Timeline for Removing outliers from circular average data
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 8, 2017 at 14:10 | comment | added | Iosif Pinelis | I have added a remark (highlighted) explaining why one should choose a finite negative $q$ to measure the distances from potential outliers to the rest of the data. | |
| Dec 8, 2017 at 14:07 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Dec 8, 2017 at 3:31 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Dec 8, 2017 at 3:11 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Dec 8, 2017 at 3:02 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Dec 7, 2017 at 23:16 | comment | added | Iosif Pinelis | Good point! Yes, this looks really unbelievable, until one looks at it a bit more closely. :-) | |
| Dec 7, 2017 at 23:14 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Dec 7, 2017 at 22:44 | comment | added | fedja | A disadvantage of this definition is that it depends on the choice of the x-axis. I thought so too until I started typing my answer. :-) The real problem is that $\tan x=\tan(\pi+x)$, so sometimes the OP gets the diametrically opposite point to the one he's looking for. However, up to this funny effect the formula is rotation invariant (unbelievable, isn't it?) | |
| Dec 7, 2017 at 22:10 | comment | added | Iosif Pinelis | I have modified the previous approach. The new approach is logically simpler. It is also substantially lighter computationally, and it likely has similar (quasi-)optimality properties. | |
| Dec 7, 2017 at 22:04 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Dec 7, 2017 at 17:42 | history | answered | Iosif Pinelis | CC BY-SA 3.0 |