python安全地删除列表元素的方法

a = [1, 2, 2, 3, 4]
def even(x):
    return x % 2 == 0

a = [1, 2, 2, 3, 4]
def even(x):
    return x % 2 == 0
 
# 列表推导，但创建了一个新的变量a
a = [x for x in a if not even(x)]
# --> a = [1, 3]
print(a)

a = [1, 2, 2, 3, 4]
def even(x):
    return x % 2 == 0
 
# 列表推导，但赋值给a[:] 就地改变列表
a[:] = [x for x in a if not even(x)]
# --> a = [1, 3]
print(a)

a = [1, 2, 2, 3, 4]
def even(x):
    return x % 2 == 0
 
# 通过itertools 快速过滤
from itertools import filterfalse
a[:] = filterfalse(even, a)
# --> a = [1, 3]
print(a)

a = [1, 2, 2, 3, 4]
def even(x):
    return x % 2 == 0
 
# 注意是在列表副本a[:] 上循环
for item in a[:]:
    if even(item):
        a.remove(item)
# --> a = [1, 3]
print(a)

a = [1, 2, 2, 3, 4]
def even(x):
    return x % 2 == 0
 
# 直接在变量a上进行循环，没有在副本上
for item in a:
    if even(item):
        a.remove(item)
# --> a = [1, 2, 3] !!!
print(a)

a = [1, 2, 2, 3, 4]
def even(x):
    return x % 2 == 0
 
# 试图在循环在改变索引i，但出错！
for i in range(len(a)):
    if even(a[i]):
        del a[i]
        i -= 1
# --> IndexError: list index out of range
print(a)