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olga55 [171]
3 years ago
8

#Write a function called find_median. find_median #should take as input a string representing a filename. #The file correspondin

g to that filename will be a list #of integers, one integer per line. find_median should #return the medi
Computers and Technology
1 answer:
azamat3 years ago
5 0

Answer:

Explanation:

The following is written in Python. It takes in a file, it then reads all of the elements in the file and adds them to a list called myList. Then it sorts the list and uses the elements in that list to calculate the median. Once the median is calculated it returns it to the user. The code has been tested and the output can be seen in the image below.

def find_median(file):

   file = open(file, 'r')

   mylist = []

   for number in file:

       mylist.append(int(number))

   numOfElements = len(mylist)

   mylist.sort()

   print(mylist)

   if numOfElements % 2 == 0:

       m1 = numOfElements / 2

       m2 = (numOfElements / 2) + 1

       m1 = int(m1) - 1

       m2 = int(m2) - 1

       median = (mylist[m1] + mylist[m2]) / 2

   else:

       m = (numOfElements + 1) / 2

       m = int(m) - 1

       median = mylist[m]

   return median

print("Median: " + str(find_median('file1.txt')))

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(a) Store last 7 digits of your student ID in a vector (7 element row or column vector). Write a MATLAB code which creates a 7x7
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Answer:

MATLAB code explained below with appropriate comments for better understanding

Explanation:

clc

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%(a)

A = zeros(7);

for i=1:7

 

A(i,i)= ID(i);

 

end

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%(b)

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%(c)

if(A(3,3)>0)

fprintf('A(3,3) is positive\n');

else if(A(3,3)<0)

fprintf('A(3,3) is negative\n');

else

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%(d)

fprintf('\nRequired series : ');

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while n>=0

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if(n>0)

fprintf(', ');

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fprintf('\n');

%(e)

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F = 1;

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y = [4 3 9 1];

A = 0;

C = x(1);

for i=1:length(x)

A = A + x(i)*y(i);

B(i) = x(i)/y(i);

if(min(x(i),y(i))<C)

C = min(x(i),y(i));

end

end

C = 1/C;

fprintf('\nA = %i\n',A);

fprintf('B = ');

disp(B)

fprintf('C = %0.5g\n',C);

%(g)

clear all

A = randi([5 25],[1,10]);

maxA = A(1);

for i = 2:10

if(maxA<A(i))

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minA = A(1);

i = 2;

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fprintf('\nA = ');

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fprintf('maxA = %i\n',maxA);

fprintf('minA = %i\n',minA);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%output

A =

1 0 0 0 0 0 0

0 2 0 0 0 0 0

0 0 3 0 0 0 0

0 0 0 4 0 0 0

0 0 0 0 5 0 0

0 0 0 0 0 6 0

0 0 0 0 0 0 7

B =

1 0 0 0 0 0 0

0 2 0 0 0 0 0

0 0 3 0 0 0 0

0 0 0 4 0 0 0

0 0 0 0 5 0 0

0 0 0 0 0 6 0

0 0 0 0 0 0 7

Both A and B are same

A(6,6) is even

A(3,3) is positive

Required series : 35, 30, 25, 20, 15, 10, 5, 0

Input an integer : 6

6! = 720

A = 52

B = 0.7500 2.3333 0.2222 1.0000

C = 1

A = 25 14 7 10 13 17 10 17 19 9

maxA = 25

minA = 7

>>

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