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How to find single index values in a matrix?

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Dalton Houghton-Schaffer
Dalton Houghton-Schaffer 2019 年 9 月 9 日
回答済み: Walter Roberson 2019 年 9 月 9 日
B1 = [2 4 6 8; 10 12 14 16; 18 20 22 24; 26 28 30 32]
idx_8=find(B1==8)
[row,column]=find(B1~=8)
RowColumn = [row:column]
Find the single index values for 26, 4, and 28?
How is a matrix indexed with single indexing values?
  3 件のコメント
Dalton Houghton-Schaffer
Dalton Houghton-Schaffer 2019 年 9 月 9 日
Is that the code to find single index values? And do I insert anything to the matrix [was_found]?
Walter Roberson
Walter Roberson 2019 年 9 月 9 日
[was_found, idx] = ismember([26 4 28], B1);
will assign to two variables: was_found and idx. was_found will be true for each element of [26 4 28] that was located somewhere in B1, and will be false for any element that was not found in B1. idx will be 0 for any element that was not found, and otherwise will be the index of the "first" location of the value in B1. idx will be a "linear index"

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madhan ravi
madhan ravi 2019 年 9 月 9 日
編集済み: madhan ravi 2019 年 9 月 9 日
Linear_indices = find(ismember(B1,[26 4 28])); % you mean linear indices by saying single indices
B1(Linear_indices) % would give [26 4 28]
  5 件のコメント
madhan ravi
madhan ravi 2019 年 9 月 9 日
Ok so don't expect me to delete my answer though.
Bruno Luong
Bruno Luong 2019 年 9 月 9 日
編集済み: Bruno Luong 2019 年 9 月 9 日
No IMO you should keep it. The difference is interesting to highlight.

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その他の回答 (1 件)

Walter Roberson
Walter Roberson 2019 年 9 月 9 日
[was_found, idx] = ismember([26 4 28], B1);
will assign to two variables: was_found and idx. was_found will be true for each element of [26 4 28] that was located somewhere in B1, and will be false for any element that was not found in B1. idx will be 0 for any element that was not found, and otherwise will be the index of the "first" location of the value in B1. idx will be a "linear index"

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