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174 lines
7.0 KiB
Python
174 lines
7.0 KiB
Python
from typing import Optional
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import pandas as pd
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from freqtrade.exchange import timeframe_to_minutes
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def merge_informative_pair(
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dataframe: pd.DataFrame,
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informative: pd.DataFrame,
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timeframe: str,
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timeframe_inf: str,
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ffill: bool = True,
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append_timeframe: bool = True,
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date_column: str = "date",
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suffix: Optional[str] = None,
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) -> pd.DataFrame:
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"""
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Correctly merge informative samples to the original dataframe, avoiding lookahead bias.
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Since dates are candle open dates, merging a 15m candle that starts at 15:00, and a
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1h candle that starts at 15:00 will result in all candles to know the close at 16:00
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which they should not know.
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Moves the date of the informative pair by 1 time interval forward.
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This way, the 14:00 1h candle is merged to 15:00 15m candle, since the 14:00 1h candle is the
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last candle that's closed at 15:00, 15:15, 15:30 or 15:45.
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Assuming inf_tf = '1d' - then the resulting columns will be:
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date_1d, open_1d, high_1d, low_1d, close_1d, rsi_1d
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:param dataframe: Original dataframe
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:param informative: Informative pair, most likely loaded via dp.get_pair_dataframe
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:param timeframe: Timeframe of the original pair sample.
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:param timeframe_inf: Timeframe of the informative pair sample.
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:param ffill: Forwardfill missing values - optional but usually required
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:param append_timeframe: Rename columns by appending timeframe.
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:param date_column: A custom date column name.
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:param suffix: A string suffix to add at the end of the informative columns. If specified,
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append_timeframe must be false.
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:return: Merged dataframe
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:raise: ValueError if the secondary timeframe is shorter than the dataframe timeframe
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"""
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informative = informative.copy()
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minutes_inf = timeframe_to_minutes(timeframe_inf)
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minutes = timeframe_to_minutes(timeframe)
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if minutes == minutes_inf:
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# No need to forwardshift if the timeframes are identical
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informative["date_merge"] = informative[date_column]
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elif minutes < minutes_inf:
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# Subtract "small" timeframe so merging is not delayed by 1 small candle
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# Detailed explanation in https://github.com/freqtrade/freqtrade/issues/4073
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if not informative.empty:
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if timeframe_inf == "1M":
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informative["date_merge"] = (
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informative[date_column] + pd.offsets.MonthBegin(1)
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) - pd.to_timedelta(minutes, "m")
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else:
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informative["date_merge"] = (
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informative[date_column]
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+ pd.to_timedelta(minutes_inf, "m")
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- pd.to_timedelta(minutes, "m")
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)
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else:
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informative["date_merge"] = informative[date_column]
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else:
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raise ValueError(
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"Tried to merge a faster timeframe to a slower timeframe."
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"This would create new rows, and can throw off your regular indicators."
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)
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# Rename columns to be unique
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date_merge = "date_merge"
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if suffix and append_timeframe:
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raise ValueError("You can not specify `append_timeframe` as True and a `suffix`.")
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elif append_timeframe:
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date_merge = f"date_merge_{timeframe_inf}"
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informative.columns = [f"{col}_{timeframe_inf}" for col in informative.columns]
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elif suffix:
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date_merge = f"date_merge_{suffix}"
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informative.columns = [f"{col}_{suffix}" for col in informative.columns]
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# Combine the 2 dataframes
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# all indicators on the informative sample MUST be calculated before this point
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if ffill:
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# https://pandas.pydata.org/docs/user_guide/merging.html#timeseries-friendly-merging
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# merge_ordered - ffill method is 2.5x faster than separate ffill()
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dataframe = pd.merge_ordered(
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dataframe,
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informative,
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fill_method="ffill",
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left_on="date",
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right_on=date_merge,
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how="left",
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)
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else:
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dataframe = pd.merge(
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dataframe, informative, left_on="date", right_on=date_merge, how="left"
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)
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dataframe = dataframe.drop(date_merge, axis=1)
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return dataframe
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def stoploss_from_open(
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open_relative_stop: float, current_profit: float, is_short: bool = False, leverage: float = 1.0
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) -> float:
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"""
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Given the current profit, and a desired stop loss value relative to the trade entry price,
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return a stop loss value that is relative to the current price, and which can be
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returned from `custom_stoploss`.
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The requested stop can be positive for a stop above the open price, or negative for
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a stop below the open price. The return value is always >= 0.
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`open_relative_stop` will be considered as adjusted for leverage if leverage is provided..
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Returns 0 if the resulting stop price would be above/below (longs/shorts) the current price
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:param open_relative_stop: Desired stop loss percentage, relative to the open price,
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adjusted for leverage
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:param current_profit: The current profit percentage
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:param is_short: When true, perform the calculation for short instead of long
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:param leverage: Leverage to use for the calculation
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:return: Stop loss value relative to current price
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"""
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# formula is undefined for current_profit -1 (longs) or 1 (shorts), return maximum value
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_current_profit = current_profit / leverage
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if (_current_profit == -1 and not is_short) or (is_short and _current_profit == 1):
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return 1
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if is_short is True:
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stoploss = -1 + ((1 - open_relative_stop / leverage) / (1 - _current_profit))
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else:
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stoploss = 1 - ((1 + open_relative_stop / leverage) / (1 + _current_profit))
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# negative stoploss values indicate the requested stop price is higher/lower
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# (long/short) than the current price
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return max(stoploss * leverage, 0.0)
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def stoploss_from_absolute(
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stop_rate: float, current_rate: float, is_short: bool = False, leverage: float = 1.0
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) -> float:
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"""
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Given current price and desired stop price, return a stop loss value that is relative to current
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price.
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The requested stop can be positive for a stop above the open price, or negative for
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a stop below the open price. The return value is always >= 0.
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Returns 0 if the resulting stop price would be above the current price.
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:param stop_rate: Stop loss price.
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:param current_rate: Current asset price.
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:param is_short: When true, perform the calculation for short instead of long
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:param leverage: Leverage to use for the calculation
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:return: Positive stop loss value relative to current price
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"""
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# formula is undefined for current_rate 0, return maximum value
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if current_rate == 0:
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return 1
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stoploss = 1 - (stop_rate / current_rate)
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if is_short:
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stoploss = -stoploss
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# negative stoploss values indicate the requested stop price is higher/lower
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# (long/short) than the current price
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# shorts can yield stoploss values higher than 1, so limit that as well
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return max(min(stoploss, 1.0), 0.0) * leverage
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