Python开发中常见的10个陷阱你踩过几个(附解决方案)

def append_to(element, to=[]):
    to.append(element)
    return to

print(append_to(1))  # 输出: [1]
print(append_to(2))  # 输出: [1, 2] 而不是预期的[2]

def append_to(element, to=None):
    if to is None:
        to = []
    to.append(element)
    return to

lst = [1, 2, 3, 4]
for item in lst:
    if item % 2 == 0:
        lst.remove(item)
print(lst)  # 输出: [1, 3] 而不是预期的[1, 3]

lst = [1, 2, 4, 5]
for item in lst:
    if item % 2 == 0:
        lst.remove(item)
print(lst)  # 输出: [1, 4, 5] 因为删除2后，4的索引变成了1被跳过了

lst = [x for x in lst if x % 2 != 0]
# 或
for item in lst[:]:
    if item % 2 == 0:
        lst.remove(item)

funcs = []
for i in range(3):
    def func():
        return i
    funcs.append(func)
    
print([f() for f in funcs])  # 输出: [2, 2, 2] 而不是预期的[0, 1, 2]

for i in range(3):
    def func(i=i):
        return i
    funcs.append(func)

a = [1, 2, 3]
b = a
c = [1, 2, 3]

print(a == b)  # True
print(a is b)  # True
print(a == c)  # True
print(a is c)  # False

a = 256
b = 256
print(a is b)  # True

a = 257
b = 257
print(a is b)  # False (在交互式环境中)

class A:
    x = 1
    
class B(A):
    pass
    
class C(A):
    pass

B.x = 2
print(A.x, B.x, C.x)  # 输出: 1 2 1

A.x = 3
print(A.x, B.x, C.x)  # 输出: 3 2 3

a = "hello"
b = "hello"
print(a is b)  # True

a = "hello world"
b = "hello world"
print(a is b)  # False (取决于实现)

gen = (x for x in range(3))
print(list(gen))  # [0, 1, 2]
print(list(gen))  # [] 因为生成器已耗尽

lst = [1, [2, 3]]
lst2 = lst.copy()
lst2[1][0] = 4
print(lst)  # [1, [4, 3]] 原始列表也被修改了

import copy
lst2 = copy.deepcopy(lst)

def func():
    try:
        return 1
    finally:
        return 2

print(func())  # 输出: 2