学习自华为云学院白泽老师《Python入门篇》课程
生成器
(i for i in range(10))
at 0x000001F6CB7E1AC0>
相互引用产生死循环
a = [1,2,3]
b = [3,4,5]
a.append(b)
b.append(a)
print(a)
print(b)
[1, 2, 3, [3, 4, 5, [...]]]
[3, 4, 5, [1, 2, 3, [...]]]
a = [""]*3
b = [a]*3
b[1][1] = 0
print(a)
print(b)
['', 0, '']
[['', 0, ''], ['', 0, ''], ['', 0, '']]
a, b = {}, 5
a[b] = a, b
print(a)
{5: ({...}, 5)}
lambda表达式&排序
# 排序
l = [(1, 2), (2, 4), (2, 3), (3, 2)]
l.sort(key=lambda x:(x[0],x[1]))
l
[(1, 2), (2, 3), (2, 4), (3, 2)]
# 列表删除元素
s = [99,100,46,44,68,50,59,50,59,66,89,54]
for i in s:
if i < 60:
s.remove(i)
s #有问题 [99, 100, 44, 68, 59, 59, 66, 89]
[99, 100, 44, 68, 59, 59, 66, 89]
# 列表删除元素
s = [99,100,46,44,68,50,59,50,59,66,89,54]
for i in s:
if i < 60:
s.pop(i)
s #IndexError: pop index out of range
---------------------------------------------------------------------------
IndexError Traceback (most recent call last)
in
3 for i in s:
4 if i < 60:
----> 5 s.pop(i)
6 s
IndexError: pop index out of range
列表在每一次循环当中都会动态变化，当元素被删，循环的列表索引下标发生改变，会直接影响循环过程
解决方法1：创建一份浅拷贝
import copy
s = [99,100,68,59,50,59,66,89,54]
s1 = copy.copy(s)
for i in s:
if i < 60:
s1.remove(i)
s1
[99, 100, 68, 66, 89]
解决方法2：列表推导式
import copy
s = [99,100,46,44,68,50,59,50,59,66,89,54]
[ i for i in s if i>=60 ]
[99, 100, 68, 66, 89]
解决方法3：倒着删，删除后避免元素迁移导致索引变化
s = [99,100,46,44,68,50,59,50,59,66,89,54]
for i in range(len(s)-1, -1, -1):
print(i,":", s[i])
if s[i] < 60:
print('removing...')
s.pop(i)
s
11 : 54
removing...
10 : 89
9 : 66
8 : 59
removing...
7 : 50
removing...
6 : 59
removing...
5 : 50
removing...
4 : 68
3 : 44
removing...
2 : 46
removing...
1 : 100
0 : 99
[99, 100, 68, 66, 89]
解决方法4：filter函数 + lambda表达式
s = [99,100,46,44,68,50,59,50,59,66,89,54]
list(filter(lambda x:x>=60, s))
[99, 100, 68, 66, 89]
浅拷贝和深拷贝
l = [1,2,3,[5,4]]
l1 = l.copy() #copy方法实现浅拷贝
import copy
l2 = copy.deepcopy(l)
l == l1 == l2
True
l1 is l
False
l2 is l
False
l1[3][1]=6
l
[1, 2, 3, [5, 6]]
l1
[1, 2, 3, [5, 6]]
l2
[1, 2, 3, [5, 4]]
诡异的运算符使用方法
不推荐，仅了解
(False == False) in [False]
False
False == (False in [False])
False
False == False in [False]
True
True is False == False
False
False is False is False
True
1 > 0 < 1 #等价于 1>0 and 0<1
True
(1 > 0 ) < 1
False
1 > (0 < 1)
False
操作符优先级：**最高， and or not最小
指定i不能打断循环
循环产生迭代器，循环只不过遍历已经生成完数值
# 循环产生迭代器，循环只不过遍历已经生成完数值
for i in range(4):
print(i)
i = 10
0
1
2
3
代码效率问题
# 判断是否感染新冠
COVID = 1
if COVID:#避免没有必要的计算执行
print("请隔离治疗")
else:
print("通过")
请隔离治疗
# 排序
import random
l = [i for i in range(10)]
random.shuffle(l)
l
[7, 1, 8, 4, 5, 6, 9, 0, 2, 3]
temp = []
n = len(l)
for i in range(n):
s = l[0]
for i in range(len(l)):
if l[i] < s:
s = l[i]
print(l)
l.remove(s)
temp.append(s)
temp
[7, 1, 8, 4, 5, 6, 9, 0, 2, 3]
[7, 1, 8, 4, 5, 6, 9, 2, 3]
[7, 8, 4, 5, 6, 9, 2, 3]
[7, 8, 4, 5, 6, 9, 3]
[7, 8, 4, 5, 6, 9]
[7, 8, 5, 6, 9]
[7, 8, 6, 9]
[7, 8, 9]
[8, 9]
[9]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
空间复杂度换时间复杂度
def quick_sort(arr):
if len(arr) < 2:
return arr
# 选取基准，随便选哪个都可以，选中间的便于理解
mid = arr[len(arr) // 2]
# 定义基准值的左右两个数列
left, right = [], []
# 从原始数组中移除基准值
arr.remove(mid)
for item in arr:
#大于基准值放右边
if item >= mid:
right.append(item)
else:
left.append(item)
#使用递归迭代进行比较
return quick_sort(left) + [mid] + quick_sort(right)
quick_sort([1, 10, 9, 5, 8, 0])
[0, 1, 5, 8, 9, 10]
装饰器
import time
def func(f):
def inner(*args, **kwargs):
start_time = time.time()
f(*args, **kwargs)
end_time = time.time()
print('耗时：%s秒' % (end_time - start_time))
return inner
# 装饰器就是在不改变原本代码情况下加入新的功能
@func
def test():
time.sleep(1)
#相当于
#func(test())