在Python中按唯一列键分组数据并连接(透视表)
我对pandas和python还不熟悉,我正在尝试改变.csv中提供的数据。数据的结构应确保在同一列中具有相应收盘和日期的股票代码是连续的 例如:在Python中按唯一列键分组数据并连接(透视表),python,pandas,time-series,dataframe,financial,Python,Pandas,Time Series,Dataframe,Financial,我对pandas和python还不熟悉,我正在尝试改变.csv中提供的数据。数据的结构应确保在同一列中具有相应收盘和日期的股票代码是连续的 例如: TIC CLOSE DATE 1984-01-03 0223B 25.37500 1984-01-04 0223B 25.75000 1984-01-05 0223B 25.75000 1983-12-30 0485B 21.37500 19
TIC CLOSE
DATE
1984-01-03 0223B 25.37500
1984-01-04 0223B 25.75000
1984-01-05 0223B 25.75000
1983-12-30 0485B 21.37500
1984-01-03 0485B 21.37500
1984-01-04 0485B 22.50000
1983-12-30 0491B 17.75000
1984-01-03 0491B 17.50000
1984-01-04 0491B 17.62500
1983-12-30 3614B 74.25000
1984-01-03 3614B 73.25000
1984-01-04 3614B 76.00000
1993-07-01 3615B 47.25000
1993-07-02 3615B 47.25000
1993-07-06 3615B 46.40625
1983-12-30 3ABNKQ 4.75000
1984-01-03 3ABNKQ 5.00000
1984-01-04 3ABNKQ 5.62500
1983-12-30 3ACKH 55.25000
1984-01-03 3ACKH 54.50000
1984-01-04 3ACKH 55.25000
我想重塑数据,这样我将有一个pandas数据框,其中每个列都将是一个带有各自关闭的标记器,如果没有日期行键的数据,它将有NaN('左'连接)
我试过这样的方法:
sp = pd.read_csv('D:\Stocks.csv')
sp = pd.DataFrame(sp)
sp.columns = ['TIC', 'DATE', 'CLOSE']
sp.index = pd.to_datetime(sp['DATE'])
sp = sp[['TIC', 'CLOSE']]
unique_tickers = sp['TIC'].unique()
s0 = sp[sp['TIC'] == unique_tickers[0]]
s0 = pd.DataFrame(s0['CLOSE'])
s1 = sp[sp['TIC'] == unique_tickers[1]]
s1 = pd.DataFrame(s1['CLOSE'])
s0s1 = pd.concat([s0, s1], axis = 1)
s0s1.columns = unique_tickers[0:2]
for i in range(len(unique_tickers)):
sx = sp[sp['TIC'] == unique_tickers[i]]
sx = pd.DataFrame(sx['CLOSE'])
s0s1 = pd.concat([s0s1, sx], axis = 1)
我想我可以破解上面的代码,让它工作,但我认为有一个更优雅的解决方案。有什么想法吗
谢谢
我得到了原始问题的解决方案(感谢BrenBarn):
但是当我在一个更大的.csv上运行它时,我遇到了以下错误
ValueError:索引包含重复的条目,无法重塑
我试图找到一个解决方案,尝试sp.groupby('TIC'),然后获取所有唯一的'Date'行键,但它可以找出语法。。再次感谢您的帮助 这就是你想要的吗
>>> d.reset_index().pivot(index='DATE', columns="TIC", values="CLOSE")
TIC 0223B 0485B 0491B 3614B 3615B 3ABNKQ 3ACKH
DATE
1983-12-30 NaN 21.375 17.750 74.25 NaN 4.750 55.25
1984-01-03 25.375 21.375 17.500 73.25 NaN 5.000 54.50
1984-01-04 25.750 22.500 17.625 76.00 NaN 5.625 55.25
1984-01-05 25.750 NaN NaN NaN NaN NaN NaN
1993-07-01 NaN NaN NaN NaN 47.25000 NaN NaN
1993-07-02 NaN NaN NaN NaN 47.25000 NaN NaN
1993-07-06 NaN NaN NaN NaN 46.40625 NaN NaN
我不得不使用
reset\u index
,因为pivot
需要一列作为索引(至少在固定之前)。完美,就是这样。。我原以为它是支点,但没法开始工作。。
>>> d.reset_index().pivot(index='DATE', columns="TIC", values="CLOSE")
TIC 0223B 0485B 0491B 3614B 3615B 3ABNKQ 3ACKH
DATE
1983-12-30 NaN 21.375 17.750 74.25 NaN 4.750 55.25
1984-01-03 25.375 21.375 17.500 73.25 NaN 5.000 54.50
1984-01-04 25.750 22.500 17.625 76.00 NaN 5.625 55.25
1984-01-05 25.750 NaN NaN NaN NaN NaN NaN
1993-07-01 NaN NaN NaN NaN 47.25000 NaN NaN
1993-07-02 NaN NaN NaN NaN 47.25000 NaN NaN
1993-07-06 NaN NaN NaN NaN 46.40625 NaN NaN