Python重构此函数以将其认知复杂度从19减少到15

Question

我从sonarlint看到这个消息，并试图弄清楚如何降低这个功能的认知复杂性。任何帮助都提前得到赞赏。

import os
import json
import click
import hcl

cfn = [".json", ".template", ".yaml", ".yml"]
tf  = ["tf"]

def file_handler(dir):
    for root, dirs, files in os.walk(dir):
        for file in files:
            if file.endswith(tuple(cfn)):
                with open(os.path.join(root, file), 'r') as fin:
                    try:
                        file = fin.read()
                        if "AWSTemplateFormatVersion" in file:
                            data = json.dumps(file)
                            print(data)

                    except ValueError as e:
                        raise SystemExit(e)

            elif file.endswith(tuple(tf)):
                with open(os.path.join(root, file), 'r') as file:
                    try:
                        obj  = hcl.load(file)
                        data = json.dumps(obj)
                        print(data)
                    except ValueError as e:
                        raise SystemExit(e)
    return data

Answer 1

提取函数以生成文件路径：

def _file_paths(directory):
    for root, dirs, filenames in os.walk(directory):
        for filename in filenames:
            file_path = os.path.join(root, filename)
            if os.path.isfile(file_path):
                yield file_path

统一异常处理：

from contextlib import contextmanager

@contextmanager
def _translate_error(from_error, to_error):
    try:
        yield
    except from_error as error:
        raise to_error(error)

提取函数以处理文件类型：

def _handle_cfn(file_object):
    file_contents = file_object.read()
    if "AWSTemplateFormatVersion" in file_contents:
        data = json.dumps(file_contents)
        print(data)


def _handle_tf(file_object):
    obj  = hcl.load(file_oject)
    data = json.dumps(obj)
    print(data)

为文件扩展名不匹配时创建一个空处理程序：

def _null_handler(file_object):
    pass

将文件扩展名映射到处理程序：

_extension_handlers = {'.json': _handle_cfn,
            '.template': _handle_cfn,
            '.yaml': _handle_cfn,
            '.yml': _handle_cfn,
            '.tf': _handle_tf}

把它放在一起：

import os
import json
import click
import hcl

def file_handler(dir):
    for file_path in _file_paths(dir):
        base, extension = os.path.splitext(file_path)
        handler = _extension_handlers.get(extension, _null_handler)
        with open(file_path) as file_object:
            with _translate_error(ValueError, SystemExit):
                handler(file_object)

您可以通过提取处理每个文件的函数来进一步：

def _handle_file(file_path):
    base, extension = os.path.splitext(file_path)
    handler = _extension_handlers.get(extension, _null_handler)
    with open(file_path) as file_object:
        with _translate_error(ValueError, SystemExit):
            handler(file_object)

然后你的主要功能是：

def file_handler(dir):
    for file_path in _file_paths(dir):
        _handle_file(file_path)

Answer 2

您可以通过使用两个生成器表达式来删除一级缩进：按扩展名筛选文件：

def file_handler(dir):
    for root, dirs, files in os.walk(dir):
        cfn_files = (file for file in files if file.endswith(tuple(cfn)))
        tf_files = (file for file in files if file.endswith(tuple(tf)))
        for file in cfn_files:
            with open(os.path.join(root, file), 'r') as fin:
                try:
                    file = fin.read()
                    if "AWSTemplateFormatVersion" in file:
                        data = json.dumps(file)
                        print(data)
                except ValueError as e:
                    raise SystemExit(e)
        for file in tf_files:
            with open(os.path.join(root, file), 'r') as file:
                try:
                    obj  = hcl.load(file)
                    data = json.dumps(obj)
                    print(data)
                except ValueError as e:
                    raise SystemExit(e)
    return data

Answer 3

您应该考虑使用glob进行递归文件查找，尤其是当您知道要查找的文件扩展名时：

import glob
import json
import os

import click
import hcl

def file_handler(dir):
    for extension in cfn:
        # eg: /path/to/dir/**/*.json
        glob_search = os.path.join(dir, "**/*{}".format(extension))  
        filenames = glob.glob(glob_search, recursive=True)

        for filename in filenames:
            with open(filename, 'r') as fin:
                try:
                    file = fin.read()
                    if "AWSTemplateFormatVersion" in file:
                        data = json.dumps(file)
                        print(data)

                except ValueError as e:
                    raise SystemExit(e)

    for extension in tf:
        # eg: /path/to/dir/**/*tf
        glob_search = os.path.join(dir, "**/*{}".format(extension))
        filenames = glob.glob(glob_search, recursive=True)

        for filename in filenames:
            with open(filename, 'r') as file:
                try:
                    obj = hcl.load(file)
                    data = json.dumps(obj)
                    print(data)
                except ValueError as e:
                    raise SystemExit(e)

仅供参考：关于使用glob（Use a Glob() to find files recursively in Python?）的问题