我有一个历史股票交易的数据框.框架包含['ticker','date','cusip','profit','security_type']等列.原来:
trades['cusip'] = np.nan
trades['security_type'] = np.nan
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我有历史配置文件,我可以加载到具有['ticker','cusip','date','name','security_type','primary_exchange']等列的框架中.
我想用配置中的cusip和security_type更新交易框架,但仅限于股票代码和日期匹配的地方.
我以为我可以这样做:
pd.merge(trades, config, on=['ticker', 'date'], how='left')
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但是这不会更新列,它只是将配置列添加到交易中.
以下是有效的,但我认为必须有更好的方法.如果没有,我可能会在熊猫之外做.
for date in trades['date'].unique():
config = get_config_file_as_df(date)
## config['date'] == date
for ticker in trades['ticker'][trades['date'] == date]:
trades['cusip'][
(trades['ticker'] == ticker)
& (trades['date'] == date)
] \
= config['cusip'][config['ticker'] == ticker].values[0]
trades['security_type'][
(trades['ticker'] == ticker)
& (trades['date'] == date)
] \
= config['security_type'][config['ticker'] == ticker].values[0]
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unu*_*tbu 13
假设你有这个设置:
import pandas as pd
import numpy as np
import datetime as DT
nan = np.nan
trades = pd.DataFrame({'ticker' : ['IBM', 'MSFT', 'GOOG', 'AAPL'],
'date' : pd.date_range('1/1/2000', periods = 4),
'cusip' : [nan, nan, 100, nan]
})
trades = trades.set_index(['ticker', 'date'])
print(trades)
# cusip
# ticker date
# IBM 2000-01-01 NaN
# MSFT 2000-01-02 NaN
# GOOG 2000-01-03 100 # <-- We do not want to overwrite this
# AAPL 2000-01-04 NaN
config = pd.DataFrame({'ticker' : ['IBM', 'MSFT', 'GOOG', 'AAPL'],
'date' : pd.date_range('1/1/2000', periods = 4),
'cusip' : [1,2,3,nan]})
config = config.set_index(['ticker', 'date'])
# Let's permute the index to show `DataFrame.update` correctly matches rows based on the index, not on the order of the rows.
new_index = sorted(config.index)
config = config.reindex(new_index)
print(config)
# cusip
# ticker date
# AAPL 2000-01-04 NaN
# GOOG 2000-01-03 3
# IBM 2000-01-01 1
# MSFT 2000-01-02 2
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然后你就可以在更新NaN值trades与值从config使用DataFrame.update方法.请注意,DataFrame.update根据索引匹配行(这就是set_index上面调用的原因).
trades.update(config, join = 'left', overwrite = False)
print(trades)
# cusip
# ticker date
# IBM 2000-01-01 1
# MSFT 2000-01-02 2
# GOOG 2000-01-03 100 # If overwrite = True, then 100 is overwritten by 3.
# AAPL 2000-01-04 NaN
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