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Count the number of times a value appears in a row and save the row number

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James Peach
James Peach 2020-10-6
评论： Star Strider 2020-10-6
I wanted to sort points in an array based on how many times a number less than or equal to 1.5 appears in the row. I want to save the row index number of those that have 3 or more occurences less than or equal to the target value of 1.5. I have placed the array below. If you have any ideas on how to do this that would be great. Thanks for your time.
A = [NaN 1.00000000000000 1.00000000000000 1.99994495218295 1.40991828545270 1.98465118243128 2.97995391814023 1.41286932041226 1.44150019470391 2.30703739193881
1.00000000000000 NaN 1.41618367600370 0.999999999999996 1.00000000000000 2.11349467948903 1.98589244648483 1.00000000000000 1.73544845449717 1.98611502337309
1.00000000000000 1.41618367600370 NaN 2.24512987514306 2.20570040562793 0.999999999999996 3.22884118121504 1.75480404405606 1 3.15259384660772
1.99994495218295 0.999999999999996 2.24512987514306 NaN 1.40867173007472 2.65578172203034 1.00000000000000 1.41514975278174 2.44463958996392 2.12446752335984
1.40991828545270 1.00000000000000 2.20570040562793 1.40867173007472 NaN 3.01897607767301 2.11695260794737 1.62219494319153 2.57847672580517 1.00000000000000
1.98465118243128 2.11349467948903 0.999999999999996 2.65578172203034 3.01897607767301 NaN 3.52998354037453 2.38833466800397 1.41223022448087 3.96751962825988
2.97995391814023 1.98589244648483 3.22884118121504 1.00000000000000 2.11695260794737 3.52998354037453 NaN 2.21498386821498 3.35954605769311 2.49207458085697
1.41286932041226 1.00000000000000 1.75480404405606 1.41514975278174 1.62219494319153 2.38833466800397 2.21498386821498 NaN 1.44673037867744 2.53341581840774
1.44150019470391 1.73544845449717 1 2.44463958996392 2.57847672580517 1.41223022448087 3.35954605769311 1.44673037867744 NaN 3.56150161869124
2.30703739193881 1.98611502337309 3.15259384660772 2.12446752335984 1.00000000000000 3.96751962825988 2.49207458085697 2.53341581840774 3.56150161869124 NaN];

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采纳的回答

Star Strider
Star Strider 2020-10-6
Try this:
rowsum = sum((A <= 1.5), 2);
result = find(rowsum >= 3)
producing:
result =
1
2
3
4
5
8
9
Those could be combined into one line:
result = find(sum((A <= 1.5), 2) >= 3)
I kept them separate for clarity.
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James Peach
James Peach 2020-10-6
Sorry Im just seeing this, but I meant the 3rd case you showed which is the complement of the original vector. Thank you again!
Star Strider
Star Strider 2020-10-6
That’s what I thought, although I wanted to explain the other options.
As always, my pleasure!

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