So i know there's a better, more efficient, way to do this. Does anyone have any ideas?

Question

continents = [
    'Asia',
    'South America',
    'North America',
    'Africa',
    'Europe',
    'Antarctica',
    'Australia',
]
Your code here
for n in continents:  
    if n == continents[0]:
        print("* " + continents[0]) 
    elif n == continents[3]:
        print("* " + continents[3])
    elif n == continents[5]:
        print("* " + continents[5])
    elif n == continents[6]:
        print("* " + continents[6])
```continents.py
continents = [
    'Asia',
    'South America',
    'North America',
    'Africa',
    'Europe',
    'Antarctica',
    'Australia',
]
Your code here
for n in continents:  
    if n == continents[0]:
        print("* " + continents[0]) 
    elif n == continents[3]:
        print("* " + continents[3])
    elif n == continents[5]:
        print("* " + continents[5])
    elif n == continents[6]:
        print("* " + continents[6])
```

amandae · Answer

If I remember correctly, this is supposed to check for words that start with the letter A.
Something like this might work.  Each word is a string, which is iterable, so we can call it by index. It is something like a list of lists.  If you weren't sure about the case, it would be good to use a .upper or .lower method, but these have consistent formatting, so it doesn't matter.    This is the same thing jb30 said, just in different words...
python
for continent in continents:
    if continent[0] == A:
        print(continent)

jb30 · Answer

You could treat n as an array and check if n[0] is 'a'.

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Conor Mosier

Conor Mosier

Así que sé que hay una forma mejor y más eficiente de hacer esto. ¿Alguien tiene alguna idea?

Tu código aquí

2 Respuestas

jb30

jb30

amandae

amandae