Alternatives of repeatedly using elif statements

Question

Im trying to return the 'students' grades according to what they got on python. I used a bunch of elif statements but i dont feel like this is efficient. I was wondering if there are any other ways i can present this as? possibly smaller. Ive attempted using for loops and dictionaries, but im not getting it, maybe cuz i dont understand dictionaries well loll

def grade_select(grade):
    if grade <  40: return "D-"
    elif grade <= 40 and grade < 45: return "D"
    elif grade <= 45 and grade < 50: return "D+"
    elif grade <= 50 and grade < 55: return "C-"
    elif grade <= 55 and grade < 60: return "C"
    elif grade <= 60 and grade < 65: return "C+"
    elif grade <= 65 and grade < 70: return "B-"
    elif grade <= 70 and grade < 75: return "B"
    elif grade <= 75 and grade < 80: return "B+"
    elif grade <= 80 and grade < 85: return "A-"
    elif grade <= 85 and grade < 90: return "A" 
    elif grade >= 90: return "A+"

@d6stringer Python doesn't really have a switch statement, it has pattern matching though, but that doesn't really give you any advantage over this `if...elif...elif` branching — juanpa.arrivillaga, Sep 19 '22 at 18:20
***Python 3.10*** provides a `match ... case` syntax you could try out. — Just5MoreMinutes, Sep 19 '22 at 18:21
@juanpa.arrivillaga right, my bad.. they just introduced "match" and "case" though which is the same idea. — d6stringer, Sep 19 '22 at 18:22
@Just5MoreMinutes How would i use match and cases? in this case and does it not take the same amount of space as if/elifs? — quadingle the 2nd, Sep 19 '22 at 18:24
@quadinglethe2nd it does. Again, your code is fine (I would just format it according to PEP8) — juanpa.arrivillaga, Sep 19 '22 at 18:25
your comparitions make no sense: `grade <= 40 and grade < 45` means the same as `grade <= 40` - you need `>=` instead of `<=`. to get `grade >= 40 and grade < 45` which means `40 <= grade < 45` but with this order of `if/elif` you can skip `grade >= 40` — furas, Sep 19 '22 at 18:29
@juanpa.arrivillaga as far as I know it's faster than your regular `if ... elif ... else` statement — Just5MoreMinutes, Sep 19 '22 at 18:29
Please read [ask] and note well that this is **not a discussion forum**. We expect questions to be written properly here, as if you were helping to build a library of searchable questions (that is the purpose of posting), not as if you were talking to a friend about the code on Discord. — Karl Knechtel, Sep 20 '22 at 06:57

Paulo Pereira · Answer 1 · 2022-09-19T18:52:42.853

since you return when a condition is matched, you do not need to use elif just if.

moreover, the previous condition eliminates the need of checking if the value is "between".

def grade_select(grade):
    if grade < 40: return "D-"
    if grade < 45: return "D"
    if grade < 50: return "D+"
    if grade < 55: return "C-"
    if grade < 60: return "C"
    if grade < 65: return "C+"
    if grade < 70: return "B-"
    if grade < 75: return "B"
    if grade < 80: return "B+"
    if grade < 85: return "A-"
    if grade < 90: return "A" 
    return "A+"

an alternative would be to declare a dictionary and calculate where the grade fits.

grades = { 40: "D-", 45: "D", 50: "D+", 55: "C-", 60: "C", 65: "C+", 70: "B-", 75: "B", 80: "B+", 85: "A-", 90: "A", 100: "A+" }

def grade_select(grade):
    grade = (grade // 5 * 5) + 5

    if grade < 40:
        grade = 40
    elif grade > 90:
        grade = 100

    return grades[grade]

Matiiss · Answer 2 · 2022-09-19T18:30:10.650

1

You can use a dictionary to store the range and it's according grade value:

def grade_select(grade):
    dct = {
        range(0, 40): "D-",
        range(40, 45): "D",
        range(45, 50): "D+",
        ...
        range(90, 100 + 1): "A+"
    }
    dct = {g: m for r, m in dct.items() for g in r}
    return dct[grade]

edited Sep 19 '22 at 18:30

answered Sep 19 '22 at 18:23

Matiiss

5,970
2
12
29

1

How is this better or more efficient? This forces you to iterate over every item, it doesn't even exit early – juanpa.arrivillaga Sep 19 '22 at 18:24
@juanpa.arrivillaga it's an alternative, also I'd say it's more readable – Matiiss Sep 19 '22 at 18:26
1

If you are going to do this, then the dictionary should probably be a global variable – juanpa.arrivillaga Sep 19 '22 at 18:26
yeah, could be, but unless it's used a bunch elsewhere (which I doubt because otherwise what's the point of that function), it's not like it would be a bottleneck anyways – Matiiss Sep 19 '22 at 18:27
True, it is all fundamentally constant time – juanpa.arrivillaga Sep 19 '22 at 18:27

score 1 · Answer 3 · answered Sep 19 '22 at 18:32

1

Well, your condition is kinda weird, because

elif grade <= 40 and grade < 50

has no sense (if grade <= 40 then grade < 50 of course)

But I think you wanna write something like this:

if 40 <= grade < 50:
    pass

You can use syntax above, it could make it a little bit better. Also, you can check this article

answered Sep 19 '22 at 18:32

Vladyslav

97
7

1

oh oops yeah thats on me, i meant elif grade >=40 and grade < 50 – quadingle the 2nd Sep 19 '22 at 18:36

Steven Rumbalski · Answer 4 · 2022-09-19T19:25:59.977

Here's the fastest version that calculates an index based on the grade value. This method only works if your grade ranges are all five wide.

_lettergrades = ['D-', 'D', 'D+', 'C-', 'C', 'C+', 'B-', 'B', 'B+', 'A-', 'A', 'A+']

def grade_select(grade):  
    return _lettergrades[(max(min(grade, 90), 35) - 35) // 5]

If your ranges weren't consistently five wide you could use bisect.bisect_right to get an index for the letter grade. The search uses bisection, which should be faster than a linear search. (The lists _thresholds and _lettergrades are global to avoid recreating them for every call to grade_select().)

from bisect import bisect_right
    
_thresholds = [40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90]
_lettergrades = ['D-', 'D', 'D+', 'C-', 'C', 'C+', 'B-', 'B', 'B+', 'A-', 'A', 'A+']

def grade_select(grade):
    return _lettergrades[bisect_right(_thresholds, grade)]

Note: I doubt speed matters as much as clarity here. For clarity I like Paulo Pereira's answer best.

Dmitriy Neledva · Answer 5 · 2022-09-20T06:47:51.917

-1

I guess using dictionary is the best way

def grade_select(grade):
    d = {
        'D-':range(0,40),
        'D':range(45,50),
        'D+':range(50,55),
        'C-':range(55,60) #, etc.
    }

    return [k for k,v in d.items() if grade in v][0]

edited Sep 20 '22 at 06:47

answered Sep 19 '22 at 18:31

Dmitriy Neledva

867
4
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Returning a list is wrong. You could use a plain for-loop instead: `for k,v in d.items(): if grade in v: return k` – wjandrea Sep 19 '22 at 23:02
@wjandrea you're right, I was inattentive. but I guess list comprehention is still better (and faster) than for loop, we just need to add `[0]` to get rid of list. `return [k for k,v in d.items() if grade in v][0]` – Dmitriy Neledva Sep 20 '22 at 05:30
Better how? It's harder to read and it can't return early. The plain for-loop is in fact better. – wjandrea Sep 20 '22 at 16:56
1

I'm wrong for loop would be better for this situation – Dmitriy Neledva Sep 20 '22 at 19:35

Alternatives of repeatedly using elif statements

5 Answers5