내가 작성한 Python 코드는 동료들로부터 칭찬을 받았습니다.

>>> s = "programming is awesome"
>>> print(s.title())
Programming Is Awesome

>>> a + b
[1, 2, 3, 4, 5, 6]

>>> a.extend(b)
>>> a
[1, 2, 3, 4, 5, 6]

>>> a = [1, 2, 3, 4, 2, 3]
>>> list(set(a))
[1, 2, 3, 4]

# sort()
>>> a = [1, 2, 3, 4, 2, 3]
>>> a.sort()
>>> a
[1, 2, 2, 3, 3, 4]
>>>
>>> a = [1, 2, 3, 4, 2, 3]
>>> a.sort(reverse=True)
>>> a
[4, 3, 3, 2, 2, 1]

# sorted()
>>> a = [1, 2, 3, 4, 2, 3]
>>> sorted(a)
[1, 2, 2, 3, 3, 4]
>>> a = [1, 2, 3, 4, 2, 3]
>>> sorted(a, reverse=True)
[4, 3, 3, 2, 2, 1]

>>> a = ['python', 'go', 'java']
>>> for i, v in enumerate(a):
... print(i, v)

# output
0 python
1 go
2 java

>>> a = [1, 2, 3, 4, 3, 4, 5, 4, 4, 2]
>>> b = max(set(a), key=a.count)
>>> b
4

>>> a = [1, 2]
>>> b = max(set(a), key=a.count)
>>> b
1

>>> from collections import Counter
>>>
>>> a = [1, 2, 3, 4, 3, 4, 5, 4, 4, 2]
>>> Counter(a)
Counter({4: 4, 2: 2, 3: 2, 1: 1, 5: 1})

>>> a = ['one', 'tow', 'three']
>>> b = [1, 2, 3]
>>> dict(zip(a, b))
{'one': 1, 'tow': 2, 'three': 3}

# list_operate.py

def main():
list_a = [1, 2, 3, 4, 5]
list_b = [4, 5, 6, 7, 8]

# 求交集的两种方式
res_a = [i for i in list_a if i in list_b]
res_b = list(set(list_a).intersection(set(list_b)))

print(f"res_a is: {res_a}")
print(f"res_b is: {res_b}")

# 求并集
res_c = list(set(list_a).union(set(list_b)))
print(f"res_c is: {res_c}")

# 求差集的两种方式，在B中但不在A中
res_d = [i for i in list_b if i not in list_a]
res_e = list(set(list_b).difference(set(list_a)))

print(f"res_d is: {res_d}")
print(f"res_e is: {res_e}")


if __name__ == '__main__':
main()

# 1、创建空字典
a = {}
b = dict()

# 2、有初始值，从输入的便利程度来说，我更喜欢第二种
a = {'a': 1, 'b': 2, 'c': 3}
b = dict(a=1, b=2, c=3)

# 3、key 来自一个列表，而 value 相同, 使用 fromkeys，那是相当的优雅
keys = ['a', 'b', 'c']
value = 100
d = dict.fromkeys(keys, value)

# 4、key 来自一个列表，而 value 也是一个列表，使用 zip
keys = ['a', 'b', 'c']
values = [1, 2, 3]
d = dict(zip(keys, values))

m = {'a': 1}
n = {'b': 2, 'c': 3}

# 合并，两种方式
# 1、使用 update
m.update(n)
# 2、使用 **
{**m, **n}

d = {'a': 1, 'b': 2}
if 'a' in d:
print('hello')

d = {'a': 1, 'b': 2}

# 1、直接用 key 取值，但这种方式不好，如果 key 不存在会报错，推荐使用 get
a = d['a']

# 2、使用 get，如果 key 不存在还可以赋默认值
a = d.get('a')
c = d.get('c', 3)

d = {'a': 1, 'b': 2, 'c': 3}

# 遍历 key
for key in d.keys():
pass

# 遍历 value
for value in d.values():
pass

# 遍历 key 和 value
for key, value in d.items():
pass

l = [1, 2, 3]
{n: n * n for n in l}
{1: 1, 2: 4, 3: 9}

d = {'a': 1, 'b': 2, 'e': 9, 'c': 5, 'd': 7}

# 按 key 排序
sorted(d.items(), key=lambda t: t[0])
# 按 key 倒序
sorted(d.items(), key=lambda t: t[0], reverse=True)

# 按 value 排序
sorted(d.items(), key=lambda t: t[1])