finding the value in a table

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Lorraine Williams
Lorraine Williams 2015 年 10 月 16 日
コメント済み: Lorraine Williams 2015 年 10 月 16 日
I have a table and I wish to store the value of the 'rating pts' column in a variable when the column 'yds per att'= 7.21 (there are two values in the table (see example).
What command do I use to get the value in the last column when another column is a value??
Thanks

採用された回答

Peter Perkins
Peter Perkins 2015 年 10 月 16 日
You said "table". Your screenshot looks a whole lot like a spreadsheet. It would help to be more clear.
Assuming you mean that you used readtable to read that spreadsheet into a table, then you probably ended up with variables in the table named YrdsPerAtt and RatingPts, so do this:
T.RatingPts(T.YrdsPerAtt == 7.21)
Again, there is documentation for this, this page is a good starting point.
  1 件のコメント
Lorraine Williams
Lorraine Williams 2015 年 10 月 16 日
Hi Peter
Thank you - this is exactly what I needed. You were right - I did read it into a table as you stated.
Sorry for not being precise enough - this is all new to me and so, reading the help (which I'd already found) didn't help me much either on this particular issue.

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その他の回答 (2 件)

Naga A
Naga A 2015 年 10 月 16 日
Suppose if you store that table in the matrix named " table" then you can use this one:
table(find(table(:,column number of "yds per att")=="your required 'yds per att"),column_number of "Rating Pts")

Walter Roberson
Walter Roberson 2015 年 10 月 16 日
There are no entries which will have value exactly 7.21 . Finite binary floating point is not able to represent 0.21 exactly, because 1/10 = 0.1 (and so 0.01 and so 0.21) requires an infinite binary expansion, the same way that in decimal, 1/7 requires an infinite decimal expansion.
You need to test for values that are "close enough" to 7.21 for your purpose. abs(x-7.21) < tolerance for some positive tolerance such as 1E-10

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