Wolfram Function Repository
Instant-use add-on functions for the Wolfram Language
Function Repository Resource:
Get a list of subsets of a list along with their complements
ResourceFunction["SubsetsWithComplements"][list] gives a List of all possible subsets of list along with their complements. | |
ResourceFunction["SubsetsWithComplements"][list,n] gives all subsets containing at most n elements along with their corresponding complements. | |
ResourceFunction["SubsetsWithComplements"][list,{n}] gives all subsets containing exactly n elements. | |
ResourceFunction["SubsetsWithComplements"][list,{nmin,nmax}] gives all subsets containing between nmin and nmax elements. | |
ResourceFunction["SubsetsWithComplements"][list,nspec,s] limits the result to the first s subsets. | |
ResourceFunction["SubsetsWithComplements"][list,nspec,{s}] gives, if possible, the sth subset and complement. |
Get all the subsets of a List of characters, paired with their complements:
In[1]:= | ![]() |
Out[1]= | ![]() |
Take a List of three characters and get all the one-element subsets thereof and their complements:
In[2]:= | ![]() |
Out[2]= | ![]() |
Get all subsets from a List of five characters with at most two elements, along with their complements:
In[3]:= | ![]() |
Out[3]= | ![]() |
Take a List of five characters and get all subsets with between two and three elements and their complements:
In[4]:= | ![]() |
Out[4]= | ![]() |
Get the first three two-element subsets and complements from a list with six elements:
In[5]:= | ![]() |
Out[5]= | ![]() |
Get the seventh two-element subset and complements of a list with six elements:
In[6]:= | ![]() |
Out[6]= | ![]() |
Find and show conditions under which each expression in a List will have a value greater than that of any other expression in the list:
In[7]:= | ![]() |
In[8]:= | ![]() |
Out[8]= | ![]() |
In[9]:= | ![]() |
In[10]:= | ![]() |
Out[10]= | ![]() |
In[11]:= | ![]() |
Out[11]= | ![]() |
Do similarly for the lowest value of any expression in the List:
In[12]:= | ![]() |
In[13]:= | ![]() |
Out[13]= | ![]() |
A certain classifier predicts the probabilities that a person will be in each of three possible classes. A misclassification results in a utility of –1, except when a person is really in the first class but is misclassified as being a member of the second class. This latter case results in a utility of –u, where u is some non-negative number. For each value of u, find and show the set of classifier probabilities {p1,(1–p1)p2,1–(p1+(1–p1)p2)} that result in each prediction being the one that maximizes expected utility:
In[14]:= | ![]() |
Out[14]= | ![]() |
In[15]:= | ![]() |
Out[15]= | ![]() |
In[16]:= | ![]() |
Out[16]= | ![]() |
This work is licensed under a Creative Commons Attribution 4.0 International License