Python中是否有SciPy函数或NumPy函数或模块来计算给定特定窗口的1D数组的运行平均值?
当前回答
更新:已经提出了更有效的解决方案,scipy的uniform_filter1d可能是“标准”第三方库中最好的,还有一些更新的或专门的库可用。
你可以用np。卷积得到:
np.convolve(x, np.ones(N)/N, mode='valid')
解释
The running mean is a case of the mathematical operation of convolution. For the running mean, you slide a window along the input and compute the mean of the window's contents. For discrete 1D signals, convolution is the same thing, except instead of the mean you compute an arbitrary linear combination, i.e., multiply each element by a corresponding coefficient and add up the results. Those coefficients, one for each position in the window, are sometimes called the convolution kernel. The arithmetic mean of N values is (x_1 + x_2 + ... + x_N) / N, so the corresponding kernel is (1/N, 1/N, ..., 1/N), and that's exactly what we get by using np.ones(N)/N.
边缘
np的模态参数。Convolve指定如何处理边缘。我在这里选择有效模式,因为我认为这是大多数人期望的运行方式,但您可能有其他优先级。下面是一个图表,说明了模式之间的差异:
import numpy as np
import matplotlib.pyplot as plt
modes = ['full', 'same', 'valid']
for m in modes:
plt.plot(np.convolve(np.ones(200), np.ones(50)/50, mode=m));
plt.axis([-10, 251, -.1, 1.1]);
plt.legend(modes, loc='lower center');
plt.show()
其他回答
对于一个简短、快速的解决方案,在一个循环中完成所有事情,没有依赖关系,下面的代码工作得很好。
mylist = [1, 2, 3, 4, 5, 6, 7]
N = 3
cumsum, moving_aves = [0], []
for i, x in enumerate(mylist, 1):
cumsum.append(cumsum[i-1] + x)
if i>=N:
moving_ave = (cumsum[i] - cumsum[i-N])/N
#can do stuff with moving_ave here
moving_aves.append(moving_ave)
更新:下面的例子展示了老熊猫。Rolling_mean函数,该函数在最近版本的pandas中已被删除。该函数调用的现代等价函数将使用pandas.Series.rolling:
In [8]: pd.Series(x).rolling(window=N).mean().iloc[N-1:].values
Out[8]:
array([ 0.49815397, 0.49844183, 0.49840518, ..., 0.49488191,
0.49456679, 0.49427121])
pandas比NumPy或SciPy更适合这一点。它的函数rolling_mean很方便地完成了这项工作。当输入是一个数组时,它还返回一个NumPy数组。
使用任何定制的纯Python实现都很难在性能上击败rolling_mean。下面是针对两个提议的解决方案的性能示例:
In [1]: import numpy as np
In [2]: import pandas as pd
In [3]: def running_mean(x, N):
...: cumsum = np.cumsum(np.insert(x, 0, 0))
...: return (cumsum[N:] - cumsum[:-N]) / N
...:
In [4]: x = np.random.random(100000)
In [5]: N = 1000
In [6]: %timeit np.convolve(x, np.ones((N,))/N, mode='valid')
10 loops, best of 3: 172 ms per loop
In [7]: %timeit running_mean(x, N)
100 loops, best of 3: 6.72 ms per loop
In [8]: %timeit pd.rolling_mean(x, N)[N-1:]
100 loops, best of 3: 4.74 ms per loop
In [9]: np.allclose(pd.rolling_mean(x, N)[N-1:], running_mean(x, N))
Out[9]: True
关于如何处理边缘值,也有很好的选项。
我觉得使用瓶颈可以很好地解决这个问题
参见下面的基本示例:
import numpy as np
import bottleneck as bn
a = np.random.randint(4, 1000, size=100)
mm = bn.move_mean(a, window=5, min_count=1)
“mm”是“a”的移动平均值。 “窗口”是考虑移动均值的最大条目数。 "min_count"是考虑移动平均值的最小条目数(例如,对于前几个元素或如果数组有nan值)。
好在瓶颈有助于处理nan值,而且非常高效。
上述所有的解决方案都很差,因为它们缺乏
由于本机python而不是numpy向量化实现, 数值稳定性,由于numpy使用不当。cumsum或 由于O(len(x) * w)实现为卷积的速度。
鉴于
import numpy
m = 10000
x = numpy.random.rand(m)
w = 1000
注意x_[:w].sum()等于x[:w-1].sum()。因此,对于第一个平均值,numpy.cumsum(…)加上x[w] / w(通过x_[w+1] / w),并减去0(从x_[0] / w)。结果是x[0:w].mean()
通过cumsum,您将通过添加x[w+1] / w并减去x[0] / w来更新第二个平均值,从而得到x[1:w+1].mean()。
这将一直进行,直到到达x[-w:].mean()。
x_ = numpy.insert(x, 0, 0)
sliding_average = x_[:w].sum() / w + numpy.cumsum(x_[w:] - x_[:-w]) / w
这个解是向量化的,O(m),可读且数值稳定。
或用于python计算的模块
在我在Tradewave.net的测试中,TA-lib总是赢:
import talib as ta
import numpy as np
import pandas as pd
import scipy
from scipy import signal
import time as t
PAIR = info.primary_pair
PERIOD = 30
def initialize():
storage.reset()
storage.elapsed = storage.get('elapsed', [0,0,0,0,0,0])
def cumsum_sma(array, period):
ret = np.cumsum(array, dtype=float)
ret[period:] = ret[period:] - ret[:-period]
return ret[period - 1:] / period
def pandas_sma(array, period):
return pd.rolling_mean(array, period)
def api_sma(array, period):
# this method is native to Tradewave and does NOT return an array
return (data[PAIR].ma(PERIOD))
def talib_sma(array, period):
return ta.MA(array, period)
def convolve_sma(array, period):
return np.convolve(array, np.ones((period,))/period, mode='valid')
def fftconvolve_sma(array, period):
return scipy.signal.fftconvolve(
array, np.ones((period,))/period, mode='valid')
def tick():
close = data[PAIR].warmup_period('close')
t1 = t.time()
sma_api = api_sma(close, PERIOD)
t2 = t.time()
sma_cumsum = cumsum_sma(close, PERIOD)
t3 = t.time()
sma_pandas = pandas_sma(close, PERIOD)
t4 = t.time()
sma_talib = talib_sma(close, PERIOD)
t5 = t.time()
sma_convolve = convolve_sma(close, PERIOD)
t6 = t.time()
sma_fftconvolve = fftconvolve_sma(close, PERIOD)
t7 = t.time()
storage.elapsed[-1] = storage.elapsed[-1] + t2-t1
storage.elapsed[-2] = storage.elapsed[-2] + t3-t2
storage.elapsed[-3] = storage.elapsed[-3] + t4-t3
storage.elapsed[-4] = storage.elapsed[-4] + t5-t4
storage.elapsed[-5] = storage.elapsed[-5] + t6-t5
storage.elapsed[-6] = storage.elapsed[-6] + t7-t6
plot('sma_api', sma_api)
plot('sma_cumsum', sma_cumsum[-5])
plot('sma_pandas', sma_pandas[-10])
plot('sma_talib', sma_talib[-15])
plot('sma_convolve', sma_convolve[-20])
plot('sma_fftconvolve', sma_fftconvolve[-25])
def stop():
log('ticks....: %s' % info.max_ticks)
log('api......: %.5f' % storage.elapsed[-1])
log('cumsum...: %.5f' % storage.elapsed[-2])
log('pandas...: %.5f' % storage.elapsed[-3])
log('talib....: %.5f' % storage.elapsed[-4])
log('convolve.: %.5f' % storage.elapsed[-5])
log('fft......: %.5f' % storage.elapsed[-6])
结果:
[2015-01-31 23:00:00] ticks....: 744
[2015-01-31 23:00:00] api......: 0.16445
[2015-01-31 23:00:00] cumsum...: 0.03189
[2015-01-31 23:00:00] pandas...: 0.03677
[2015-01-31 23:00:00] talib....: 0.00700 # <<< Winner!
[2015-01-31 23:00:00] convolve.: 0.04871
[2015-01-31 23:00:00] fft......: 0.22306
推荐文章
- Numpy Max vs amax vs maximum
- 我应该在.gitignore文件中添加Django迁移文件吗?
- 每n行有熊猫
- 实例属性attribute_name定义在__init__之外
- 如何获取在Python中捕获的异常的名称?
- 第一次出现的值大于现有值的Numpy
- 如何从Python函数中返回两个值?
- 前一个月的Python日期
- Python中方括号括起来的列表和圆括号括起来的列表有什么区别?
- Python日志记录不输出任何东西
- 每n秒运行特定代码
- SQLAlchemy是否有与Django的get_or_create等价的函数?
- 如何将python datetime转换为字符串,具有可读格式的日期?
- 美丽的汤和提取div及其内容的ID
- 在Python中重置生成器对象