如何仅显示文件中的前 5 个高分？

Question

This is the file with the scores:

Ishaan - 72
Jack - 84
Bob - 23
Louis - 77
Arnav - 56
Ben - 48
Ted - 39

So far I have sorted the file but I don't know how to display just the top 5 scores from the file.

ScoresFile2 = "/Users/KADAM BOYS/Desktop/Ishaan's Folder/Homework (Year 10)/Computing/Mock Project/Scores.txt"
ScoresWithNames = []
with open(ScoresFile2) as file2:
    for line in file2:
        ScoresWithNames.append(line.strip())
ScoresWithNames.sort(key=lambda x: int(x.split(" - ")[-1]))

Answer 1

This also has the same answer as your previous question, but modified:

l = ['Ishaan - 72', 'Jack - 84', 'Bob - 23', 'Louis - 77']
[' - '.join(k for k in j[::-1]) for j in sorted([i.split(' - ')[::-1] for i in l],reverse = True,key=lambda x: int(x[0]))][:5]

So if your sorted list is list1 :

top5 = list1[:5]

If you are using lambda:

ScoresWithNames.sort(key=lambda x: int(x.split(" - ")[-1]),reverse = True)
print(ScoresWithNames[:5])

---

Now if you want to print it with a new-line, you have two ways:

for i in top5:
    print(i)

or:

print('\n'.join(i for i in top5)) # or scoreswithnames 

Answer 2

An easy way to get the 5 highest scores is with most_common from collections.Counter :

from collections import Counter

with open("data.txt") as f:
    counts = Counter()
    for line in f:
        name, score = line.split(" - ")
        counts[name] += int(score)

    print(counts.most_common(5))

Output:

[('Jack', 84), ('Louis', 77), ('Ishaan', 72), ('Arnav', 56), ('Ben', 48)]

If you want your scores formatted back into "name - score" format:

print([f"{name} - {score}" for name, score in counts.most_common(5)])
# ['Jack - 84', 'Louis - 77', 'Ishaan - 72', 'Arnav - 56', 'Ben - 48']