我写的 Python 代码，同事都说好

>>> 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)# output0 python1 go2 java

>>> a = [1, 2, 3, 4, 3, 4, 5, 4, 4, 2]>>> b = max(set(a), key=a.count)>>> b4

>>> a = [1, 2]>>> b = max(set(a), key=a.count)>>> b1

>>> 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.pydef 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 = 100d = dict.fromkeys(keys, value)# 4、key 来自一个列表，而 value 也是一个列表，使用 zipkeys = ['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 updatem.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 geta = 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 keyfor key in d.keys():pass# 遍历 valuefor value in d.values():pass# 遍历 key 和 valuefor 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])

l = [{'name': 'a', 'count': 4}, {'name': 'b', 'count': 1}, {'name': 'd', 'count': 2}, {'name': 'c', 'count': 6}]sorted(l, key=lambda e: e.__getitem__('count'))# 倒序sorted(l, key=lambda e: e.__getitem__('count'), reverse=True)