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transposition based on input method #67

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karoraw1 opened this issue Feb 9, 2018 · 3 comments
Open

transposition based on input method #67

karoraw1 opened this issue Feb 9, 2018 · 3 comments

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@karoraw1
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karoraw1 commented Feb 9, 2018

[[email protected]@login-node03 bhtsne]$ echo -e '1.0\t0.0\n0.0\t1.0' 
1.0	0.0
0.0	1.0
[[email protected]@login-node03 bhtsne]$ echo -e '1.0\t0.0\n0.0\t1.0' | ./bhtsne.py -d 2 -p 0.1
-2227.32653069	6608.48958328
2227.32653069	-6608.48958328
[[email protected]@login-node03 bhtsne]$ echo -e '1.0\t0.0\n0.0\t1.0' > a_file.txt
[[email protected]@login-node03 bhtsne]$ cat a_file.txt 
1.0	0.0
0.0	1.0
[[email protected]@login-node03 bhtsne]$ ./bhtsne.py -d 2 -p 0.1 -i a_file.txt 
-6863.21277159	-1236.73732294
6863.21277159	1236.73732294
@karoraw1
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karoraw1 commented Feb 9, 2018

[[email protected]@login-node03 bhtsne]$ ./bhtsne.py -d 2 -p 0.1 -i a_file.txt | python -c 'import numpy;  from sys import stdin, stdout; d = numpy.loadtxt(stdin); d -= d.min(axis=0); d /= d.max(axis=0); numpy.savetxt(stdout, d, fmt="%.8f", delimiter="\t")'
0.00000000	0.00000000
1.00000000	1.00000000
[[email protected]@login-node03 bhtsne]$ echo -e '1.0\t0.0\n0.0\t1.0' | ./bhtsne.py -d 2 -p 0.1 | python -c 'import numpy;  from sys import stdin, stdout; d = numpy.loadtxt(stdin); d -= d.min(axis=0); d /= d.max(axis=0); numpy.savetxt(stdout, d, fmt="%.8f", delimiter="\t")'
1.00000000	0.00000000
0.00000000	1.00000000

@karoraw1
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karoraw1 commented Feb 9, 2018

i don't think it makes a difference. i just wondered why it happens.

@lvdmaaten
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By definition, the loss of any multidimensional scaling method (that is, a loss that measures to what extent pairwise distances are preserved) is invariant under rotations and inflections of the embedding. I suppose that is what you're seeing here?

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