jayphe:
[left]
Collinsanele Taofeekdboy kensmoney Daddyshome Zabiboy

Write a Python program that accepts the name of a student, matric number, course code and the score he/she got in the course. E.g
Name: John Alfred
Matric. No.: MUC/CS/014/0882
Course Code: CMP 212
Score: 100

def printStudentDetails():
    grade = getGrade(int(score))

    print(f'{name}, {matNo}, {courseCode}, {grade}')

def getGrade(s):
    if s<49:
        return 'F'
    elif s<59:
        return 'E'
    elif s<69:
        return 'D'
    elif s<84:
        return 'C'
    elif s<94:
        return 'B'
    else:
        return 'A'

name = input('Enter name:\t')
matNo = input('Enter mat. no.:\t')
courseCode = input('Enter course code:\t')
score = input('Enter score: ')
grade = ''

printStudentDetails()

nwele2017:


Using Pandas read it in as CSV then convert it to xlsx using data name .to_excel.(ie. name of data .yo_excel("name.xlsx"

SKYQUEST:
what python code would you run on a table with data separated by spaces as shown in the image below to convert it in into a readable excel table; the raw data is also given in the attachment below (ben2):

text = input('Enter a string: ')
for i in range(len(text)):
	if text[i] == a:
		print(i, end=',')

end = ','

end = ' ' #there is a space between both quotes.

Shepherdd:
@ jayphe Collinsanele gbolly1151 kensmoney StevDesmond
Please let's roll in more exciting challenges. The thread is winding down.

HappyPagan:
@iCode2

The comma is there because you use

end = ','

On line 4.

If you want to remove the final comma, you can save the final output in a string and use one of the string methods to remove the last comma.

If you want to remove all the commas, maybe you prefer spaces, use

end = ' ' #there is a space between both quotes.

iCode2:
Thanks a lot. I'll try that string method and see.

'astring'.rstrip(',')

gbolly1151:


Write a recursive function to reverse a list

def reverse(data):
	if len(data) == 0:
		return data
	else:
		return [data[-1]] + reverse(data[:-1])

or 

def reverse(data):
	return [data[-1]] + reverse(data[:-1]) if len(data) > 0 else data

Shepherdd:


Nice one

def reverse(data):
	if len(data) == 0:
		return data
	else:
		return [data[-1]] + reverse(data[:-1])

or 

def reverse(data):
	return [data[-1]] + reverse(data[:-1]) if len(data) > 0 else data

gbolly1151:


Can you share screenshot of your run

HappyPagan:

No wahala. I think rstrip will work.

'astring'.rstrip(',')

I'm not on my PC now. Will try it once I get home.

Taofeekdboy:

I saw this late, I have been busy with my project and it took a lot of my time.
I will leave it here for who is interested to check..
It is a social media clone I made.
http://mysocialapp.pythonanywhere.com

Looking forward to your comments

iCode2:
I've tried my hands on this though. I want to see if someone can answer it a different way. Maybe yours will be better.

Write a program that asks the user to enter a list of integers. Do the following:
(a) Print the total number of items in the list.
(b) Print the last item in the list.
(c) Print the list in reverse order.
(d) Print Yes if the list contains a 5 and No otherwise.
(e) Print the number of fives in the list.
(f) Remove the first and last items from the list, sort the remaining items, and print the result.
(g) Print how many integers are less than 5

class Solution:

    def __init__(self, n):
        self.n = list(n)

    def prog(self):
        print("Total Number: ", len(self.n))
        print("Last Item: ", self.n[-1])
        print("Reverse List: ", self.n[::-1])
        l = filter(lambda x: x == 5, self.n)
        print("YES List Contains 5" if 5 in list(l) else "NO List does not contain 5" )
        print("Number of fives in list: ", self.n.count(5))
        _, *b, _ = self.n
        print("Sorted list minus first and last elements: ", sorted([*b]))
        y = filter(lambda x: x < 5, self.n)
        print("Number less than five: ", len(list(y)))
        
list_elements = eval(input("Enter desired integers separated by comma: " ))
sol = Solution(list_elements)
sol.prog()

modash:


The site looks real good.

scarplanet:

class Solution:

    def __init__(self, n):
        self.n = list(n)

    def prog(self):
        print("Total Number: ", len(self.n))
        print("Last Item: ", self.n[-1])
        print("Reverse List: ", self.n[::-1])
        l = filter(lambda x: x == 5, self.n)
        print("YES List Contains 5" if 5 in list(l) else "NO List does not contain 5" )
        print("Number of fives in list: ", self.n.count(5))
        _, *b, _ = self.n
        print("Sorted list minus first and last elements: ", sorted([*b]))
        y = filter(lambda x: x < 5, self.n)
        print("Number less than five: ", len(list(y)))
        
list_elements = eval(input("Enter desired integers separated by comma: " ))
sol = Solution(list_elements)
sol.prog()

myList = eval(input('Enter a list of integers)) #Enter the numbers as well as the square brackets.

print(len(myList))
print(myList[-1])
myList.reverse()
if 5 in myList:
	print(' Yes') 
else:
	print('It doesn\' t contain a 5')

count = 0
for i in range(len(myList)):
	if myList[i] < 5:
		count += 1
	print('There are', count, 'integers less than 5')

del myList[0]
del myList[-1]
myList.sort()
print(myList)

LessThanFive = [] 
for i in range (len(myList)):
	if 5 == myList[i]:
		LessThanFive.append(myList.count(i))
print('5 appeared', len(LessThanFive), 'times.') 

gbolly1151:
Guy...this is advance question and it is base on graph data structure,i myself haven't try to solve it but you can give it a try and test your data structure skill



Write a program that builds the routing tables for the nodes in a computer network, based on shortest-path routing, where path distance is measured by hop count, that is, the number of edges in a path. The input for this problem is the connectivity information for all the nodes in the network, as in the following example:
241.12.31.14:
241.12.31.15
241.12.31.18
241.12.31.19

which indicates three network nodes that are connected to 241.12.31.14, that is, three nodes that are one hop away. The routing table for the node at address A is a set of pairs (B,C), which indicates that, to route a message from A to B, the next node to send to (on the shortest path from A to B) is C. Your program should output the routing table for each node in the network, given an input list of node connectivity lists, each of which is input in the syntax as shown above, one per line.

jayphe:

Exercise 1

Write a python program that obtains the domain name (thapar.edu) from the URL
http://www.thapar.edu/index.php/about-us/mission

 import re

 pattern = r"(\w+)(\.)(\w+)(\.(\w+))"
 str = 'http://www.thapar.edu/index.php/about-us/mission'
 match = re.search(pattern, str)
 if match:
 	print(match.group())
 else:
 	print('Not found!')

 

gbolly1151:
Give a recursive algorithm to compute the product of two positive integers, m and n, using only addition and subtraction.

Jummate:


def product(m, n):

if n == 0:

return 0

else:

return m + product(m, n-1)

Jummate:


I indented lines 2 and 4 to be under the function declaration.

I indented lines 3 to be under the if statement and line 5 to be under the else statement but I don't know why these are not reflecting as such........

 print('hello word')

IamGoke:



If you are writing code, you need to format your text in code format.

[Code]

#Your code here

[/Code]

Change the C to c

Example below:

 print('hello word')

 def appreciation():
       print("Thanks bro" )

gbolly1151:
Guy...this is advance question and it is base on graph data structure,i myself haven't try to solve it but you can give it a try and test your data structure skill



Write a program that builds the routing tables for the nodes in a computer network, based on shortest-path routing, where path distance is measured by hop count, that is, the number of edges in a path. The input for this problem is the connectivity information for all the nodes in the network, as in the following example:
241.12.31.14:
241.12.31.15
241.12.31.18
241.12.31.19

which indicates three network nodes that are connected to 241.12.31.14, that is, three nodes that are one hop away. The routing table for the node at address A is a set of pairs (B,C), which indicates that, to route a message from A to B, the next node to send to (on the shortest path from A to B) is C. Your program should output the routing table for each node in the network, given an input list of node connectivity lists, each of which is input in the syntax as shown above, one per line.

routing_table = {
    'a': {'c': 5, 'b': 1},
    'b': {'a': 1, 'd': 4, 'c': 2},
    'c': {'e': 3, 'b': 2, 'd': 2},
    'd': {'b': 4, 'c': 2, 'e': 8, 'g': 1},
    'e': {'c': 3, 'f': 2, 'd': 8, 'g': 1},
    'f': {'e': 2, 'g': 1},
    'g': {'d': 1, 'e': 1, 'f': 1},
}

def  grade_route (vertex, links, total_cost, previous):
    for destination, cost in links.items():
        previous_cost = total_cost[destination]
        potential_cost = cost + total_cost[vertex]
        
        if potential_cost < previous_cost:
            total_cost[destination] = potential_cost
            previous[destination] = vertex

def shortest_path(graph, source, destination):
    previous = {}
    total_cost = {vertex: (lambda x : float('inf') if x != source else 0)(vertex) for vertex in graph.keys()}
    route = [destination]
    
    for vertex, links in graph.items():
        grade_route (vertex, links, total_cost, previous)
    
    trace = destination
    while True:
        pth = previous[trace]
        route.append(pth)
        trace = pth
        if pth == source:
            return [total_cost[destination], route[::-1]]

print(shortest_path(routing_table , 'a', 'c')) 
### ===> [3, ['a', 'b', 'c']]

Shepherdd:


Although this question does not provide the routing table I made some assumptions.

routing_table = {
    'a': {'c': 5, 'b': 1},
    'b': {'a': 1, 'd': 4, 'c': 2},
    'c': {'e': 3, 'b': 2, 'd': 2},
    'd': {'b': 4, 'c': 2, 'e': 8, 'g': 1},
    'e': {'c': 3, 'f': 2, 'd': 8, 'g': 1},
    'f': {'e': 2, 'g': 1},
    'g': {'d': 1, 'e': 1, 'f': 1},
}

def  grade_route (vertex, links, total_cost, previous):
    for destination, cost in links.items():
        previous_cost = total_cost[destination]
        potential_cost = cost + total_cost[vertex]
        
        if potential_cost < previous_cost:
            total_cost[destination] = potential_cost
            previous[destination] = vertex

def shortest_path(graph, source, destination):
    previous = {}
    total_cost = {vertex: (lambda x : float('inf') if x != source else 0)(vertex) for vertex in graph.keys()}
    route = [destination]
    
    for vertex, links in graph.items():
        grade_route (vertex, links, total_cost, previous)
    
    trace = destination
    while True:
        pth = previous[trace]
        route.append(pth)
        trace = pth
        if pth == source:
            return [total_cost[destination], route[::-1]]

print(shortest_path(routing_table , 'a', 'c')) 
### ===> [3, ['a', 'b', 'c']]

Although their might be multiple route to a destination I avoided implementing this for brevity. DS & Algo questions sometimes aren't easy!!!