Conditional not evaluating to true

user_13555490 0
Viewed 64

According to the question, If the difference between the grade and the next multiple of 5 is less than 3, round grade up to the next multiple of 5 . If the value of grade is less than 38, no rounding occurs as the result will still be a failing grade.

Here's my solution for it,

def gradingStudents(grades):

    for i in grades:
        if (5 * round(1 + i/5) - i) < 3 and i>= 38:
            print (5 * round(1 + i/5))
        else:
            print (i)



grades_count = int(input().strip())
grades = []

for p in range(grades_count):
    g = int(input(''))
    grades.append(g)


result = gradingStudents(grades)

However, on examining the output I noticed that the if condition is not working as the output generates the same grade as the input.

python
3 Answers

I think something like this should work

def gradingStudents(grades):
  output =[]
  for g in filter(lambda x: x>38, grades):
     step = g + (5 - g % 5)
     output.append(step < 3 : g + step ? g)
  output.extend(filter(lambda x: x<=38, grades)):
  return output
# ... Your other code
print(gradingStudents(grades))
user_34874410

Simplify your approach

def finalGrade(grade, multiple_of=5, limit=38):
    if not grade < limit:
        grade = int(round(grade / multiple_of) * multiple_of)
    return grade

grades = [20, 34, 37, 38, 39, 41, 45, 48, 52]
finalGrades = [finalGrade(grade) for grade in grades]
# [20, 34, 37, 40, 40, 40, 45, 50, 50]
user_11702070

Using modulo % 5 should help you. Modulo is the remainder got after the division. Example 8 % 5=3 Also the // operator automatically rounds down for you after the division

grades = [1, 9, 37, 38, 43, 47, 49, 99, 91]
rounded_grades = [grade if grade < 38 or grade % 5 in [0, 1, 2] else 5 * (grade // 5 + 1) for grade in grades]

print(grades)
print(rounded_grades)

#Output
[1, 9, 37, 38, 43, 47, 49, 99, 91]
[1, 9, 37, 40, 45, 47, 50, 100, 91]
user_54046940
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