在忽略NaN的同时取np.average吗? [英] Taking np.average while ignoring NaN's?

问题描述

我有一个形状为(64,17)的矩阵对应于时间&纬度.我想获得一个加权的纬度平均值，我知道np.average可以做到这一点，因为与我用来平均经度的np.nanmean不同，可以在参数中使用权重.但是，np.average不会像np.nanmean那样忽略NaN，因此每行的前5个条目都包含在纬度平均中，并使整个时间序列充满NaN.

I have a matrix with shape (64,17) correspond to time & latitude. I want to take a weighted latitude average, which I know np.average can do because, unlike np.nanmean, which I used to average the longitudes, weights can be used in the arguments. However, np.average doesn't ignore NaN like np.nanmean does, so my first 5 entries of each row are included in the latitude averaging and make the entire time series full of NaN.

是否可以在不将NaN包括在计算中的情况下进行加权平均值计算?

Is there a way I can take a weighted average without the NaN's being included in the calculation?

file = Dataset("sst_aso_1951-2014latlon_seasavgs.nc")
sst = file.variables['sst']
lat = file.variables['lat']

sst_filt = np.asarray(sst)
missing_values_indices = sst_filt < -8000000   #missing values have value -infinity
sst_filt[missing_values_indices] = np.nan      #all missing values set to NaN

weights = np.cos(np.deg2rad(lat))
sst_zonalavg = np.nanmean(sst_filt, axis=2)
print sst_zonalavg[0,:]
sst_ts = np.average(sst_zonalavg, axis=1, weights=weights)
print sst_ts[:]

输出:

[ nan nan nan nan nan
 27.08499908 27.33333397 28.1457119 28.32899857 28.34454346
 28.27285767 28.18571472 28.10199928 28.10812378 28.03411865
 28.06411552 28.16529465]

[ nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan
 nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan
 nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan
 nan nan nan nan]

推荐答案

您可以创建如下的蒙版数组:

You can create a masked array like this:

data = np.array([[1,2,3], [4,5,np.NaN], [np.NaN,6,np.NaN], [0,0,0]])
masked_data = np.ma.masked_array(data, np.isnan(data))
# calculate your weighted average here instead
weights = [1, 1, 1]
average = np.ma.average(masked_data, axis=1, weights=weights)
# this gives you the result
result = average.filled(np.nan)
print(result)

这将输出:

[ 2.   4.5  6.   0. ]

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