从pandas DataFrame计算RSI指标？
我的问题
我在 Github 上尝试了很多库，但是它们都没有为 TradingView 生成匹配的结果，因此我按照此link上的公式计算了 RSI 指示器。我使用 Excel 进行了计算，并使用 TradingView 整理了结果。我知道它是绝对正确，但是我没有找到使用 Pandas 进行计算的方法。
公式
100
RSI = 100 - --------
1 + RS
RS = Average Gain / Average Loss
The very first calculations for average gain and average loss are simple
14-period averages:
First Average Gain = Sum of Gains over the past 14 periods / 14.
First Average Loss = Sum of Losses over the past 14 periods / 14
The second, and subsequent, calculations are based on the prior averages
and the current gain loss:
Average Gain = [(previous Average Gain) x 13 + current Gain] / 14.
Average Loss = [(previous Average Loss) x 13 + current Loss] / 14.
预期结果
close change gain loss avg_gian avg_loss rs \
0 4724.89 NaN NaN NaN NaN NaN NaN
1 4378.51 -346.38 0.00 346.38 NaN NaN NaN
2 6463.00 2084.49 2084.49 0.00 NaN NaN NaN
3 9838.96 3375.96 3375.96 0.00 NaN NaN NaN
4 13716.36 3877.40 3877.40 0.00 NaN NaN NaN
5 10285.10 -3431.26 0.00 3431.26 NaN NaN NaN
6 10326.76 41.66 41.66 0.00 NaN NaN NaN
7 6923.91 -3402.85 0.00 3402.85 NaN NaN NaN
8 9246.01 2322.10 2322.10 0.00 NaN NaN NaN
9 7485.01 -1761.00 0.00 1761.00 NaN NaN NaN
10 6390.07 -1094.94 0.00 1094.94 NaN NaN NaN
11 7730.93 1340.86 1340.86 0.00 NaN NaN NaN
12 7011.21 -719.72 0.00 719.72 NaN NaN NaN
13 6626.57 -384.64 0.00 384.64 NaN NaN NaN
14 6371.93 -254.64 0.00 254.64 931.605000 813.959286 1.144535
15 4041.32 -2330.61 0.00 2330.61 865.061786 922.291480 0.937948
16 3702.90 -338.42 0.00 338.42 803.271658 880.586374 0.912201
17 3434.10 -268.80 0.00 268.80 745.895111 836.887347 0.891273
18 3813.69 379.59 379.59 0.00 719.730460 777.109680 0.926163
19 4103.95 290.26 290.26 0.00 689.053999 721.601845 0.954895
20 5320.81 1216.86 1216.86 0.00 726.754428 670.058856 1.084613
21 8555.00 3234.19 3234.19 0.00 905.856968 622.197509 1.455899
22 10854.10 2299.10 2299.10 0.00 1005.374328 577.754830 1.740140
rsi_14
0 NaN
1 NaN
2 NaN
3 NaN
4 NaN
5 NaN
6 NaN
7 NaN
8 NaN
9 NaN
10 NaN
11 NaN
12 NaN
13 NaN
14 53.369848
15 48.399038
16 47.704239
17 47.125561
18 48.083322
19 48.846358
20 52.029461
21 59.281719
22 63.505515
我的代码
导入
import pandas as pd
import numpy as np
加载数据
df = pd.read_csv("rsi_14_test_data.csv")
close = df['close']
print(close)
0 4724.89
1 4378.51
2 6463.00
3 9838.96
4 13716.36
5 10285.10
6 10326.76
7 6923.91
8 9246.01
9 7485.01
10 6390.07
11 7730.93
12 7011.21
13 6626.57
14 6371.93
15 4041.32
16 3702.90
17 3434.10
18 3813.69
19 4103.95
20 5320.81
21 8555.00
22 10854.10
Name: close, dtype: float64
更改
计算每行的更改
change = close.diff(1)
print(change)
0 NaN
1 -346.38
2 2084.49
3 3375.96
4 3877.40
5 -3431.26
6 41.66
7 -3402.85
8 2322.10
9 -1761.00
10 -1094.94
11 1340.86
12 -719.72
13 -384.64
14 -254.64
15 -2330.61
16 -338.42
17 -268.80
18 379.59
19 290.26
20 1216.86
21 3234.19
22 2299.10
Name: close, dtype: float64
盈亏
从变化中获得收益和损失
is_gain, is_loss = change > 0, change < 0
gain, loss = change, -change
gain[is_loss] = 0
loss[is_gain] = 0
gain.name = 'gain'
loss.name = 'loss'
print(loss)
0 NaN
1 346.38
2 0.00
3 0.00
4 0.00
5 3431.26
6 0.00
7 3402.85
8 0.00
9 1761.00
10 1094.94
11 0.00
12 719.72
13 384.64
14 254.64
15 2330.61
16 338.42
17 268.80
18 0.00
19 0.00
20 0.00
21 0.00
22 0.00
Name: loss, dtype: float64
计算拳头平均损益
前n行的平均值
n = 14
avg_gain = change * np.nan
avg_loss = change * np.nan
avg_gain[n] = gain[:n+1].mean()
avg_loss[n] = loss[:n+1].mean()
avg_gain.name = 'avg_gain'
avg_loss.name = 'avg_loss'
avg_df = pd.concat([gain, loss, avg_gain, avg_loss], axis=1)
print(avg_df)
gain loss avg_gain avg_loss
0 NaN NaN NaN NaN
1 0.00 346.38 NaN NaN
2 2084.49 0.00 NaN NaN
3 3375.96 0.00 NaN NaN
4 3877.40 0.00 NaN NaN
5 0.00 3431.26 NaN NaN
6 41.66 0.00 NaN NaN
7 0.00 3402.85 NaN NaN
8 2322.10 0.00 NaN NaN
9 0.00 1761.00 NaN NaN
10 0.00 1094.94 NaN NaN
11 1340.86 0.00 NaN NaN
12 0.00 719.72 NaN NaN
13 0.00 384.64 NaN NaN
14 0.00 254.64 931.605 813.959286
15 0.00 2330.61 NaN NaN
16 0.00 338.42 NaN NaN
17 0.00 268.80 NaN NaN
18 379.59 0.00 NaN NaN
19 290.26 0.00 NaN NaN
20 1216.86 0.00 NaN NaN
21 3234.19 0.00 NaN NaN
22 2299.10 0.00 NaN NaN
对于平均增益和平均损耗的第一个计算是可以的，但是我不知道如何对第二个应用pandas.core.window.Rolling.apply，以及随后的应用，因为它们位于许多行和不同的列中。 可能是这样的：
avg_gain[n] = (avg_gain[n-1]*13 + gain[n]) / 14
我的愿望-我的问题
- 计算和使用技术指标的最佳方法？
- 以“ Pandas样式”完成上述代码。
- 与熊猫相比，传统的循环编码方式是否会降低性能？
4 个答案:
答案 0 :(得分：2)
平均损益由递归公式计算，该公式不能用numpy向量化。但是，我们可以尝试找到一种 analytical （即非递归）解决方案来计算各个元素。然后可以使用numpy来实现这种解决方案。
将平均增益表示为y，将当前增益表示为x，得到y[i] = a*y[i-1] + b*x[i]，其中a = 13/14和b = 1/14代表n = 14 。展开递归会导致：
（抱歉图片，键入起来很麻烦）
可以使用cumsum以有效的方式以numpy计算（rma =运行移动平均线）：
import pandas as pd
import numpy as np
df = pd.DataFrame({'close':[4724.89, 4378.51,6463.00,9838.96,13716.36,10285.10,
10326.76,6923.91,9246.01,7485.01,6390.07,7730.93,
7011.21,6626.57,6371.93,4041.32,3702.90,3434.10,
3813.69,4103.95,5320.81,8555.00,10854.10]})
n = 14
def rma(x, n, y0):
a = (n-1) / n
ak = a**np.arange(len(x)-1, -1, -1)
return np.append(y0, np.cumsum(ak * x) / ak / n + y0 * a**np.arange(1, len(x)+1))
df['change'] = df['close'].diff()
df['gain'] = df.change.mask(df.change < 0, 0.0)
df['loss'] = -df.change.mask(df.change > 0, -0.0)
df.loc[n:,'avg_gain'] = rma( df.gain[n+1:].values, n, df.loc[:n, 'gain'].mean())
df.loc[n:,'avg_loss'] = rma( df.loss[n+1:].values, n, df.loc[:n, 'loss'].mean())
df['rs'] = df.avg_gain / df.avg_loss
df['rsi_14'] = 100 - (100 / (1 + df.rs))
df.round(2)的输出：
close change gain loss avg_gain avg_loss rs rsi rsi_14
0 4724.89 NaN NaN NaN NaN NaN NaN NaN NaN
1 4378.51 -346.38 0.00 346.38 NaN NaN NaN NaN NaN
2 6463.00 2084.49 2084.49 0.00 NaN NaN NaN NaN NaN
3 9838.96 3375.96 3375.96 0.00 NaN NaN NaN NaN NaN
4 13716.36 3877.40 3877.40 0.00 NaN NaN NaN NaN NaN
5 10285.10 -3431.26 0.00 3431.26 NaN NaN NaN NaN NaN
6 10326.76 41.66 41.66 0.00 NaN NaN NaN NaN NaN
7 6923.91 -3402.85 0.00 3402.85 NaN NaN NaN NaN NaN
8 9246.01 2322.10 2322.10 0.00 NaN NaN NaN NaN NaN
9 7485.01 -1761.00 0.00 1761.00 NaN NaN NaN NaN NaN
10 6390.07 -1094.94 0.00 1094.94 NaN NaN NaN NaN NaN
11 7730.93 1340.86 1340.86 0.00 NaN NaN NaN NaN NaN
12 7011.21 -719.72 0.00 719.72 NaN NaN NaN NaN NaN
13 6626.57 -384.64 0.00 384.64 NaN NaN NaN NaN NaN
14 6371.93 -254.64 0.00 254.64 931.61 813.96 1.14 53.37 53.37
15 4041.32 -2330.61 0.00 2330.61 865.06 922.29 0.94 48.40 48.40
16 3702.90 -338.42 0.00 338.42 803.27 880.59 0.91 47.70 47.70
17 3434.10 -268.80 0.00 268.80 745.90 836.89 0.89 47.13 47.13
18 3813.69 379.59 379.59 0.00 719.73 777.11 0.93 48.08 48.08
19 4103.95 290.26 290.26 0.00 689.05 721.60 0.95 48.85 48.85
20 5320.81 1216.86 1216.86 0.00 726.75 670.06 1.08 52.03 52.03
21 8555.00 3234.19 3234.19 0.00 905.86 622.20 1.46 59.28 59.28
22 10854.10 2299.10 2299.10 0.00 1005.37 577.75 1.74 63.51 63.51
关于性能的最后一个问题： python / pandas中的显式循环很糟糕，请尽可能避免使用。如果不能，请尝试cython or numba。
为了说明这一点，我对numpy解决方案与dimitris_ps'loop solution进行了较小的比较：
import pandas as pd
import numpy as np
import timeit
mult = 1 # length of dataframe = 23 * mult
number = 1000 # number of loop for timeit
df0 = pd.DataFrame({'close':[4724.89, 4378.51,6463.00,9838.96,13716.36,10285.10,
10326.76,6923.91,9246.01,7485.01,6390.07,7730.93,
7011.21,6626.57,6371.93,4041.32,3702.90,3434.10,
3813.69,4103.95,5320.81,8555.00,10854.10] * mult })
n = 14
def rsi_np():
# my numpy solution from above
return df
def rsi_loop():
# loop solution https://stackoverflow.com/a/57008625/3944322
# without the wrong alternative calculation of df['avg_gain'][14]
return df
df = df0.copy()
time_np = timeit.timeit('rsi_np()', globals=globals(), number = number) / 1000 * number
df = df0.copy()
time_loop = timeit.timeit('rsi_loop()', globals=globals(), number = number) / 1000 * number
print(f'rows\tnp\tloop\n{len(df0)}\t{time_np:.1f}\t{time_loop:.1f}')
assert np.allclose(rsi_np(), rsi_loop(), equal_nan=True)
结果（毫秒/循环）：
rows np loop
23 4.9 9.2
230 5.0 112.3
2300 5.5 1122.7
因此，即使对于8行（第15 ... 22行），循环求解所花费的时间也是numpy解决方案的两倍。 Numpy可很好地扩展，而循环解决方案不适用于大型数据集。
答案 1 :(得分：1)
这里是一个选项。
我只会触摸你的第二颗子弹
# libraries required
import pandas as pd
import numpy as np
# create dataframe
df = pd.DataFrame({'close':[4724.89, 4378.51,6463.00,9838.96,13716.36,10285.10,
10326.76,6923.91,9246.01,7485.01,6390.07,7730.93,
7011.21,6626.57,6371.93,4041.32,3702.90,3434.10,
3813.69,4103.95,5320.81,8555.00,10854.10]})
df['change'] = df['close'].diff(1) # Calculate change
# calculate gain / loss from every change
df['gain'] = np.select([df['change']>0, df['change'].isna()],
[df['change'], np.nan],
default=0)
df['loss'] = np.select([df['change']<0, df['change'].isna()],
[-df['change'], np.nan],
default=0)
# create avg_gain / avg_loss columns with all nan
df['avg_gain'] = np.nan
df['avg_loss'] = np.nan
n = 14 # what is the window
# keep first occurrence of rolling mean
df['avg_gain'][n] = df['gain'].rolling(window=n).mean().dropna().iloc[0]
df['avg_loss'][n] = df['loss'].rolling(window=n).mean().dropna().iloc[0]
# Alternatively
df['avg_gain'][n] = df.loc[:n, 'gain'].mean()
df['avg_loss'][n] = df.loc[:n, 'loss'].mean()
# This is not a pandas way, looping through the pandas series, but it does what you need
for i in range(n+1, df.shape[0]):
df['avg_gain'].iloc[i] = (df['avg_gain'].iloc[i-1] * (n - 1) + df['gain'].iloc[i]) / n
df['avg_loss'].iloc[i] = (df['avg_loss'].iloc[i-1] * (n - 1) + df['loss'].iloc[i]) / n
# calculate rs and rsi
df['rs'] = df['avg_gain'] / df['avg_loss']
df['rsi'] = 100 - (100 / (1 + df['rs'] ))
答案 2 :(得分：0)
还有一种更简单的方法，即talib软件包。
import talib
close = df['close']
rsi = talib.RSI(close, timeperiod=14)
如果您希望Bollinger Bands使用RSI也很容易。
upperBB, middleBB, lowerBB = talib.BBANDS(close, timeperiod=20, nbdevup=2, nbdevdn=2, matype=0)
您可以在RSI上使用布林带，而不是固定参考水平70和30。
upperBBrsi, MiddleBBrsi, lowerBBrsi = talib.BBANDS(rsi, timeperiod=50, nbdevup=2, nbdevdn=2, matype=0)
最后，您可以使用％b钙化对RSI进行归一化。
normrsi = (rsi - lowerBBrsi) / (upperBBrsi - lowerBBrsi)
关于Talib的信息 https://mrjbq7.github.io/ta-lib/
答案 3 :(得分：0)
好的，这里是圣诞老人： 这是 rsi 代码，替换所有带有“aa”的东西：
import pandas as pd
rsi_period = 14
df = pd.Series(coinaalist)
chg = df.diff(1)
gain = chg.mask(chg<0,0)
loss = chg.mask(chg>0,0)
avg_gain = gain.ewm(com = rsi_period-1,min_periods=rsi_period).mean()
avg_loss = loss.ewm(com = rsi_period-1,min_periods=rsi_period).mean()
rs = abs(avg_gain / avg_loss)
crplaa = 100 - (100/(1+rs))
coinaarsi = crplaa.iloc[-1]
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