如果列全部为nan，则使numpy.nanargmin返回nan [英] making numpy.nanargmin return nan if column is all nan

问题描述

是否可以使用numpy.nanargmin，以便在其中仅包含nan的列上返回numpy.nan.现在，当发生这种情况时，它会引发一个ValueError.而且我不能使用numpy.argmin，因为当列中只有少数Nan时，它将失败.

Is it possible to use numpy.nanargmin, so that it returns numpy.nan, on columns where there are only nans in them. Right now, it raises a ValueError, when that happens. And i cant use numpy.argmin, since that will fail when there are only a few nans in the column.

http://docs.scipy.org/doc/numpy/reference/generated/numpy.nanargmin.html 表示ValueError对于全南切片都适用.在那种情况下，我希望它返回numpy.nan(只是为了进一步用nans掩盖非数据")

http://docs.scipy.org/doc/numpy/reference/generated/numpy.nanargmin.html says that the ValueError is raised for all-nan slices. In that case, i want it to return numpy.nan (just to further mask the "non-data" with nans)

这下一点可以做到这一点，但是它非常慢，而且不是真正的pythonic:

this next bit does this, but is super-slow and not really pythonic:

for i in range(R.shape[0]):
    bestindex = numpy.nanargmin(R[i,:])
    if(numpy.isnan(bestindex)):
        bestepsilons[i]=numpy.nan
    else:
        bestepsilons[i]=epsilon[bestindex]

下一点也可以使用，但前提是不涉及all-nan列:

This next bit works too, but only if no all-nan columns are involved:

ar = numpy.nanargmin(R, axis=1)
bestepsilons = epsilon[ar]

所以理想情况下，我希望最后一点也能与全南列一起使用

So ideally i would want this last bit to work with all-nan columns as well

推荐答案

找到了解决方案:

# makes everything nan to start with
bestepsilons1 = numpy.zeros(R.shape[0])+numpy.nan 
# finds the indices where the entire column would be nan, so the nanargmin would raise an error
d0 = numpy.nanmin(R, axis=1) 
# on the indices where we do not have a nan-column, get the right index with nanargmin, and than put the right value in those points
bestepsilons1[~numpy.isnan(d0)] = epsilon[numpy.nanargmin(R[~numpy.isnan(d0),:], axis=1)]

这基本上是一种解决方法，只将nanargmin放在不会产生错误的地方，因为在那些地方，我们希望所得的索引始终是nan

This basically is a workaround, by only taking the nanargmin on the places where it will not give an error, since at those places we want the resulting index to be a nan anyways

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