finding the value in a table

Question

Lorraine Williams 2015-10-16

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/248871-finding-the-value-in-a-table

评论： Lorraine Williams 2015-10-16

Screen Shot 2015-10-15 at 5.01.41 PM.png

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

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Peter Perkins 2015-10-16

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/248871-finding-the-value-in-a-table#answer_196138

在 MATLAB Online 中打开

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 个评论
显示 -1更早的评论隐藏 -1更早的评论

Lorraine Williams 2015-10-16

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.

请先登录，再进行评论。

Answer 2

Naga A 2015-10-16

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/248871-finding-the-value-in-a-table#answer_196128

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")

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 3

Walter Roberson 2015-10-16

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/248871-finding-the-value-in-a-table#answer_196140

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