Replace empty element in table located in different rows

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Song Decn 2019-10-26

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https://ww2.mathworks.cn/matlabcentral/answers/487541-replace-empty-element-in-table-located-in-different-rows

评论： Song Decn 2019-10-26

ann = {'20191025';'20191025';'20190829'};
fann = {'20191025';'20191025';'20190829'};
eann = {'20190930';'20190930';'20190630'};
rec = {877134853.120000;877134853.120000;846524470.030000};
ass = {0;2692456847.78000;2712484638.45000};
cip = {0;88281897.8000000;63035038.2900000};
t = table(ann, fann, eann, rec, ass, cip);

Suppose I have a table like this

t =

3×6 table

ann fann eann rec ass cip

__________ __________ __________ ____________ ____________ ____________

'20191025' '20191025' '20190930' [8.7713e+08] [ 0] [ 0]

'20191025' '20191025' '20190930' [8.7713e+08] [2.6925e+09] [8.8282e+07]

'20190829' '20190829' '20190630' [8.4652e+08] [2.7125e+09] [6.3035e+07]

What I need to get is that if any value of "rec" "ass" "cip" is 0, then it should be replaced with non-zero element of another row with identical "ann" "fann" "eann" properties.

So the results for t would be

ann fann eann rec ass cip

__________ __________ __________ ____________ ____________ ____________

'20191025' '20191025' '20190930' [8.7713e+08] [2.6925e+09] [8.8282e+07]

'20190829' '20190829' '20190630' [8.4652e+08] [2.7125e+09] [6.3035e+07]

Is there a way to achieve this in a fast way?

Thank u very much

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Guillaume 2019-10-26

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Loops won't be needed and replace or delete is probably just as easy.

Again, in your example, rec, ass, cip are cell arrays (where each cell contain a 1x1 double array. Compare the difference between

ann = {'20191025';'20191025';'20190829'};
fann = {'20191025';'20191025';'20190829'};
eann = {'20190930';'20190930';'20190630'};
rec = {877134853.120000;877134853.120000;846524470.030000};
ass = {0;2692456847.78000;2712484638.45000};
cip = {0;88281897.8000000;63035038.2900000};
t1 = table(ann, fann, eann, rec, ass, cip)
class(t1.rec)
rec = [877134853.120000;877134853.120000;846524470.030000];
ass = [0;2692456847.78000;2712484638.45000];
cip = [0;88281897.8000000;63035038.2900000];
t2 = table(ann, fann, eann, rec, ass, cip)
class(t2.rec)

The different storage makes a lot of difference. The latter one will make your life much easier. (And, since your first three columns appear to be dates, storing that as datetime would also make your life easier).

So, I ask again: are the numeric columns really stored as cell arrays (as you construct them in your example) or are they actually stored more sensibly as numeric vectors?

Song Decn 2019-10-26