Python根据值是奇数还是偶数来更改列表元素

Question

Supposed to change a value of a list based on if the value is odd or even. Error:

list assignment index out of range 

Code:

def list_mangler(list_in):
    for i in list_in:
        if i % 2 == 0:
            list_in[i] = i * 2
        else:
            list_in[i] = i * 3
    return list_in



list_mangler([1, 2, 3, 4])

Answer 1

the problem is that for i in list_in yields items in the list, not indices . To get the indices, use enumerate :

for i, val in enumerate(list_in):
    if i % 2 == 0:
        list_in[i] = val * 2
    ...

If you wanted to return a new list (rather than mutating the old list), a list comprehension with a conditional expression could do nicely:

[val * 3 if val % 2 else val * 2 for val in list_in]

And of course, you can use this with slice assignment to mutate list_in if you really need to:

list_in[:] = [val * 3 if val % 2 else val * 2 for val in list_in]

---

^{As a point of style, if your function's purpose is to modify the list in place, don't return the list. That makes it clear that the input was mutated rather than returning a new list that was somehow constructed from the data in the original.}

Answer 2

You need use enumerate , because for i in list returns the element not the index.

for index, val in enumerate(list_in):
        if val % 2 == 0:
            list_in[index] = val* 2
        else:
            list_in[index] = val * 3
    return list_in