Python 2.7 如何减少Python中If-Else语句的代码重复
我是一个学生,有一个很坏的习惯,就是到处复制代码,这是我想要改变的 下面是我正在编写的函数的一段代码。快速解释:代码将为一个人查看人力资源网站,并返回他管理的员工的信息(假设他管理任何人) 有两种类型的员工:正式员工和合同工。在网站上,经理下面的正式员工将全部列在员工名单中,承包商将列在承包商名单中Python 2.7 如何减少Python中If-Else语句的代码重复,python-2.7,code-duplication,Python 2.7,Code Duplication,我是一个学生,有一个很坏的习惯,就是到处复制代码,这是我想要改变的 下面是我正在编写的函数的一段代码。快速解释:代码将为一个人查看人力资源网站,并返回他管理的员工的信息(假设他管理任何人) 有两种类型的员工:正式员工和合同工。在网站上,经理下面的正式员工将全部列在员工名单中,承包商将列在承包商名单中 response = opener.open('myFakeOrgHierarchy.com/JohnSmith_The_Manager') allDataFromPage = (response.
response = opener.open('myFakeOrgHierarchy.com/JohnSmith_The_Manager')
allDataFromPage = (response.read())
jsonVersionOfAllData = json.loads(allDataFromPage)
listOfAllReports = []
numOfEmployeeDirectReports = len(jsonVersionOfAllData['employeeList']['list'])
numOfContractWorkerReports = len(jsonVersionOfAllData['contractWorkerList']['list'])
if numOfEmployeeDirectReports != 0:
for i in range(0, numOfEmployeeDirectReports, 1):
workerInfo = {}
workerInfo['empLname'] = jsonVersionOfAllData['employeeList']['list'][i]['lastName']
workerInfo['empFname'] = jsonVersionOfAllData['employeeList']['list'][i]['firstName']
listOfAllReports.append(workerInfo)
if numOfContractWorkerReports != 0:
for i in range(0, numOfContractWorkerReports, 1):
workerInfo = {}
workerInfo['empLname'] = jsonVersionOfAllData['contractWorkerList']['list'][i]['lastName']
workerInfo['empFname'] = jsonVersionOfAllData['contractWorkerList']['list'][i]['firstName']
listOfAllReports.append(workerInfo)
如您所见,我有几行代码与其他代码几乎相同,只是有一些小的变化。有没有办法检查contractWorkerList和employeeList是否为空,并且(假设它们不是空的)检查contractWorkerList和employeeList并获取值而不复制代码
(因为我是一个相对的初学者,如果您能提供任何简单的示例和建议,我将不胜感激)避免代码重复的最常见方法是使用该代码定义函数
def checkIfEmpty(numOfReports, listName):
if numOfReports != 0:
for i in range(0, numOfReports, 1):
workerInfo = {}
workerInfo['empLname'] = jsonVersionOfAllData[listName]['list'][i]['lastName']
workerInfo['empFname'] = jsonVersionOfAllData[listName]['list'][i]['firstName']
listOfAllReports.append(workerInfo)
因此,您将得到简单易读的代码:
checkIfEmpty(numOfEmployeeDirectReports, 'employeeList')
checkIfEmpty(numOfContractWorkerReports, 'contractWorkerList')
在此特定场景中,您可以执行以下操作:
for var, key in [(numOfEmployeeDirectReports, 'employeeList'),
(numOfContractWorkerReports, 'contractWorkerList')]:
if var != 0:
for i in range(0, var, 1):
workerInfo = {}
workerInfo['empLname'] = jsonVersionOfAllData[key]['list'][i]['lastName']
workerInfo['empFname'] = jsonVersionOfAllData[key]['list'][i]['firstName']
listOfAllReports.append(workerInfo)
对于初学者,每次看到重复的内容时,都要考虑预先创建一个变量并使用它。之后,您可以决定应该将哪些因素分解到函数中。下面,我刚刚删除了大部分重复的项目
response = opener.open('myFakeOrgHierarchy.com/JohnSmith_The_Manager')
allDataFromPage = (response.read())
jsonVersionOfAllData = json.loads(allDataFromPage)
listOfAllReports = []
for listType in ('employeeList', 'contractWorkerList'):
json_ver = jsonVersionOfAllData[listType]['list']
directReports = len(json_ver)
if directReports != 0:
for i in range(0, directReports, 1):
workerInfo = {}
for wi_name, json_name in (('empLname', 'lastName'), ('empFname', 'firstName')):
workerInfo[wi_name] = json_ver[i][json_name]
listOfAllReports.append(workerInfo)
将其分离为一个函数,该函数接受一个参数,如JSONVersionFallData和一个字符串,如
employeeList
(可能还接受一些其他参数,具体取决于返回的内容)。