Python lambda匿名函数，递归应用

Posted 2021-01-17

import os

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Lambda函数能接收任何数量的参数但只能返回一个表达式的值
匿名函数不能直接调用print，因为lambda需要一个表达式
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sum = lambda x, y: x + y
print(‘x+y=‘, sum(2, 6))

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匿名函数应用
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def test_nm(a, b, func):
result = func(a, b)
return result

print(test_nm(20, 30, lambda x, y: x * y))
print(test_nm(20, 30, lambda x, y: x + y))

stus = [{"name": "zs", "age": 20}, {"name": "tom", "age": 32}, {"name": "jack", "age": 19}]

通过匿名lambda函数排序，reverse=True 降序。否则为升序

匿名函数作为参数传递

stus.sort(key=lambda x: x["age"], reverse=True)
print(stus)

递归读取文件目录信息

py_list = []
p_dir = r‘E:2016qBook2018‘
file_name = r‘PythonCode‘

def read_contain_key_word(parent_dir, file_name, key_word):
full_path = os.path.join(parent_dir, file_name)
if os.path.isdir(full_path):
for f in os.listdir(full_path):
read_contain_key_word(full_path, f, key_word)
else:
if full_path.endswith(‘.py‘):
if read_key_word(full_path, key_word):
py_list.append(full_path)

def read_key_word(full_path, key_word):
read_flag = False
f = open(full_path, ‘r‘, encoding=‘utf-8‘)
while True:
line = f.readline()
if line == "":
break
elif key_word in line:
read_flag = True
break
f.close()
return read_flag

read_contain_key_word(p_dir, file_name, ‘sort‘)
print(py_list)