我無(wú)法將一個(gè) numpy 數(shù)組從一個(gè) numpy 數(shù)組的元組分配給一個(gè) pandas 數(shù)據(jù)框。有人能幫我嗎？我正在使用第三方庫(kù)來(lái)獲取數(shù)據(jù)并生成輸出 numpy 數(shù)組。代碼：import timeimport numpy as npimport pandas as pdimport ccxt # crypto exchange libraryimport tulipy as ti # financial indicatorsdef currentTimeMillis(self): return int(round(time.time() * 1000))currentMillisMinus250min = currentTimeMillis() - 30000000bitmex = ccxt.bitmex()ohlcvDf = pd.DataFrame( bitmex.fetch_ohlcv(symbol="BTC/USD", limit=500, timeframe="1m", since=currentMillisMinus250min), columns = ["timestamp", "open", "high", "low", "close", "volume"])macdNp = ti.macd(ohlcvDf["close"].to_numpy(), 12, 26, 9)ohlcvDf["macd"] = macdNp[0]ohlcvDf["signal"] = macdNp[1]錯(cuò)誤：ValueError: Length of values does not match length of index字符元組：( array([ 6.68695914e+00, 6.12108219e+00, 5.64479926e+00, 4.94431782e+00, 4.33794071e+00, 3.81237487e+00, 3.46875952e+00, 3.15956840e+00]), array([ 6.68695914, 6.57378375, 6.38798685, 6.09925304, 5.74699058, 5.36006744, 4.98180585, 4.61735836]), array([ 0.00000000e+00, -4.52701560e-01, -7.43187590e-01, -1.15493523e+00, -1.40904987e+00, -1.54769256e+00, -1.51304634e+00, -1.45778996e+00]))
2 回答
12345678_0001
TA貢獻(xiàn)1802條經(jīng)驗(yàn) 獲得超5個(gè)贊
import time
import numpy as np
import pandas as pd
import ccxt # crypto exchange library
import tulipy as ti # financial indicators
def currentTimeMillis():
return int(round(time.time() * 1000))
currentMillisMinus250min = currentTimeMillis() - 30000000
bitmex = ccxt.bitmex()
ohlcvDf = pd.DataFrame(
bitmex.fetch_ohlcv(symbol="BTC/USD", limit=500, timeframe="1m", since=currentMillisMinus250min),
columns = ["timestamp", "open", "high", "low", "close", "volume"])
macdNp = ti.macd(ohlcvDf["close"].to_numpy(), 12, 26, 9)
nan_arr = np.nan * np.ones(shape=(25,))
ohlcvDf["macd"] = np.append(nan_arr, macdNp[0])
ohlcvDf["signal"] = np.append(nan_arr, macdNp[1])
在您的代碼中，您使用了self,self.macdPeriodFast, self.macdPeriodSlow, self.macdSignal，但沒(méi)有在任何地方引用它，因?yàn)槲覄倓傞喿x了這些庫(kù)并嘗試生成元組（可能與您的不同）。但我能夠生成元組并打印這些數(shù)組。
您能否提供在此特定代碼中使用的任何其他內(nèi)容！
我在這里讀過(guò)圖書(shū)館
慕的地8271018
TA貢獻(xiàn)1796條經(jīng)驗(yàn) 獲得超4個(gè)贊
更好的方法是使用 np.array 而不僅僅是數(shù)組。
import numpy as np
arrays = (
np.array([ 6.68695914e+00, 6.12108219e+00, 5.64479926e+00, 4.94431782e+00,
4.33794071e+00, 3.81237487e+00, 3.46875952e+00, 3.15956840e+00]),
np.array([ 6.68695914, 6.57378375, 6.38798685, 6.09925304,
5.74699058, 5.36006744, 4.98180585, 4.61735836]),
np.array([ 0.00000000e+00, -4.52701560e-01, -7.43187590e-01, -1.15493523e+00,
-1.40904987e+00, -1.54769256e+00, -1.51304634e+00, -1.45778996e+00])
)
print(arrays[0])
輸出：
[6.68695914 6.12108219 5.64479926 4.94431782 4.33794071 3.81237487
3.46875952 3.1595684 ]
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