using Break with list / dictionary comprehension in Python

Question

I want to get the grades of student based on there marks, while using for and if i can use break when the first time numbers are greater the specified grade the program is

          numb = int(input("Enter the numb:"))
          my_dict = {90: 'A+', 80: 'A',  70: 'B', 60:'C'}
          for gr in my_dict.keys():
          if numb > gr:
               grade = my_dict.get(gr)
               break
          print(grade)

I want to implement it using list comprehension

Any suggestion don't want to import any modules just using basic python

this is what i wrote:

         grade = [my_dict.get(gr) for gr in my_dict.keys() if numb > gr ]

but this would obviously give me a list of all grades when number are greater than specific value of number.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. — Community, Jul 16 '23 at 07:45
It looks like you don't want a list comprehension. Perhaps a generator expression where you only take the first result. — quamrana, Jul 16 '23 at 07:59

Chris · Answer 1 · 2023-07-16T23:36:27.267

You don't break from within a list comprehension, but you can stop at the first match using next to retrieve the first value from a generator expression.

>>> my_dict = {90: 'A+', 80: 'A',  70: 'B', 60:'C'}
>>> numb = int(input("Enter the numb:"))
Enter the numb:78
>>> numb
78
>>> next(
...   (v 
...    for k, v in my_dict.items() 
...    if numb > k), 
...   'F'
... )
'B'

The 'F' is provided as a default in case there is no next match.

This is similar to generating a list of all of the matches and then grabbing the first result, except that it's lazy (it doesn't have to create an entire list) and provides a handy mechanism for a default value.

>>> lst = [v for k, v in my_dict.items() if numb > k]
>>> lst
['B', 'C']
>>> lst[0] if lst else 'F'
'B'

using Break with list / dictionary comprehension in Python

1 Answers1