# [✓] Generate a list with its elements in the order of a second list?

GROUPS:
 This seems like it should be easy, but I've spend a half day trying unsuccessfully to solve it. I have three variables a,b,c (not in a list). The values are integers, For example a=6, b=21, c=7. I also have a list of the permuted numbers 1, 2 and 3. For example, let this list be be listA={3,1,2}. ListA could be any of the other five possible permutations of 1, 2 & 3, and I don't know what listA is ahead of time other than that it contains only a permutation of the numbers 1, 2 & 3..So I want to form a new list containing a,b,and c in the order specified by the permutation list. Thus, the desired output would be listB={c,a,b}={7,6,21}. I've tried a lot of ways and nothing worked, probably because i was not using certain functions correctly, but maybe I just don't know which function to use.I would appreciate any help you can give.Kenneth Bures
26 days ago
7 Replies
 Use Ordering to find the order of list, and apply it to the other one: order = Ordering[list1] list2 = list2[[order]] 
26 days ago
 Use the function Part: In[1]:= a = 6; b = 21; c = 7; In[2]:= listA = {3, 1, 2}; In[3]:= Part[{a, b, c}, listA] Out[3]= {7, 6, 21} Shorthand: In[4]:= {a, b, c}[[listA]] Out[4]= {7, 6, 21} 
26 days ago
 I'm repeating my reply because somehow the formatting of my response got changed in the publishing, and it is hard to read. Thank you for the two responses. The solution that uses the Ordering function does not seem to work. You can see from the code below that when I apply Ordering to list1, it changes the order from {3,1,2} t o {2,3,1). Then when I apply the Ordering to list2 I get {b,c,a} rather than {c,a,b}. Is it possible that I need to use Sort or SortBy along with Ordering? I haven't had a chance to play around with this.list1 = {3, 1, 2} list2 = {a, b, c} order = Ordering[list1] list2 = list2[[order]]{3, 1, 2}{a, b, c}{2, 3, 1}{b, c, a}The second solution uses Part, as in Part[list2,list1] or the shortcut list2[[list1]], and this simple solution does give the correct result.Thanks again to both of you for taking the time to respond.
25 days ago
 Can i place items in the list as unique and related to the list from data within each object in the list? Can I do an interpolated display 2d, 3d, movie?
 Any list can be permuted. In general, a list is still a list even if all the elements are not of the same type. An example: sourceList = {ExampleData[{"TestImage", "House"}], Style[Red, 80], 77, Plot[Sin@x, {x, 0, 2*\[Pi]}]} But as a special case a list that defines a permutation needs to have all its elements from a collection of consecutive integers, starting from the number 1. WL documentation: permutationList = {2, 4, 1, 3}; The source list permuted: sourceList[[permutationList]] 
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