英文:
DataFrame most efficient way update row value less than 40% to NaN?
问题
我有一个大型数据框，需要找到每一行中所有小于40%的元素并将其设置为NaN，元素未排序，需要为每一行重复这个操作。
我可以强制计算，但你可以想象它不太高效，有没有更高效的方法？
这里的40%意味着将行元素按升序排序，并将低排序的40%元素设为NaN，不包括本身为NaN的元素。
如果我有十个元素：1,21,20,4,5,6,7,9,10,11，应该对它进行排序，变成1,4,5,6,7,9,10,11,20,21，然后移除1,4,5,6，最终变成NaN, 21, 20, NaN, NaN, NaN, 7, 9, 10, 11。
英文:
I have big dataframe, need to find all element less than 40% in a row set to NaN, element not sorted, repeat this for each row.
I can force the calculation, but you can imagine it's not very efficient, there is no efficient way to do it?
40% mean row element order asc, and set low order 40% element to nan, does not contain an element that is itself a nan.
If I have ten element : 1,21,20,4,5,6,7,9,10,11, should sort it to 1,4,5,6,7,9,10,11,20,21 and remove 1,4,5,6, finally become nan,21,20,nan,nan,nan,7,9,10,11.
1 21 20 4 5 6 7 9 10 11
to
NaN 21 20 NaN NaN NaN 7 9 10 11
答案1
得分: 2
使用DataFrame.count来获取每行非缺失值的数量，然后通过双重numpy.argsort排序值的位置进行比较，最后根据掩码设置缺失值：
print (df)0 1 2 3 4 5 6 7 8 9 100 1 2 3 10 5 6 7 NaN 9 4 11.01 1 21 20 4 5 6 7 9.0 10 11 NaNcounts = df.count(axis=1).mul(0.4).to_numpy()[:, None]arr = np.argsort(np.argsort(df.to_numpy()))df[arr < counts] = np.nanprint (df)0 1 2 3 4 5 6 7 8 9 100 NaN NaN NaN 10.0 5.0 6.0 7 NaN 9 NaN 11.01 NaN 21.0 20.0 NaN NaN NaN 7 9.0 10 11.0 NaN
英文:
Use DataFrame.count for get number of non missing values per rows, then compare by positions of sorted values by double numpy.argsort and last set missing values by mask:
print (df)0 1 2 3 4 5 6 7 8 9 100 1 2 3 10 5 6 7 NaN 9 4 11.01 1 21 20 4 5 6 7 9.0 10 11 NaNcounts = df.count(axis=1).mul(0.4).to_numpy()[:, None]arr = np.argsort(np.argsort(df.to_numpy()))df[arr < counts] = np.nanprint (df)0 1 2 3 4 5 6 7 8 9 100 NaN NaN NaN 10.0 5.0 6.0 7 NaN 9 NaN 11.01 NaN 21.0 20.0 NaN NaN NaN 7 9.0 10 11.0 NaN
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