python - 如何改变列表中的每三个元素？

Question

<分区>

Answer 1

不要想太多。在这种情况下，列表推导式、函数等都是矫枉过正的。您需要遍历列表并就地修改它，对吗？所以你没有修改你正在迭代的结构。迭代目标实际上是列表的索引。

xs = [1, 2, 3, 4, 5, 6, 7, 8, 9]

print(xs)

for idx in range(2, len(xs), 3):
    if xs[idx] % 2 == 1:
        xs[idx] -= 1

print(xs)

这相当于 for(i = 2; i < len(xs); i++)在许多低级语言中循环。

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与另一个答案中提出的列表理解解决方案相比，for循环稍微快一些:

from timeit import timeit


def in_place(xs):
    for idx in range(2, len(xs), 3):
        if xs[idx] % 2 == 1:
            xs[idx] -= 1
    return xs


def list_comp(xs):
    xs[2::3] = [i - 1 if (i % 3 == 0 and i % 2 != 0) else i for i in xs[2::3]]
    return xs


# This answer is an improvement, as it eliminates one modulus
def list_comp2(xs):
    xs[2::3] = [x - 1 if x % 2 else x for x in xs[2::3]]
    return xs


context = {"globals": globals(), "setup": "xs = [1, 2, 3, 4, 5, 6, 7, 8, 9]"}
print(f"For loop: {timeit('in_place(xs)', **context)}")
print(f"List comprehension:  {timeit('list_comp(xs)', **context)}")
print(f"List comprehension 2:  {timeit('list_comp2(xs)', **context)}")

在我的机器上，这会产生:

For loop: 0.608657514
List comprehension:  0.7510721879999999
List comprehension 2:  0.641639047