Sorting list of lists with string and int

[Otbredbaron]

So I'm trying to solve this exercise: https://open.kattis.com/problems/cups

Although the problem is simple and does not require much work I'd like to improve my code because it looks pretty bad:

#!/usr/bin/python3 n = int(input()) a = [input().split() for i in range(n)] for i in a: if(i[0].isdigit()): i[0] = int(i[0])/2 i[1], i[0] = i[0], i[1] elif(i[1].isdigit()): i[1] = int(i[1]) a.sort(key=lambda x: x[1]) [print(i[0]) for i in a]

The main concern is that I don't know if the string will be in a form int string or string int so I need to check for it.

[Gribouillis]

You could use

for i in a: try: i[1] = int(i[1]) except ValueError: i[0], i[1] = i[1], int(i[0]) / 2

[Otbredbaron]

Why do we need to be explicit and write except ValueError?

[Gribouillis]

Why do we need to be explicit and write except ValueError?

I'm not sure I understand exactly what you mean, but here I know that if i[1] contains anything that is not convertible to an integer the call to int() will raise ValueError. In such a case, I always catch the exact exception because I don't want to catch anything else. For example if for some reason there is no i[1], python will throw KeyError and I don't want to catch this at this point. For debugging purposes, it is always better to let unexpected exceptions propagate.

[buran]

actually, the easiest way would be to add sorted to your line 5 list comprehension

a = [sorted(input().split()) for i in range(n)]

this way you always get the number as element with index 0 because when sorting '0' to '9' always go before 'A'

def parse_data(item): return (int(item[0]), item[1]) data = ['red 10', '10 blue', 'green 7'] data = [parse_data(sorted(item.split())) for item in data] output = [cup for radius, cup in sorted(data)] print('\n'.join(output))

note that sample output in your assignment is NOT correct

[Otbredbaron]

actually, the easiest way would be to add sorted to your line 5 list comprehension

a = [sorted(input().split()) for i in range(n)]

this way you always get the number as element with index 0 because when sorting '0' to '9' always go before 'A'

def parse_data(item): return (int(item[0]), item[1]) data = ['red 10', '10 blue', 'green 7'] data = [parse_data(sorted(item.split())) for item in data] output = [cup for radius, cup in sorted(data)] print('\n'.join(output))

note that sample output in your assignment is NOT correct

It is correct because if it's in the format number color, then number need to be divided by 2 so it will be 5.

Why do we need to be explicit and write except ValueError?

I'm not sure I understand exactly what you mean, but here I know that if i[1] contains anything that is not convertible to an integer the call to int() will raise ValueError. In such a case, I always catch the exact exception because I don't want to catch anything else. For example if for some reason there is no i[1], python will throw KeyError and I don't want to catch this at this point. For debugging purposes, it is always better to let unexpected exceptions propagate.

I see now, I need to get used to this try except thing.

[buran]

if it's in the format number color, then number need to be divided by 2 so it will be 5.

sorry, didn't notice that - my mistake

---

def parse_data(item): try: return int(item[0])/2, item[1] except ValueError: return (int(item[1]), item[0]) data = ['red 10', '10 blue', 'green 7'] data = [parse_data(item.split()) for item in data] output = sorted([cup for radius, cup in data]) print('\n'.join(output))

and if you are comfortable with

def parse_data(item): item = item.split() try: return int(item[0])/2, item[1] except ValueError: return (int(item[1]), item[0]) data = ['red 10', '10 blue', 'green 7'] print('\n'.join(sorted([cup for radius, cup in map(parse_data, data)])))