我有一个可重现的代码如下:
import pandas as pd
import datetime
foo = pd.read_csv("http://m.uploadedit.com/bbtc/1545406250692.txt", header=None, names=["Stock","Date","Time", "Open", "High", "Low", "Close", "Volume", "OI"], dtype={"Stock":"category"}, parse_dates= [['Date', 'Time']], index_col="Date_Time")
foo.sort_index(inplace=True)
bar = foo.between_time('09:00:00', '15:30:00') #Dropping post and pre market data i.e. from index 15:31 - 16:35
#resampling the data by 120 Minutes (2 hours)
twohour = bar.loc["2018-11-22 09:08:00":].resample('120Min',closed = 'right',label = 'left', base=75).agg({'Open': 'first', 'High': 'max', 'Low': 'min','Close': 'last'}).dropna()
twohour.head(7)
Out[]:
Close High Open Low
Date_Time
2018-11-22 07:15:00 321.3 321.30 321.30 321.30
2018-11-22 09:15:00 324.5 326.90 320.10 320.00
2018-11-22 11:15:00 323.2 324.85 324.60 322.20
2018-11-22 13:15:00 319.9 324.35 323.20 319.50
2018-11-22 15:15:00 320.0 320.35 319.85 319.15
2018-11-26 07:15:00 324.90 324.90 324.90 324.90
2018-11-26 09:15:00 311.35 324.40 323.10 309.60
我希望时间09:15:00的索引中的Open列中的每个值都被索引中的Close列的值替换为时间07:15:00.
简而言之,我需要这个输出:
Out[]:
Close High Open Low
Date_Time
2018-11-22 07:15:00 321.3 321.30 321.30 321.30
2018-11-22 09:15:00 324.5 326.90 321.30 320.00
2018-11-22 11:15:00 323.2 324.85 324.60 322.20
2018-11-22 13:15:00 319.9 324.35 323.20 319.50
2018-11-22 15:15:00 320.0 320.35 319.85 319.15
2018-11-26 07:15:00 324.90 324.90 324.90 324.90
2018-11-26 09:15:00 311.35 324.40 324.90 309.60
我尝试通过将DateTimeindex转换为字典然后替换值来使用.loc.但字典没有排序,所以需要对字典进行排序,代码变得越来越难看.
任何帮助将不胜感激.
最佳答案 您可以在比较之前将索引转换为timdelta或字符串:
# timedelta option, vectorised & efficient
mask_bool = (df.index - df.index.normalize()) == '09:15:00'
# string alternative, inefficient
mask_bool = df.index.strftime('%H:%M') == '09:15'
然后通过loc或mask分配:
# Option 1: assign conditionally via loc
df.loc[mask_bool, 'Open'] = df['Close'].shift(1)
# Option 2: mask with pd.Series.mask
df['Open'] = df['Open'].mask(mask_bool, df['Close'].shift(1))
结果:
print(df)
Close High Open Low
Date_Time
2018-11-22 07:15:00 321.30 321.30 321.30 321.30
2018-11-22 09:15:00 324.50 326.90 321.30 320.00
2018-11-22 11:15:00 323.20 324.85 324.60 322.20
2018-11-22 13:15:00 319.90 324.35 323.20 319.50
2018-11-22 15:15:00 320.00 320.35 319.85 319.15
2018-11-26 07:15:00 324.90 324.90 324.90 324.90
2018-11-26 09:15:00 311.35 324.40 324.90 309.60
绩效基准
对于较大的数据帧,timedelta矢量化版本应该是高效的,但请注意这将取决于系统和设置:
# Python 3.6.5, Pandas 0.23, NumPy 1.14.3
import pandas as pd
from datetime import time
df = pd.DataFrame.from_dict({'Date_Time': ['2018-11-22 07:15:00', '2018-11-22 09:15:00',
'2018-11-22 11:15:00', '2018-11-22 13:15:00',
'2018-11-22 15:15:00', '2018-11-26 07:15:00',
'2018-11-26 09:15:00'],
'Close': [321.3, 324.5, 323.2, 319.9, 320.0, 324.9, 311.35],
'High': [321.3, 326.9, 324.85, 324.35, 320.35, 324.9, 324.4],
'Open': [321.3, 321.3, 324.6, 323.2, 319.85, 324.9, 324.9],
'Low': [321.3, 320.0, 322.2, 319.5, 319.15, 324.9, 309.6]})
df['Date_Time'] = pd.to_datetime(df['Date_Time'])
df = df.set_index('Date_Time')
df = pd.concat([df]*10**4)
%timeit (df.index - df.index.normalize()) == '09:15:00' # 8.67 ms
%timeit df.index.strftime('%H:%M') == '09:15' # 651 ms
%timeit df.index.time == time(9, 15) # 28.3 ms