我们如何动态地保存csv文件名，我有一个dataframe最终结果我将它保存为csv文件，并基于一些变量使用动态名称

Question

currentYear=int(time.strftime("%y", time.localtime()))-1
baseQuarter=4 #(it can be 1 or 2 or 3 or 4)

filename = "Q"+str(baseQuarter)+"-"+str(currentYear)+" ChronicNonFunnel"
filename=Q4-19 ChronicNonFunnel

final_result_Chronic(let say final_result_chronic is a dataframe)
final_result_Chronic=final_result_Chronic.to_csv('/home/developer/CustRetAnuragData/filename.csv'index = None)

我希望我的csv文件名与Q4-19 ChronicnUnfunnel一起保存

Answer 1

import os
.
.
filename = 'Q4-19 ChronicNonFunnel'
.
.
final_result_Chronic.to_csv(os.path.join('/home/developer/CustRetAnuragData',"{}.csv".format(filename)))

Answer 2

尝试：

final_result_Chronic.to_csv(f'/home/developer/CustRetAnuragData/{filename}.csv')