Re: [MORPHMET] Doubts about Error Measurement
Thank you very much Carmelo for your help. Finally I think that understand the equation. It is some basic statistical concepts relate to the variance components calculation. Now, I think that it is better way to report the error: as the statistical books show it. In my case (I'm sorry but is in Spanish): Modelo general: *Yijk = **m** + **a**i** + **b**j(i)** + **e**ijk* Y para el caso de la sumas de los cuadrados: SCT = SCA + SCB (A) + SCE Donde: SCA = suma de cuadrados entre individuos. SCB(A) = suma de cuadrados para las imágenes anidadas en los individuos. SCE = suma de cuadrados para las digitalizaciones anidadas en los individuos y las imágenes. El valor F correspondió a: F factor A (individuo): división del cuadrado medio de A (CMA) por el cuadrado medio de B o la imagen (CMA/CMB) F factor B (imagen): división del cuadrado medio de B por el cuadrado medio del error o digitalización (CME). Los cuadrados medios fueron obtenidos por la división de la suma de cuadrados de cada efecto por sus grados de libertad (g. l.). g. l. factor A = (a-1) (k – 4)*[1]* <#_ftn1> g. l. factor B = a (b - 1) (k – 4) g. l. factor E = ab(r-1) (k – 4) g. l. total = abr (k – 4) - 1 Donde: a = número de individuos digitalizados b = número de imágenes repetidas (= 2) r = número de digitalizaciones de cada imagen (= 2) k = número de coordenadas = # puntos anatómicos - # semilandmarks (ya que cada semilandmark sólo aporta un grado de libertad. Las esperanzas de los cuadrados fueron: *E(MCA)* = *s**2 **+ r**s**2**b* *+ br**s**2**a* *E(MCB(A))* = *s**2 **+ r**s**2**b* *E(MCE)* = *s**2 * Y las estimas de los componentes de varianza: S2*a* = [CMA ind – CMB(A)imag] / ( b * r) S2*b* = [CMB(A)imag – CME dig] / r S2dig = CME Estos efectos representaron las diferencias reales entre individuos, el error metodológico e instrumental y el error personal, respectivamente. Esta estimación de las varianzas corresponde a la teoría clásica estadística de modelos anidados con efectos aleatorios, la cual es retomada por Bailey y Birnes (1990). -- [1] <#_ftnref1> Los cuatro grados de libertad que se le restan a k en todos los cálculos corresponden a los que se pierden durante el ajuste de Procrustes. Again, thanks a lot for the attention and help. Liu 2016-03-17 1:57 GMT-03:00 Carmelo Fruciano : > Hi Liu, > I'm not sure I understand your question. > Those are the percentages of the total sum of squares due to each of those > terms and, yes, as you can see from the numbers, they all add up to 100%. > > Of course, you can quantify only what you measure, and you cannot exclude > that there are other sources of error than the ones you tested (I say this > because of your wording "total error"). > > I hope this helps, > Best, > Carmelo > > > > > Liu Idárraga ha scritto: > > And respect to the total error, the sum of the % SSimage + % SSdig? >> >> Liu >> >> 2016-03-15 21:38 GMT-03:00 Carmelo Fruciano : >> >> Liu Idárraga ha scritto: >>> >>> >>> Appreciated colleagues. >>> I have some doubts about the correct calculation of error in my data. I had understood that the error corresponde to the proportion that represent the SS of each effect (image / digitalization) in the SStotal; but I found in the classical paper of Bayley & Byrnes 1990. Syst. Zool. 39(2):126 that the way to calculate the percent of the measurement error is: % ME = 100% s2 within / (s2 within - s2 among) s2 within = MS within but s2 among = (MS among -MS within)/# replicates I have the followed results for 71 ind in 2D, and 44 coord (66 - 4 - 18 semi landmarks) Procrustes ANOVA para la forma: con g.l. corregidos Efecto SS SS (%) MS df F p (param.) Individuo0,26857296 92,84 8,59709E-05 3124 25,97 <.0001 Imagen 0,01048868 3,63 3,31082E-06 3168 2,05 <.0001 Digitalización 0,01021206 3,53 1,61175E-06 6336 Total 0,28927370 100,00 My questions are: 1) Is correct to say that 3,63 and 3,53 are the errors of image and digitalization? 2) Is the total error equal to 3,63 + 3,53 or I should calculate it with the equations of Bayley & Byrnes 1990? In advance, thanks a lot for the help. Liu >>> Dear Liu, >>> Perhaps the best way to report this would be to say that 3.63% of the >>> total sum of squares is accounted for by/due to a certain term >>> (digitization, preparation)? >>> Best, >>> Carmelo >>> >>> >>> >>> -- >>> Carmelo Fruciano >>> Postdoctoral Fellow - Queensland University of Technology - Brisbane, >>> Australia >>> Honorary Fellow - University of Catania - Catania, Italy >>> e-mail c.fruci...@unict.it >>> http://www.fruciano.it/research/ >>> >>> -- >>> MORPHMET may be accessed via its webpage at http://www.morphometrics.org >>> ---You received this message because you are subscribed to the Google >>> Groups "MO
Re: [MORPHMET] Doubts about Error Measurement
Hi Liu, I'm not sure I understand your question. Those are the percentages of the total sum of squares due to each of those terms and, yes, as you can see from the numbers, they all add up to 100%. Of course, you can quantify only what you measure, and you cannot exclude that there are other sources of error than the ones you tested (I say this because of your wording "total error"). I hope this helps, Best, Carmelo Liu Idárraga ha scritto: And respect to the total error, the sum of the % SSimage + % SSdig? Liu 2016-03-15 21:38 GMT-03:00 Carmelo Fruciano : Liu Idárraga ha scritto: Appreciated colleagues. I have some doubts about the correct calculation of error in my data. I had understood that the error corresponde to the proportion that represent the SS of each effect (image / digitalization) in the SStotal; but I found in the classical paper of Bayley & Byrnes 1990. Syst. Zool. 39(2):126 that the way to calculate the percent of the measurement error is: % ME = 100% s2 within / (s2 within - s2 among) s2 within = MS within but s2 among = (MS among -MS within)/# replicates I have the followed results for 71 ind in 2D, and 44 coord (66 - 4 - 18 semi landmarks) Procrustes ANOVA para la forma: con g.l. corregidos Efecto SS SS (%) MS df F p (param.) Individuo0,26857296 92,84 8,59709E-05 3124 25,97 <.0001 Imagen 0,01048868 3,63 3,31082E-06 3168 2,05 <.0001 Digitalización 0,01021206 3,53 1,61175E-06 6336 Total 0,28927370 100,00 My questions are: 1) Is correct to say that 3,63 and 3,53 are the errors of image and digitalization? 2) Is the total error equal to 3,63 + 3,53 or I should calculate it with the equations of Bayley & Byrnes 1990? In advance, thanks a lot for the help. Liu Dear Liu, Perhaps the best way to report this would be to say that 3.63% of the total sum of squares is accounted for by/due to a certain term (digitization, preparation)? Best, Carmelo -- Carmelo Fruciano Postdoctoral Fellow - Queensland University of Technology - Brisbane, Australia Honorary Fellow - University of Catania - Catania, Italy e-mail c.fruci...@unict.it http://www.fruciano.it/research/ -- MORPHMET may be accessed via its webpage at http://www.morphometrics.org ---You received this message because you are subscribed to the Google Groups "MORPHMET" group. To unsubscribe from this group and stop receiving emails from it, send an email to morphmet+unsubscr...@morphometrics.org. -- *Liu Idárraga* Estudiante doctoral Facultad de Ciencias Naturales y Museo Universidad de La Plata - Argentina -- MORPHMET may be accessed via its webpage at http://www.morphometrics.org --- You received this message because you are subscribed to the Google Groups "MORPHMET" group. To unsubscribe from this group and stop receiving emails from it, send an email to morphmet+unsubscr...@morphometrics.org. -- Carmelo Fruciano Postdoctoral Fellow - Queensland University of Technology - Brisbane, Australia Honorary Fellow - University of Catania - Catania, Italy e-mail c.fruci...@unict.it http://www.fruciano.it/research/ -- MORPHMET may be accessed via its webpage at http://www.morphometrics.org --- You received this message because you are subscribed to the Google Groups "MORPHMET" group. To unsubscribe from this group and stop receiving emails from it, send an email to morphmet+unsubscr...@morphometrics.org.
Re: [MORPHMET] Sliding Semilandmarks
Hi all, I just stumbled across this discussion a bit too late, so I am joining the party after everybody has left. > If the semilandmarks slide a lot relative to the local curvature, they > get off the curve. Of course, they can be projected back, but the > following trick often is sufficient: Instead of the full amount of > sliding, let all the semilandmarks slide just a fraction of the > computed distance, say 20% (multiply T by 0.2 in equation of 4 of Gunz > et al. 2005). Then update the tangents and let the semilandmarks slide > again a fraction of the computed distance, etc. This requires more > iterations but keeps the semilandmarks closer to the curve or surface. As Philipp wrote, this can really make the difference for problematic cases (and especially when minimizing ProcD: even with reprojection, this can lead to distorted shapes if the coordinates have slid away on the tangent planes to find a minimum) For those who want to test the effect: In Morpho::slider3d you can control this dampening by specifying stepsize (e.g. to 0.2 according to Philipps example). Best Stefan On 18/02/16 21:18, mitte...@univie.ac.at wrote: > > As Michael described, the average shape configuration affects the > sliding when used as reference for the TPS; the final configurations > thus are sample-dependent. However, if the curves/surfaces are covered > densely enough by the semilandmarks (e.g., to avoid that a > semilandmark can slide away from a relevant region), Procrustes > distances are quite stable. Dense sampling can also improve the > estimation of the tangents. > > If the semilandmarks slide a lot relative to the local curvature, they > get off the curve. Of course, they can be projected back, but the > following trick often is sufficient: Instead of the full amount of > sliding, let all the semilandmarks slide just a fraction of the > computed distance, say 20% (multiply T by 0.2 in equation of 4 of Gunz > et al. 2005). Then update the tangents and let the semilandmarks slide > again a fraction of the computed distance, etc. This requires more > iterations but keeps the semilandmarks closer to the curve or surface. > > Also when minimizing Procrustes distance instead of BE, these > distances are reduced relative to the sample average. But as for the > superimposition itself, the sample configuration has only limited > effect on the final configurations for small to moderate shape > variation. (If variation is very large, the analysis is problematic > anyway.) Note that the full sample must be slid together for a joint > analysis (i.e., don't slide each population separately and then > analyze them together). > > The choice of the minimization criterion (Proc dist versus BE) can > lead to different configurations. For most datasets, this difference > is negligible, but in some situations it can matter. For example, when > minimizing Proc dist semilandmarks can change their order or slide > across a real landmark, whereas this is almost impossible for > minimizing BE (changing order would have a very high BE). On the other > hand, minimizing BE does not minimize affine shape variation (because > it has zero BE). If affine shape variation is not constrained by real > landmarks, this can lead to strange results. For instance, I had a > dataset of mandibular cross-sections, which were U-shaped with real > landmarks only at the two upper ends and semilandmarks in-between. > Affine variation thus was not properly controlled. After BE sliding, > the group differences comprised a lot of (meaningless) affine > differences. I thus decided for minimizing Proc dist. Usually, though, > I prefer minimizing BE because its is closer to our biological > understanding of homology, including the preservation of landmark > order and large scale shape features. Minimizing BE leads to smoother > TPS deformation grids, whereas miminizing Proc dists leads to smaller > sum of squares. > > Note that when updating the reference configuration in each iteration, > the algorithm can converge to quite undesired minima (e.g. all > semilandmarks collapse to a single point). This can be avoided by > iterating just a few times, which is usually enough, or by keeping the > reference constant at some point in the algorithm. In general, the > more the semilandmarks are constrained by real landmarks and the > smoother the curves, the more stable is the algorithm. > > Because of these issues, it is important to apply the semilandmark > algorithm carefully, especially for 3D surfaces. Always check the > tangents and how the semilandmarks slide along these tangents. Check > how the total sliding reduces from one iteration to the next, and > interpret the final pattern of shape variation in the light of the > property being minimized. > > Best wishes, > > Philipp Mitteroecker > > > > > > > Am Donnerstag, 18. Februar 2016 18:41:44 UTC+1 schrieb Collyer, Michael: > > Andrea, > > I like to think of semilandmark sliding as iteratively finding
