Matfyz Wiki
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| nz2ycp , [url=http://mapzddyawftv.com/]mapzddyawftv[/url], [link=http://abssodnekdye.com/]abssodnekdye[/link], http://shzpnihwnsea.com/ | {{Předmět|Dobývání znalostí|Doc. RNDr. Mrázová Iveta, CSc.|NDBI023}} |
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| == pPyPxRMMtEzK == | ## Požadavky 12/13 ## |
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| Personal service, as reiuqred in the bill, would establish a higher standard of service for paternity actions than all other civil actions. This higher standard does not directly address paternity fraud or prevent fraud in the future but instead would adversely impact the establishment of paternities. Thousands of paternity judgments are established timely each year by serving individuals by substitute service or by mail. The bill’s requirement of personal delivery service would severely delay this process, but more important, would provide biological fathers the ability to evade service of process [my emphasis], preventing the establishment of paternity in the majority of these cases and allowing the avoidance of parental responsibilities. This would directly impact child support collections and would jeopardize California’s ability to meet federally reiuqred performance measures putting California at risk of losing up to $40 million in federal funds. [emphasis added]In addition, AB 2240 has substantial federal compliance problems that would adversely affect California. The bill’s requirement of a paternity questionnaire, signed by themother, would prevent the filing of a paternity action in cases against the father if the mother is deceased or unavailable, or if she simply refuses to cooperate. This would prevent moving ahead on cases even if other evidence establishes paternity. This would also apply to foster care cases where federal law requires the establishment of paternity and child support. AB 2240 would prevent California from proceeding on a large number, if not most, of foster care cases, putting California out of compliance with federal law.This is about money. Many states depend upon federal grants to support their child support systems. If the states can't declare the guys these whores are married to to be the fathers of those bastard children, then they can't declare any guy to be the father of those bastard children when the actual father isn't known. That means states lose tens of millions of dollars in federal grants - which means those states have to raise taxes to pay for the welfare that those women will surely need...because those women are going to need it when their husbands divorce them and they're left to pay for those kids on their own.This is about money. It's that simple. And that money is going to come from somewhere. And, as former-governor Gray Davis made clear, the rights of the victims don't matter. | * výsledná známka = 40% body ze zkoušky, 20% body z dvou písemek, 40% body z úkolů ze cvičení a projektu (včetně bodů ze cvičení, všechny úkoly jsou povinné) * vypracovat projekt * celkem je potreba splnit aspon 56% * terminy v září budou jeden až dva |
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| == SCPRkuSHjypArRkrsq == | ## Zdroje k předmětu ## |
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| PDWfnd <a href="http://nhtztxjxmyce.com/">nhtztxjxmyce</a> | * [Stránky předmětu](http://ksvi.mff.cuni.cz/%7Emraz/datamining/index.html) * <NDBI023%20Shrnutí> |
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| == USCHpeYE == | * [Wikipedia:Datamining](Wikipedia:Datamining) * [Statistics Tutorials](http://home.okstate.edu/homepages.nsf/toc/onlinetutorialsstatistics) * [How to choose a statistical test](http://www.graphpad.com/www/book/choose.htm) * [Statistical Data Mining Tutorials](http://www.autonlab.org/tutorials/) * [Data mining in Matlab](http://matlabdatamining.blogspot.cz/) |
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| 9n6gkT , [url=http://srmclvtqgyqs.com/]srmclvtqgyqs[/url], [link=http://nbhvcrhgtzso.com/]nbhvcrhgtzso[/link], http://oshlquvnvqvk.com/ | ### Fisherův test ### |
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| == jMFqXVtJ == | * http://en.wikipedia.org/wiki/Fisher%27s_exact_test#Example - odvození, příklad jednostranného * http://oldweb.izip.cz/ds3/hypertext/JZAAA.htm - příklad česky * http://www.quantpsy.org/fisher/fisher.htm - hezké vysvětlení jednostranných a oboustranných testů, kalkulačka * http://graphpad.com/quickcalcs/contingency2/ - ještě jedna kalkulačka pro kontrolu :) * http://www.stahroun.me.cz/interstat/kategorialni/asociace/fisher/index.htm - spousta zdrojů |
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| oGzqIo , [url=http://owfgjechtwzg.com/]owfgjechtwzg[/url], [link=http://gncmcpaoumhh.com/]gncmcpaoumhh[/link], http://ayvucyttrqws.com/ | ### chi-kvadrát test ### |
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| == yMHHcphFbNu == | * http://mathhelpforum.com/advanced-statistics/44800-two-tailed-chi-squared-tests.html - trošku světla na oboustranné testy * http://stats.stackexchange.com/questions/22347/chi-squared-always-a-one-sided-test - a jeste jednou oboustranné testy * http://itl.nist.gov/div898/handbook/eda/section3/eda3674.htm - tabulky |
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| sRQhWL , [url=http://jrtbglbwbfps.com/]jrtbglbwbfps[/url], [link=http://cvonmmjhgkjz.com/]cvonmmjhgkjz[/link], http://qwdjjutlcafb.com/ | ## Písemky ## |
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| == OSIMikBYKDyBGVwUtA == | [NDBI023_Písemky](NDBI023_Písemky) |
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| SEN1fw , [url=http://tmliusqnazys.com/]tmliusqnazys[/url], [link=http://srxfatqvibun.com/]srxfatqvibun[/link], http://xhrbxllegxqs.com/ | ## Příklady ze cvičení ## |
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| == LtUIGTSYrmXgxaYgVwp == | ### Algoritmus TDIDT/ID3 ### |
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| l83mwJ , [url=http://wwkkgfklpumm.com/]wwkkgfklpumm[/url], [link=http://jgoxfeoewgyf.com/]jgoxfeoewgyf[/link], http://dydxssqnqrjh.com/ | * http://ksvi.mff.cuni.cz/~mraz/datamining/Cvicenie_DT.pdf |
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| == pohGwvQVZIpAHa == | Zvol jeden atribut jako kořen podstromu |
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| efn94r <a href="http://ivjjhobeabhp.com/">ivjjhobeabhp</a> | #### Kořen #### ##### Vitamín ##### * B (2+, 3-) $-\frac{n_+(Vit(B))}{n(Vit(B))}.log_2\frac {n_+(Vit(B))}{n(Vit(B))} - \frac{n_-(Vit(B))}{n(Vit(B))}.log_2\frac {n_-(Vit(B))}{n(Vit(B))} = - \frac{2}{5}.log_2\frac {2}{5} - \frac{3}{5}.log_2\frac {3}{5} = 0.97$ * C (4+, 0-) $\dots = 0.0$ (z "definície"...) * D (3+, 2-) $\dots = 0.97$ Entropia: $H(Vit) = \frac{0.971*5 + 0*4 + 0.971*5}{14} = 0.6936$ (vážený priemer) ##### Velikost rodiny ##### * veľká (2+, 2-) $-\frac{n_+(Rodina(velka))}{m(Rodina(velka))}.log_2\frac {n_+(Rodina(velka))}{n(Rodina(velka))} - \frac{n_-(Rodina(velka))}{n(Rodina(velka))}.log_2\frac {n_-(Rodina(velka))}{n(Rodina(velka))} = 1$ * stredná (4+, 2-) $\dots = 0.9183$ * malá (3+, 1-) $\dots = 0.8113$ Entropia: $H(Rodina)=0.9111$ ##### Cvičil ##### * pravidelně (3+, 4-) $\dots = - \frac{3}{7}.log_2\frac {3}{7} - \frac{4}{7}.log_2\frac {4}{7} = 0.985$ * málo (6+, 1-) $\dots = - \frac{6}{7}.log_2\frac {6}{7} - \frac{1}{7}.log_2\frac {1}{7} = 0.592$ Entropia: $H(Cvicil) = \frac{0.985*7 + 0.592*7}{14} = 0.7885$ ##### Bypass ##### Entropia: $H(Bypass)=0.8922$ ... budeme teda pokračovať podľa atribútu Vitamín (najmenšia entropia). Množinu rozdelíme na 3 skupiny (B,C,D). C-čko všetci prežili, máme dve skupiny pre Bcomplex a D - spočítame znovu strednú entropiu (aby sme vedeli, podľa čoho ďalej štiepiť). #### B #### vyberieme riadky s Bcomplex a robíme to isté :) ##### Velikost rodiny ##### * veľká (0, 2-) $\dots = 0$ (z "definice"...) * stredná (1+, 1-) $\dots = 1$ * malá (1+, 0) $\dots = 0$ (z "definice"...) Entropia: $H(Rodina) = \frac{0*2 + 1*2 + 0*1}{5} = 0.4$ ##### Cvicil ##### $H(Cvicil)=0$ ##### Bypass ##### $H(Bypass)=0.9183$ #### C #### ...je jasné, tam nik nezomrel... :) #### D #### $H(Rodina)=0.9183$ $H(Cvicil)=0.9183$ $H(Bypass)=0$ ...ďalej sa teda bude štiepiť v Bčku podľa "Cvicil" a v Dčku podľa "Bypass" Vitamin / | \ B/ C| \D Cviceni + Bypassprav./ m.| a| n\ - + - + ### Perceptron ### * uloha: naucit perceptron rozoznavat body v 2D na 2 skupiny... * popis perceptronu v matlabe: $p=\[ \begin{matrix} w_1 & w_2 & prah \end{matrix} ]$ * vyhodnoti sa potom... $x=\[\begin{matrix} x_1 & x_2 \end{matrix} ]$ => $perc_{recall} = w_1*x_1 + w_2*x_2 + prah*1$ * ...rozsireny vstupny vektor $x_1'=\[\begin{matrix} x_1 & x_2 & 1 \end{matrix} ]$ * $vystup=hardlim1(p*x_1')$ (hardlim1(<=0)=0; hardlim1(>0)=1) * inicializacia : $p=\[ \begin{matrix} 1 1 1 \end{matrix} ]$ * vstup: $A=\left\[ \begin{matrix} 1 & 1 & 2 & 3 \\ 1 & 3 & 2 & 1 \end{matrix} \right]$ * vystup (chceme): $ c = \[ \begin{matrix} 1 & 1 & 0 & 0 \end{matrix} ] $ * uciaca konstanta $a=0.2$ * ucenie prebieha takto: * vezme sa vzor, ak sedi, nic nerobime * ak najdeme chybu - pricitame/odcitame (pozadovany-skutocny vystup - napr. v priklade $y=hardlim1(p*AA); dif=c(3)-y(3)$) dany vektor k vaham... $(c(3)-y(3))*AA(:,3)$ * dostavam novy perceptron $p1=p+a*((c(3)-y(3))*AA(:,3))'$ * $y1=hardlim1(p1*AA) = \[ \begin{matrix} 1 & 1 & 1 & 1 \end{matrix} ]$ * to je zla matica (ma byt samozrejme \[1 1 0 0])- znovu zopakujeme postup... ****$ p1=\[ \begin{matrix} 0.6 & 0.6 & -0.2 \end{matrix} ] => y1=\[ \begin{matrix} 1 & 1 & 1 & 1 \end{matrix} ]$ ****$ p2=\[ \begin{matrix} 0 & 0.4 & -0.4 \end{matrix} ] => y2=\[ \begin{matrix} 0 & 1 & 1 & 0 \end{matrix} ]$ *****teraz budeme skusat prvy vektor (ktory je nespravny) ****$ p3=\[ \begin{matrix} 0.2 & 0.6 & -0.2 \end{matrix} ] => y2=\[ \begin{matrix} 1 & 1 & 1 & 1 \end{matrix} ]$ * ... N iteracii (N=?) [Category:Informatika](Category:Informatika) |