NumPy: calculate averages with NaNs removed

Assuming you’ve also got SciPy installed:

http://www.scipy.org/doc/api_docs/SciPy.stats.stats.html#nanmean

If performance matters, you should use bottleneck.nanmean() instead:

http://pypi.python.org/pypi/Bottleneck

A masked array with the nans filtered out can also be created on the fly:

print np.ma.masked_invalid(dat).mean(1)

You can always find a workaround in something like:

numpy.nansum(dat, axis=1) / numpy.sum(numpy.isfinite(dat), axis=1)

Numpy 2.0’s numpy.mean has a skipna option which should take care of that.

This is built upon the solution suggested by JoshAdel.

Define the following function:

def nanmean(data, **args):
    return numpy.ma.filled(numpy.ma.masked_array(data,numpy.isnan(data)).mean(**args), fill_value=numpy.nan)

Example use:

data = [[0, 1, numpy.nan], [8, 5, 1]]
data = numpy.array(data)
print data
print nanmean(data)
print nanmean(data, axis=0)
print nanmean(data, axis=1)

Will print out:

[[  0.   1.  nan]
 [  8.   5.   1.]]

3.0

[ 4.  3.  1.]

[ 0.5         4.66666667]

Or you use laxarray, freshly uploaded, which is among other a wrapper for masked arrays.

import laxarray as la
la.array(dat).mean(axis=1)

following JoshAdel’s protocoll I get:

Time: 0.048791  Ratio: 1.000000   
Time: 0.062242  Ratio: 1.275689   # laxarray's one-liner

So laxarray is marginally slower (would need to check why, maybe fixable), but much easier to use and allow labelling dimensions with strings.

check out: https://github.com/perrette/laxarray

EDIT: I have checked with another module, “la”, larry, which beats all tests:

import la
la.larry(dat).mean(axis=1)

By hand, Time: 0.049013 Ratio: 1.000000
Larry,   Time: 0.005467 Ratio: 0.111540
laxarray Time: 0.061751 Ratio: 1.259889

Impressive !

How about using Pandas to do this:

import numpy as np
import pandas as pd
dat = np.array([[1, 2, 3], [4, 5, np.nan], [np.nan, 6, np.nan], [np.nan, np.nan, np.nan]])
print dat
print dat.mean(1)

df = pd.DataFrame(dat)
print df.mean(axis=1)

Gives:

0    2.0
1    4.5
2    6.0
3    NaN

One more speed check for all proposed approaches:

Python 2.7.11 |Anaconda 2.4.1 (64-bit)| (default, Jan 19 2016, 12:08:31) [MSC v.1500 64 bit (AMD64)]
IPython 4.0.1 -- An enhanced Interactive Python.

import numpy as np
from scipy.stats.stats import nanmean    
dat = np.random.normal(size=(1000,1000))
ii = np.ix_(np.random.randint(0,99,size=50),np.random.randint(0,99,size=50))
dat[ii] = np.nan
In[185]: def method1():
    mdat = np.ma.masked_array(dat,np.isnan(dat))
    mm = np.mean(mdat,axis=1)
    mm.filled(np.nan) 

In[190]: %timeit method1()
100 loops, best of 3: 7.09 ms per loop
In[191]: %timeit [np.mean([l for l in d if not np.isnan(l)]) for d in dat]
1 loops, best of 3: 1.04 s per loop
In[192]: %timeit np.array([r[np.isfinite(r)].mean() for r in dat])
10 loops, best of 3: 19.6 ms per loop
In[193]: %timeit np.ma.masked_invalid(dat).mean(axis=1)
100 loops, best of 3: 11.8 ms per loop
In[194]: %timeit nanmean(dat,axis=1)
100 loops, best of 3: 6.36 ms per loop
In[195]: import bottleneck as bn
In[196]: %timeit bn.nanmean(dat,axis=1)
1000 loops, best of 3: 1.05 ms per loop
In[197]: from scipy import stats
In[198]: %timeit stats.nanmean(dat)
100 loops, best of 3: 6.19 ms per loop

So the best is ‘bottleneck.nanmean(dat, axis=1)’
‘scipy.stats.nanmean(dat)’ is not faster then numpy.nanmean(dat, axis=1).

From numpy 1.8 (released 2013-10-30) onwards, nanmean does precisely what you need:

>>> import numpy as np
>>> np.nanmean(np.array([1.5, 3.5, np.nan]))
2.5

'''define dataMat'''
numFeat= shape(datMat)[1]
for i in range(numFeat):
     meanVal=mean(dataMat[nonzero(~isnan(datMat[:,i].A))[0],i])

# I suggest you this way:
import numpy as np
dat  = np.array([[1, 2, 3], [4, 5, np.nan], [np.nan, 6, np.nan], [np.nan, np.nan, np.nan]])
dat2 = np.ma.masked_invalid(dat)
print np.mean(dat2, axis=1)