How to replace outliers with NaN

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012786534
012786534 2019-8-22
回答： Steven Lord 2019-8-22
Hello,
I am trying to replace values above the 99th percentile (outliers) by NaN for each group (for both group A and group B) in a table t.
group = repelem(['A' 'B'], 1000)';
val = repelem(1:1000, 2)';
t = table(group, val);
unique_gr = unique(t.group);
for g = 1:length(unique_gr)
sub = t(strcmp(t.group, unique_gr(g, 1)), :);
f = filloutliers(sub.val, 'NaN', 'percentiles', [0 99])
end
Ideas ? Please note that I do not have any toolboxes.
  2 个评论
Walter Roberson
Walter Roberson 2019-8-22
Use unique with three outputs and iterate through the group numbers,
[unique_gr, ~, groupnum] = unique(t.group);
for g = 1 : size(unique_gr,1)
mask = groupnum == g;
t(mask,:) = filloutliers(t(mask,:), nan, 'percentiles', [0 99]);
end
012786534
012786534 2019-8-22
Thank you Walter, work like a charm

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回答（1 个）

Steven Lord
Steven Lord 2019-8-22
You can use grouptransform with an anonymous function that calls filloutliers. Let's use your sample data.
group = repelem(['A' 'B'], 1000)';
val = repelem(1:1000, 2)';
t = table(group, val);
This grouptransform call uses the variable group from the table t as the grouping variable. The anonymous function is the same as what you used and Walter each used in your for loops, though I chose to replace it with the double NaN rather than the text 'NaN' like Walter did.
t2 = grouptransform(t, 'group', ...
@(x) filloutliers(x, NaN, 'percentiles', [0 99]));
Let's see what values of val in t were replaced by NaN in t2.
t(isnan(t2.val), :)
By the way you built t, those do look like the top 1% of values for each group.

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