I am trying to calculate the moving average in a large numpy array that contains NaNs. Currently I am using:
import numpy as np
def moving_average(a,n=5):
ret = np.cumsum(a,dtype=float)
ret[n:] = ret[n:]-ret[:-n]
return ret[-1:]/n
When calculating with a masked array:
x = np.array([1.,3,np.nan,7,8,1,2,4,np.nan,np.nan,4,4,np.nan,1,3,6,3])
mx = np.ma.masked_array(x,np.isnan(x))
y = moving_average(mx).filled(np.nan)
print y
>>> array([3.8,3.8,3.6,nan,nan,nan,2,2.4,nan,nan,nan,2.8,2.6])
The result I am looking for (below) should ideally have NaNs only in the place where the original array, x, had NaNs and the averaging should be done over the number of non-NaN elements in the grouping (I need some way to change the size of n in the function.)
y = array([4.75,4.75,nan,4.4,3.75,2.33,3.33,4,nan,nan,3,3.5,nan,3.25,4,4.5,3])
I could loop over the entire array and check index by index but the array I am using is very large and that would take a long time. Is there a numpythonic way to do this?
I'll just add to the great answers before that you could still use cumsum to achieve this:
import numpy as np
def moving_average(a, n=5):
ret = np.cumsum(a.filled(0))
ret[n:] = ret[n:] - ret[:-n]
counts = np.cumsum(~a.mask)
counts[n:] = counts[n:] - counts[:-n]
ret[~a.mask] /= counts[~a.mask]
ret[a.mask] = np.nan
return ret
x = np.array([1.,3,np.nan,7,8,1,2,4,np.nan,np.nan,4,4,np.nan,1,3,6,3])
mx = np.ma.masked_array(x,np.isnan(x))
y = moving_average(mx)
Pandas has a lot of really nice functionality with this. For example:
x = np.array([np.nan, np.nan, 3, 3, 3, np.nan, 5, 7, 7])
# requires three valid values in a row or the resulting value is null
print(pd.Series(x).rolling(3).mean())
#output
nan,nan,nan, nan, 3, nan, nan, nan, 6.333
# only requires 2 valid values out of three for size=3 window
print(pd.Series(x).rolling(3, min_periods=2).mean())
#output
nan, nan, nan, 3, 3, 3, 4, 6, 6.3333
You can play around with the windows/min_periods and consider filling-in nulls all in one chained line of code.
You could create a temporary array and use np.nanmean() (new in version 1.8 if I'm not mistaken):
import numpy as np
temp = np.vstack([x[i:-(5-i)] for i in range(5)]) # stacks vertically the strided arrays
means = np.nanmean(temp, axis=0)
and put original nan back in place with means[np.isnan(x[:-5])] = np.nan
However this look redundant both in terms of memory (stacking the same array strided 5 times) and computation.
If I understand correctly, you want to create a moving average and then populate the resulting elements as nan
if their index in the original array was nan
.
import numpy as np
>>> inc = 5 #the moving avg increment
>>> x = np.array([1.,3,np.nan,7,8,1,2,4,np.nan,np.nan,4,4,np.nan,1,3,6,3])
>>> mov_avg = np.array([np.nanmean(x[idx:idx+inc]) for idx in range(len(x))])
# Determine indices in x that are nans
>>> nan_idxs = np.where(np.isnan(x))[0]
# Populate output array with nans
>>> mov_avg[nan_idxs] = np.nan
>>> mov_avg
array([ 4.75, 4.75, nan, 4.4, 3.75, 2.33333333, 3.33333333, 4., nan, nan, 3., 3.5, nan, 3.25, 4., 4.5, 3.])
Here's an approach using strides -
w = 5 # Window size
n = x.strides[0]
avgs = np.nanmean(np.lib.stride_tricks.as_strided(x, \
shape=(x.size-w+1,w), strides=(n,n)),1)
x_rem = np.append(x[-w+1:],np.full(w-1,np.nan))
avgs_rem = np.nanmean(np.lib.stride_tricks.as_strided(x_rem, \
shape=(w-1,w), strides=(n,n)),1)
avgs = np.append(avgs,avgs_rem)
avgs[np.isnan(x)] = np.nan
来源:https://stackoverflow.com/questions/39919050/calculate-moving-average-in-numpy-array-with-nans