grouping elements from a table
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Hi everyone:
i have a table like this:
Week Open High Low Close
---- ---- ---- --- -----
10 1.2 1.4 1.1 1.2
10 1.5 1.7 1.5 1.6
10 1.4 2.1 1.3 2
10 2.2 2.4 2 2.1
11 2 2.1 2 2
11 2.2 2.5 1.1 1.3
12 1.7 1.9 1.6 1.6
12 1.8 1.9 0.6 1.4
12 0.9 1.4 0.8 1
is there a way to get the first element(open), maximum element(high), minimum element(low) and last element (close) for each week? so the result table would be
Week Open High Low Close
---- ---- ---- --- -----
10 1.2 2.4 1.1 2.1
11 2 2.5 1.1 1.3
12 1.7 1.9 0.6 1
thanks in advance
采纳的回答
Akira Agata
2017-9-8
Assuming your table is T, the following code can generate what you want.
[group, id] = findgroups(T.Week);
func = @(p, q, r, s) [p(1), max(q), min(r), s(end)];
result = splitapply(func, T.Open, T.High, T.Low, T.Close, group)
Tout = array2table([id, result],...
'VariableNames', T.Properties.VariableNames);
3 个评论
eric capnu
2017-9-18
Walter Roberson
2017-9-18
>> findgroups([1 2.5 2 2.5 1])
ans =
1 3 2 3 1
Looks like it works for double to me.
eric capnu
2017-9-19
更多回答（1 个）
Guillaume
2017-9-8
Step 1: define a function in its own m file that applies the appropriate filter to each column (unfortunately, you can't use anonymous functions for this):
function [firstopen, maxhigh, minlow, lastclose] = filtercols(open, high, low, close)
firstopen = open(1); %first element
maxhigh = max(high);
minlow = min(low);
lastclose = close(end); %last element
end
t = array2table([
10 1.2 1.4 1.1 1.2
10 1.5 1.7 1.5 1.6
10 1.4 2.1 1.3 2
10 2.2 2.4 2 2.1
11 2 2.1 2 2
11 2.2 2.5 1.1 1.3
12 1.7 1.9 1.6 1.6
12 1.8 1.9 0.6 1.4
12 0.9 1.4 0.8 1], ...
'VariableNames', {'Week', 'Open', 'High', 'Low', 'Close'});
result = rowfun(@filtercols, t, 'GroupingVariable', 'Week', 'NumOutputs', 4, 'OutputVariableNames', {'Open', 'High', 'Low', 'Close'})
The only potential issue is that the order in which the rows are grouped together by rowfun is not documented, so it may be that the first element and last element are not the correct ones. In my version of matlab (R2017a), it is correct.
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