of wagons, seat names and availability of them(P.S this dictionary was much bigger, this is just short example).
d = {
'one': [{"Seat_name" : "A1", "isTaken" : False},
{ "Seat_name" : "A2", "isTaken" : True}],
'two': [{ "Seat_name" : "B1", "isTaken" : True},
{ "Seat_name" : "B2", "isTaken" : False}]
}
I had to offer the customer the wagon and the seat that was available and if the customer chose manually and the seat was taken or wagon was full, I had to offer the ones that was available. After the exam, I still could not find a way to solve it. So, any help, would be much appreciated.
Answers:
Your input is not a valid dictionary. For me d should be like this(correct me if i am wrong)
d={1: [{'Seat_name': 'A1', 'isTaken': False},
{'Seat_name': 'A2', 'isTaken': True}],
2: [{'Seat_name': 'B2', 'isTaken': False}]
}
Create a dictionary that will store all seats_name available for each wagon
a=dict()
for item in d:
a[item]=list()
for seat in d[item]:
if not seat.get('isTaken'):
a[item].append(seat['Seat_name'])
now if you want to check available wagon you can directly access from a dict and if value is empty in a then you can suggest some other wagon number
basically this requires just a bit of dict handling. what you could do is extract the ‘available seats’ part of the input dictionary. ‘not avialable’ is then given by exclusion. The rest is a bunch of if and print statements; I will leave the details and “prettification” to you since detailled requirements are not provided.
I took the freedom to convert your example dict to valid Python syntax and modified it a bit so we can test the behavior of the evaluation. In principle, you’d also have to check for validity of the choice, e.g. does wagon/seat exist. I skipped that for now…
def check_availability(data, choice):
# get the available seats, let each wagon be a key here as well.
availability = {}
for k, v in data.items():
# k: wagon no., v: dicts specifying seat/availability in the wagon
# check if all seats are occupied in a wagon
if not all(i['isTaken'] for i in v):
# if seats are available, append them to availability dict:
availability[k] = [i['Seat_name'] for i in v if not i['isTaken']]
# short version, actually bad style since line too long...
# availability = {k: [i['Seat_name'] for i in v if not i['isTaken']] for k, v in data.items() if not all(i['isTaken'] for i in v)}
# now there are three options we can walk through:
if choice['wagon'] not in availability.keys():
print(f"wagon {choice['wagon']} is full. available are:n{availability}")
elif choice['seat'] not in availability[choice['wagon']]:
print(f"seat {choice['seat']} is taken. available in wagon {choice['wagon']} are:n{availability[choice['wagon']]}")
else:
print(f"seat {choice['seat']} in wagon {choice['wagon']} is available!")
# testing
d = {1: [{"Seat_name": "A1", "isTaken": True},
{"Seat_name": "A2", "isTaken": True}],
2: [{"Seat_name": "B1", "isTaken": True},
{"Seat_name": "B2", "isTaken": False},
{"Seat_name": "B3", "isTaken": False}]}
choice = {'wagon': 1, 'seat': "A1"}
check_availability(d, choice)
# wagon 1 is full. available are:
# {2: ['B2', 'B3']}
choice = {'wagon': 2, 'seat': "B1"}
check_availability(d, choice)
# seat B1 is taken. available in wagon 2 are:
# ['B2', 'B3']
choice = {'wagon': 2, 'seat': "B2"}
check_availability(d, choice)
# seat B2 in wagon 2 is available!
⦁ Suppose that there is a list which contains the id number, mid and final marks of four students. The program uses a dictionary, where the key is the ID of the student and value is a list containing the mid and final result of student, like this { “R/1212/09”: [ 35, 40] ,”R/1213/09”: [ 25, 45] ,…} .
Id Quiz (10% Mid exam (30%) Final exam (60%) Total
R/1212/09 8 35 40
R/1213/09 8 25 45
R/1214/09 9 34 54
A.) Calculate the total marks of each student and append them to your original lists inside the dictionary:
Similar to {“R/1212/09”: [8, 35, 40, 75], ”R/1213/09”: [8,25, 45, 65],…}
B.) Write a python program that print top 2 students (id and total) based on total
C.) Print the dictionary in the following format
Grade distribution
Id quiz mid exam final exam total
⦁ Write a python program count all words in word.txt longer than 18 letters print these words and the number of such words.
⦁ Explain about default and named parameter with example.
⦁ Explain the main difference local and global variable with example
⦁ with Explain the main difference between list, tuple and dictionary example.
⦁ What is comment? Explain types of problem of comment with example.
⦁ What are computations? explain the problem solving steps with computer
⦁ What is cloning in graphics
This was the task in Python middle exam at my university(first course) and I could not figure it out how to complete it.
So, there was a dictionary of wagons, seat names and availability of them(P.S this dictionary was much bigger, this is just short example).
d = {
'one': [{"Seat_name" : "A1", "isTaken" : False},
{ "Seat_name" : "A2", "isTaken" : True}],
'two': [{ "Seat_name" : "B1", "isTaken" : True},
{ "Seat_name" : "B2", "isTaken" : False}]
}
I had to offer the customer the wagon and the seat that was available and if the customer chose manually and the seat was taken or wagon was full, I had to offer the ones that was available. After the exam, I still could not find a way to solve it. So, any help, would be much appreciated.
Your input is not a valid dictionary. For me d should be like this(correct me if i am wrong)
d={1: [{'Seat_name': 'A1', 'isTaken': False},
{'Seat_name': 'A2', 'isTaken': True}],
2: [{'Seat_name': 'B2', 'isTaken': False}]
}
Create a dictionary that will store all seats_name available for each wagon
a=dict()
for item in d:
a[item]=list()
for seat in d[item]:
if not seat.get('isTaken'):
a[item].append(seat['Seat_name'])
now if you want to check available wagon you can directly access from a dict and if value is empty in a then you can suggest some other wagon number
basically this requires just a bit of dict handling. what you could do is extract the ‘available seats’ part of the input dictionary. ‘not avialable’ is then given by exclusion. The rest is a bunch of if and print statements; I will leave the details and “prettification” to you since detailled requirements are not provided.
I took the freedom to convert your example dict to valid Python syntax and modified it a bit so we can test the behavior of the evaluation. In principle, you’d also have to check for validity of the choice, e.g. does wagon/seat exist. I skipped that for now…
def check_availability(data, choice):
# get the available seats, let each wagon be a key here as well.
availability = {}
for k, v in data.items():
# k: wagon no., v: dicts specifying seat/availability in the wagon
# check if all seats are occupied in a wagon
if not all(i['isTaken'] for i in v):
# if seats are available, append them to availability dict:
availability[k] = [i['Seat_name'] for i in v if not i['isTaken']]
# short version, actually bad style since line too long...
# availability = {k: [i['Seat_name'] for i in v if not i['isTaken']] for k, v in data.items() if not all(i['isTaken'] for i in v)}
# now there are three options we can walk through:
if choice['wagon'] not in availability.keys():
print(f"wagon {choice['wagon']} is full. available are:n{availability}")
elif choice['seat'] not in availability[choice['wagon']]:
print(f"seat {choice['seat']} is taken. available in wagon {choice['wagon']} are:n{availability[choice['wagon']]}")
else:
print(f"seat {choice['seat']} in wagon {choice['wagon']} is available!")
# testing
d = {1: [{"Seat_name": "A1", "isTaken": True},
{"Seat_name": "A2", "isTaken": True}],
2: [{"Seat_name": "B1", "isTaken": True},
{"Seat_name": "B2", "isTaken": False},
{"Seat_name": "B3", "isTaken": False}]}
choice = {'wagon': 1, 'seat': "A1"}
check_availability(d, choice)
# wagon 1 is full. available are:
# {2: ['B2', 'B3']}
choice = {'wagon': 2, 'seat': "B1"}
check_availability(d, choice)
# seat B1 is taken. available in wagon 2 are:
# ['B2', 'B3']
choice = {'wagon': 2, 'seat': "B2"}
check_availability(d, choice)
# seat B2 in wagon 2 is available!
⦁ Suppose that there is a list which contains the id number, mid and final marks of four students. The program uses a dictionary, where the key is the ID of the student and value is a list containing the mid and final result of student, like this { “R/1212/09”: [ 35, 40] ,”R/1213/09”: [ 25, 45] ,…} .
Id Quiz (10% Mid exam (30%) Final exam (60%) Total
R/1212/09 8 35 40
R/1213/09 8 25 45
R/1214/09 9 34 54
A.) Calculate the total marks of each student and append them to your original lists inside the dictionary:
Similar to {“R/1212/09”: [8, 35, 40, 75], ”R/1213/09”: [8,25, 45, 65],…}
B.) Write a python program that print top 2 students (id and total) based on total
C.) Print the dictionary in the following format
Grade distribution
Id quiz mid exam final exam total
⦁ Write a python program count all words in word.txt longer than 18 letters print these words and the number of such words.
⦁ Explain about default and named parameter with example.
⦁ Explain the main difference local and global variable with example
⦁ with Explain the main difference between list, tuple and dictionary example.
⦁ What is comment? Explain types of problem of comment with example.
⦁ What are computations? explain the problem solving steps with computer
⦁ What is cloning in graphics