Solucionario De Probabilidad E Inferencia Estadistica Rufino Moya Y Gregorio Saravia |LINK|

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Solucionario De Probabilidad E Inferencia Estadistica Rufino Moya Y Gregorio Saravia |LINK|





 
 
 
 
 
 
 

Solucionario De Probabilidad E Inferencia Estadistica Rufino Moya Y Gregorio Saravia

Rufino Moya Y Gregorio Saravia EPUB.[Current approach to prosthetic management of complex spine fractures].
Recent advances in radiotherapy and the occurrence of new diseases have increased the number of patients with spine fractures who require implantation of prostheses. Of all posterior spinal instrumentation, sublaminar wiring, pedicle screw fixation and transpedicular instrumentation are the most frequently used procedures and are also technically most complex. Complications are more frequent with transpedicular instrumentation, with long operations, using a long rod, and with malpositioning of the pedicle screw.Q:

How to use bsxfun to replace value in a data frame?

I have two matrices A and B, where each row corresponds to a token and each column to a category in A. I would like to replace the corresponding categories from A with those from B. For example, the input A would have zeros in the category where there is no category B, and 1 for the category where B is present. The goal is to calculate the percentage of “category B” in A. So far, I have:
%B: In A there are only 1s in the column corresponding to %B. Therefore, the answer is 100
A_bin_ind: In A, the column corresponding to category B also have zeros in the row corresponding to %B. Therefore, the answer is the first row in B_bin
%B: Now, there are zeros for %B for row 1 and 3 and 1 for row 2 (no %B)
B_bin_ind: Now, there are 1s for %B for row 2, 1s for %B for row 4 and zero for row 5, 6 and 7 (no %B)
A_bin_ind = bsxfun(@eq, A_bin_ind, B_bin_ind)

Now this will give me the following matrix:
ans =
0 0 0 0 0 0 0 0 1 1 1 1 1 1
1 1 1 1 0

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