下面有很多代碼向您展示我必須使用的技能級別才能完成此任務。初學者技術只請。如何對列表的不同列表中的某些索引執行操作,然後通過另一個索引將它們組合在一起
DEF get_monthly_averages(original_list):
#print(original_list)
daily_averages_list = [ ]
product_vol_close = [ ] # used for numerator
monthly_averages_numerator_list = [ ]
for i in range (0, len(original_list)):
month_list = original_list[i][0][0:7] #Cutting day out of the date leaving Y-M
volume_str = float(original_list[i][5]) #V
adj_close_str = float(original_list[i][6]) #C
daily_averages_sublists = [month_list,volume_str,adj_close_str] #[Date,V,C]
daily_averages_list.append(daily_averages_sublists)
for i in range (0, len(daily_averages_list)): #Attempt at operation
vol_close = daily_averages_list[i][1]*daily_averages_list[i][2]
month_help = daily_averages_list[i][0]
product_vol_sublists = [month_help,vol_close]
product_vol_close.append(product_vol_sublists)
print(product_vol_close)
for i in range (0, len(product_vol_close)): #<-------TROUBLE STARTS
for product_vol_close[i][0]==product_vol_close[i][0]: #When the month is the same
monthly_averages_numerator = product_vol_close[i][1]+product_vol_close[i][1]
# monthly_averages_numerator = sum(product_vol_close[i][1]) #tried both
month_assn = product_vol_close[i][0]
numerator_list_sublists = [month_assn,monthly_averages_numerator]
monthly_averages_numerator_list.append(numerator_list_sublists)
print(monthly_averages_numerator_list)
原版是在形式:
[['2004-08-30', '105.28', '105.49', '102.01', '102.01', '2601000', '102.01'],
['2004-08-27', '108.10', '108.62', '105.69', '106.15', '3109000', '106.15'],
['2004-08-26', '104.95', '107.95', '104.66', '107.91', '3551000', '107.91'],
['2004-08-25', '104.96', '108.00', '103.88', '106.00', '4598900', '106.00'],
['2004-08-24', '111.24', '111.60', '103.57', '104.87', '7631300', '104.87'],
['2004-08-23', '110.75', '113.48', '109.05', '109.40', '9137200', '109.40'],
['2004-08-20', '101.01', '109.08', '100.50', '108.31', '11428600', '108.31'],
['2004-08-19', '100.00', '104.06', '95.96', '100.34', '22351900', '100.34']]
的0指數的時間,第5是V,第六是C.
我需要爲每個月單獨執行下面的操作,最後有一個包含兩個元素的元組; 0代表月份年份,1代表'average_price',如下所示。我試圖最終從原始列表中的每個列表中取出第5和第6個值,並執行如下操作...(我需要爲我的班級使用初級技術...感謝您的理解)
average_price =(V1 * C1 + V2 * C2 + ... + Vn * Cn)/(V1 + V2 + ... + Vn)
(V =列表中的每個第5個元素C =名單)
我的問題是,只有在執行上述任務,一個月單獨,而不是整個列表,然後有一個結果,例如,
[('month1',average_price),('month2',average_price),...]
我做了
for i in range (0, len(product_vol_close)): #<-------TROUBLE STARTS
for product_vol_close[i][0]==product_vol_close[i][0]:
同月和今年他們組合在一起的時候。
嘗試展示我想要做的事情。我無法找到任何答案,如何讓我的工作方式,我想要的。
如果仍有混淆請評論!再次感謝您對此事的耐心,理解和幫助!
我完全失去了。