calculate exponential moving average in python

Question

I have a range of dates and a measurement on each of those dates. I'd like to calculate an exponential moving average for each of the dates. Does anybody know how to do this?

I'm new to python. It doesn't appear that averages are built into the standard python library, which strikes me as a little odd. Maybe I'm not looking in the right place.

So, given the following code, how could I calculate the moving weighted average of IQ points for calendar dates?

from datetime import date
days = [date(2008,1,1), date(2008,1,2), date(2008,1,7)]
IQ = [110, 105, 90]

(there's probably a better way to structure the data, any advice would be appreciated)

Answer 1

EDIT: It seems that mov_average_expw() function from scikits.timeseries.lib.moving_funcs submodule from SciKits (add-on toolkits that complement SciPy) better suits the wording of your question.

---

To calculate an exponential smoothing of your data with a smoothing factor alpha (it is (1 - alpha) in Wikipedia's terms):

>>> alpha = 0.5
>>> assert 0 < alpha <= 1.0
>>> av = sum(alpha**n.days * iq 
...     for n, iq in map(lambda (day, iq), today=max(days): (today-day, iq), 
...         sorted(zip(days, IQ), key=lambda p: p[0], reverse=True)))
95.0

The above is not pretty, so let's refactor it a bit:

from collections import namedtuple
from operator    import itemgetter

def smooth(iq_data, alpha=1, today=None):
    """Perform exponential smoothing with factor `alpha`.

    Time period is a day.
    Each time period the value of `iq` drops `alpha` times.
    The most recent data is the most valuable one.
    """
    assert 0 < alpha <= 1

    if alpha == 1: # no smoothing
        return sum(map(itemgetter(1), iq_data))

    if today is None:
        today = max(map(itemgetter(0), iq_data))

    return sum(alpha**((today - date).days) * iq for date, iq in iq_data)

IQData = namedtuple("IQData", "date iq")

if __name__ == "__main__":
    from datetime import date

    days = [date(2008,1,1), date(2008,1,2), date(2008,1,7)]
    IQ = [110, 105, 90]
    iqdata = list(map(IQData, days, IQ))
    print("\n".join(map(str, iqdata)))

    print(smooth(iqdata, alpha=0.5))

Example:

$ python26 smooth.py
IQData(date=datetime.date(2008, 1, 1), iq=110)
IQData(date=datetime.date(2008, 1, 2), iq=105)
IQData(date=datetime.date(2008, 1, 7), iq=90)
95.0

Answer 2

I'm always calculating EMAs with Pandas:

Here is an example how to do it:

import pandas as pd
import numpy as np

def ema(values, period):
    values = np.array(values)
    return pd.ewma(values, span=period)[-1]

values = [9, 5, 10, 16, 5]
period = 5

print ema(values, period)

More infos about Pandas EWMA:

http://pandas.pydata.org/pandas-docs/stable/generated/pandas.ewma.html

Answer 3

I did a bit of googling and I found the following sample code (http://osdir.com/ml/python.matplotlib.general/2005-04/msg00044.html):

def ema(s, n):
    """
    returns an n period exponential moving average for
    the time series s

    s is a list ordered from oldest (index 0) to most
    recent (index -1)
    n is an integer

    returns a numeric array of the exponential
    moving average
    """
    s = array(s)
    ema = []
    j = 1

    #get n sma first and calculate the next n period ema
    sma = sum(s[:n]) / n
    multiplier = 2 / float(1 + n)
    ema.append(sma)

    #EMA(current) = ( (Price(current) - EMA(prev) ) x Multiplier) + EMA(prev)
    ema.append(( (s[n] - sma) * multiplier) + sma)

    #now calculate the rest of the values
    for i in s[n+1:]:
        tmp = ( (i - ema[j]) * multiplier) + ema[j]
        j = j + 1
        ema.append(tmp)

    return ema

Answer 4

You can also use the SciPy filter method because the EMA is an IIR filter. This will have the benefit of being approximately 64 times faster as measured on my system using timeit on large data sets when compared to the enumerate() approach.

import numpy as np
from scipy.signal import lfilter

x = np.random.normal(size=1234)
alpha = .1 # smoothing coefficient
zi = [x[0]] # seed the filter state with first value
# filter can process blocks of continuous data if <zi> is maintained
y, zi = lfilter([1.-alpha], [1., -alpha], x, zi=zi)

Answer 5

I don't know Python, but for the averaging part, do you mean an exponentially decaying low-pass filter of the form

y_new = y_old + (input - y_old)*alpha

where alpha = dt/tau, dt = the timestep of the filter, tau = the time constant of the filter? (the variable-timestep form of this is as follows, just clip dt/tau to not be more than 1.0)

y_new = y_old + (input - y_old)*dt/tau

If you want to filter something like a date, make sure you convert to a floating-point quantity like # of seconds since Jan 1 1970.

Answer 6

In matplotlib.org examples (http://matplotlib.org/examples/pylab_examples/finance_work2.html) is provided one good example of Exponential Moving Average (EMA) function using numpy:

def moving_average(x, n, type):
    x = np.asarray(x)
    if type=='simple':
        weights = np.ones(n)
    else:
        weights = np.exp(np.linspace(-1., 0., n))

    weights /= weights.sum()

    a =  np.convolve(x, weights, mode='full')[:len(x)]
    a[:n] = a[n]
    return a

Answer 7

My python is a little bit rusty (anyone can feel free to edit this code to make corrections, if I've messed up the syntax somehow), but here goes....

def movingAverageExponential(values, alpha, epsilon = 0):

   if not 0 < alpha < 1:
      raise ValueError("out of range, alpha='%s'" % alpha)

   if not 0 <= epsilon < alpha:
      raise ValueError("out of range, epsilon='%s'" % epsilon)

   result = [None] * len(values)

   for i in range(len(result)):
       currentWeight = 1.0

       numerator     = 0
       denominator   = 0
       for value in values[i::-1]:
           numerator     += value * currentWeight
           denominator   += currentWeight

           currentWeight *= alpha
           if currentWeight < epsilon: 
              break

       result[i] = numerator / denominator

   return result

This function moves backward, from the end of the list to the beginning, calculating the exponential moving average for each value by working backward until the weight coefficient for an element is less than the given epsilon.

At the end of the function, it reverses the values before returning the list (so that they're in the correct order for the caller).

(SIDE NOTE: if I was using a language other than python, I'd create a full-size empty array first and then fill it backwards-order, so that I wouldn't have to reverse it at the end. But I don't think you can declare a big empty array in python. And in python lists, appending is much less expensive than prepending, which is why I built the list in reverse order. Please correct me if I'm wrong.)

The 'alpha' argument is the decay factor on each iteration. For example, if you used an alpha of 0.5, then today's moving average value would be composed of the following weighted values:

today:        1.0
yesterday:    0.5
2 days ago:   0.25
3 days ago:   0.125
...etc...

Of course, if you've got a huge array of values, the values from ten or fifteen days ago won't contribute very much to today's weighted average. The 'epsilon' argument lets you set a cutoff point, below which you will cease to care about old values (since their contribution to today's value will be insignificant).

You'd invoke the function something like this:

result = movingAverageExponential(values, 0.75, 0.0001)

Answer 8

I found the above code snippet by @earino pretty useful - but I needed something that could continuously smooth a stream of values - so I refactored it to this:

def exponential_moving_average(period=1000):
    """ Exponential moving average. Smooths the values in v over ther period. Send in values - at first it'll return a simple average, but as soon as it's gahtered 'period' values, it'll start to use the Exponential Moving Averge to smooth the values.
    period: int - how many values to smooth over (default=100). """
    multiplier = 2 / float(1 + period)
    cum_temp = yield None  # We are being primed

    # Start by just returning the simple average until we have enough data.
    for i in xrange(1, period + 1):
        cum_temp += yield cum_temp / float(i)

    # Grab the timple avergae
    ema = cum_temp / period

    # and start calculating the exponentially smoothed average
    while True:
        ema = (((yield ema) - ema) * multiplier) + ema

and I use it like this:

def temp_monitor(pin):
    """ Read from the temperature monitor - and smooth the value out. The sensor is noisy, so we use exponential smoothing. """
    ema = exponential_moving_average()
    next(ema)  # Prime the generator

    while True:
        yield ema.send(val_to_temp(pin.read()))

(where pin.read() produces the next value I'd like to consume).

Answer 9

May be shortest:

#Specify decay in terms of span
#data_series should be a DataFrame

ema=data_series.ewm(span=5, adjust=False).mean()

Answer 10

import pandas_ta as ta

data["EMA3"] = ta.ema(data["close"], length=3)

pandas_ta is a Technical Analysis Library: https://github.com/twopirllc/pandas-ta. Above code calculates the Exponential Moving Average (EMA) for a series. You can specify the lag value using 'length'. Spesifically, above code calculates '3-day EMA'.

Answer 11

Here is a simple sample I worked up based on http://stockcharts.com/school/doku.php?id=chart_school:technical_indicators:moving_averages

Note that unlike in their spreadsheet, I don't calculate the SMA, and I don't wait to generate the EMA after 10 samples. This means my values differ slightly, but if you chart it, it follows exactly after 10 samples. During the first 10 samples, the EMA I calculate is appropriately smoothed.

def emaWeight(numSamples):
    return 2 / float(numSamples + 1)

def ema(close, prevEma, numSamples):
    return ((close-prevEma) * emaWeight(numSamples) ) + prevEma

samples = [
22.27, 22.19, 22.08, 22.17, 22.18, 22.13, 22.23, 22.43, 22.24, 22.29,
22.15, 22.39, 22.38, 22.61, 23.36, 24.05, 23.75, 23.83, 23.95, 23.63,
23.82, 23.87, 23.65, 23.19, 23.10, 23.33, 22.68, 23.10, 22.40, 22.17,
]
emaCap = 10
e=samples[0]
for s in range(len(samples)):
    numSamples = emaCap if s > emaCap else s
    e =  ema(samples[s], e, numSamples)
    print e

Answer 12

I'm a little late to the party here, but none of the solutions given were what I was looking for. Nice little challenge using recursion and the exact formula given in investopedia. No numpy or pandas required.

prices = [{'i': 1, 'close': 24.5}, {'i': 2, 'close': 24.6}, {'i': 3, 'close': 24.8}, {'i': 4, 'close': 24.9},
          {'i': 5, 'close': 25.6}, {'i': 6, 'close': 25.0}, {'i': 7, 'close': 24.7}]


def rec_calculate_ema(n):
    k = 2 / (n + 1)
    price = prices[n]['close']
    if n == 1:
        return price
    res = (price * k) + (rec_calculate_ema(n - 1) * (1 - k))
    return res


print(rec_calculate_ema(3))

Answer 13

A fast way (copy-pasted from here) is the following:

def ExpMovingAverage(values, window):
    """ Numpy implementation of EMA
    """
    weights = np.exp(np.linspace(-1., 0., window))
    weights /= weights.sum()
    a =  np.convolve(values, weights, mode='full')[:len(values)]
    a[:window] = a[window]
    return a

Answer 14

I am using a list and a rate of decay as inputs. I hope this little function with just two lines may help you here, considering deep recursion is not stable in python.

def expma(aseries, ratio):
    return sum([ratio*aseries[-x-1]*((1-ratio)**x) for x in range(len(aseries))])

Answer 15

more simply, using pandas

def EMA(tw):
    for x in tw:
        data["EMA{}".format(x)] = data['close'].ewm(span=x, adjust=False).mean()
        EMA([10,50,100])

Answer 16

Papahaba's answer was almost what I was looking for (thanks!) but I needed to match initial conditions. Using an IIR filter with scipy.signal.lfilter is certainly the most efficient. Here's my redux:

Given a NumPy vector, x

import numpy as np
from scipy import signal

period = 12
b = np.array((1,), 'd')
a = np.array((period, 1-period), 'd')
zi = signal.lfilter_zi(b, a)
y, zi = signal.lfilter(b, a, x, zi=zi*x[0:1])

Get the N-point EMA (here, 12) returned in the vector y