How to count the number of non-nan values for combination of multiple variables

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Blue
Blue 2024-7-17
评论： Blue 2024-7-17
Ran in:
Hello, I would like to count the number of non-nan values in the d column for unique combinations of a, b and c (i.e I want to generate the e column in tt). If any a,b or c are NaN then the count should be nan as well
% Initial table
a = [1,1,1,2,2,2,2]';
b = [660, 661, 661, 663, 663, NaN, 663]';
c = [1,2,2,5,5,NaN,6]';
d = [11,12,NaN, 13, 14, NaN,5]';
t = table(a,b,c,d)
t = 7x4 table
a b c d _ ___ ___ ___ 1 660 1 11 1 661 2 12 1 661 2 NaN 2 663 5 13 2 663 5 14 2 NaN NaN NaN 2 663 6 5
% Desired output
a = [1,1,1,2,2,2,2]';
b = [660, 661, 661, 663, 663, NaN, 663]';
c = [1,2,2,5,5,NaN,6]';
d = [11,12,NaN, 13, 14, NaN,5]';
e = [1, 1, 1, 2, 2, NaN, 1]';
tt = table(a,b,c,d,e)
tt = 7x5 table
a b c d e _ ___ ___ ___ ___ 1 660 1 11 1 1 661 2 12 1 1 661 2 NaN 1 2 663 5 13 2 2 663 5 14 2 2 NaN NaN NaN NaN 2 663 6 5 1

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采纳的回答

Voss
Voss 2024-7-17
编辑：Voss 2024-7-17
Ran in:
% Initial table
a = [1;1;1;2;2;2;2];
b = [660; 661; 661; 663; 663; NaN; 663];
c = [1;2;2;5;5;NaN;6];
d = [11;12;NaN; 13; 14; NaN;5];
t = table(a,b,c,d)
t = 7x4 table
a b c d _ ___ ___ ___ 1 660 1 11 1 661 2 12 1 661 2 NaN 2 663 5 13 2 663 5 14 2 NaN NaN NaN 2 663 6 5
G = {'a','b','c'};
F = @(x)nnz(~isnan(x));
T = groupsummary(t,G,F,'d');
[~,~,idx] = unique(t(:,G),'rows');
e = T.fun1_d(idx);
e(any(isnan(t{:,G}),2)) = NaN;
tt = addvars(t,e)
tt = 7x5 table
a b c d e _ ___ ___ ___ ___ 1 660 1 11 1 1 661 2 12 1 1 661 2 NaN 1 2 663 5 13 2 2 663 5 14 2 2 NaN NaN NaN NaN 2 663 6 5 1

更多回答（1 个）

J. Alex Lee
J. Alex Lee 2024-7-17
编辑：J. Alex Lee 2024-7-17
Ran in:
Not sure what d and e need to be, but i think this should at least partially help
a = [1,1,1,2,2,2,2]';
b = [660, 661, 661, 663, 663, NaN, 663]';
c = [1,2,2,5,5,NaN,6]';
x = [a,b,c]
x = 7x3
1 660 1 1 661 2 1 661 2 2 663 5 2 663 5 2 NaN NaN 2 663 6
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% any(_,2) goes row-wise ORs.
any(isnan(x),2)
ans = 7x1 logical array
0 0 0 0 0 1 0

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