从Pandas DataFrame计算RSI指标? [英] Calculate RSI indicator from pandas DataFrame?
本文介绍了从Pandas DataFrame计算RSI指标?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我的问题
解决方案
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，我们得到y[i] = a*y[i-1] + b*x[i]，其中a = 13/14和b = 1/14表示n = 14.展开递归会导致:
(抱歉图片，键入起来很麻烦)
这可以使用cumsum高效地以numpy计算(rma =连续移动均线):
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或numba .
为了说明这一点，我将numpy解决方案与dimitris_ps的循环解决方案进行了比较:
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，而对于大型数据集，循环解决方案不可行.
My problem
I tried many libraries on Github but all of them did not produce matching results for TradingView so I followed the formula on this link to calculate RSI indicator. I calculated it with Excel and collated the results with TradingView. I know it's absolutely correct but, but I didn't find a way to calculate it with Pandas.
Formula
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.
Expected Results
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
My Code
Import
import pandas as pd
import numpy as np
Load data
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
Change
Calculate change every row
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
Gain and loss
get gain and loss from change
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
Calculate fist avg gain and loss
Mean of n prior rows
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
The very first calculations for average gain and the average loss is ok but I don't know how to apply pandas.core.window.Rolling.apply for the second, and subsequent because they are in many rows and different columns. It may be something like this:
avg_gain[n] = (avg_gain[n-1]*13 + gain[n]) / 14
My Wish - My Question
- The best way to calculate and work with technical indicators?
- Complete the above code in "Pandas Style".
- Does the traditional way of coding with loops reduce performance compared to Pandas?
解决方案
The average gain and loss are calculated by a recursive formula, which can't be vectorized with numpy. We can, however, try and find an analytical (i.e. non-recursive) solution for calculating the individual elements. Such a solution can then be implemented using numpy.
Denoting the average gain as y and the current gain as x, we get y[i] = a*y[i-1] + b*x[i], where a = 13/14 and b = 1/14 for n = 14. Unwrapping the recursion leads to:
(sorry for the picture, was just to cumbersome to type it)
This can be efficiently calculated in numpy using cumsum (rma = running moving average):
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))
Output of df.round(2):
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
Concerning your last question about performance: explicite loops in python / pandas are terrible, avoid them whenever you can. If you can't, try cython or numba.
To illustrate this, I made a small comparison of my numpy solution with dimitris_ps' loop solution:
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)
Results (ms / loop):
rows np loop
23 4.9 9.2
230 5.0 112.3
2300 5.5 1122.7
So even for 8 rows (rows 15...22) the loop solution takes about twice the time of the numpy solution. Numpy scales well, whereas the loop solution isn't feasable for large datasets.
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