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# How do I replace some elements of my list? {{a,b},{c,d}} ->{{1,b},{2,d}}

Posted 10 years ago
 Hi, I want replace some elements of my list "res" . res = {{{a,b},{c,d},{e,f}}} lis ={{1,2,3}}  I 'd like obtein : rep = {{{1,b},{2,d},{3,f}}}  How to do with the command ReplacePart? Thanks a lot
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Posted 10 years ago
 Hi Marco, I have applied your suggestions in my problems. I have another questions for a similar cas " ReplaceAll[res, MapThread[Rule, {res[[1, All, 1]], lis[[1]]}]]". I have GG1 = {{0.220067, 0.440134, 0.634358, 0.828582, 1.}, {0.220067, 0.440134, 0.634358, 0.828582, 1.}} res = {{{0.117779, 0.406482}, {0.235558, 0.333125}, {0.339506, 0.464985}, {0.443454, 0.544808}, {0.535196, 0.61788}}, {{0.117779, 0.406482}, {0.235558, 0.333125}, {0.339506, 0.464985}, {0.443454, 0.544808}, {0.535196, 0.61788}}}If I do : ReplaceAll[res, MapThread[Rule, {result[[All, All, 1]], GG1 }]] it doesn't run. Why? result[[All, All, 1]] and GG1 have the same dimensions. The problem occurs because res has too many parenthesis?Thank you Cheers, Margherita
Posted 10 years ago
 Thank you a lot. Cheers, Margherita
Posted 10 years ago
 Again, I am not sure whether you want MapThread here. You could try Map: Map[Select[#, # <= 2 &] &, res] or the short hand notation for that same thing: Select[#, # <= 2 &] & /@ res Cheers,Marco
Posted 10 years ago
 Hi, very good reponse. You are right, I haven't need use "ReplacePart".I notice that you have practice with Mapthread. So, I could ask you another question. If I use MapThread for Select, where is the position of select condition (I 'd like create a list where the elements are <=2) ? for exemple If I have this: "res1= {1,2,3}" the command will be "Lat = Select[ res1 , #<=2. &];" but with res= {{1,2,3},{1,2,3},{1,4,5}} . Which is the command to have: Lat = {{1,2},{1,2},{1,}} MapThread[Select, res] (where is the position for the select condition?)Thank you. Margherita
Posted 10 years ago
 Hi,here is a solution with ReplaceAll ReplaceAll[res, MapThread[Rule, {res[[1, All, 1]], lis[[1]]}]] and here is one without Replace at all: Transpose[{lis[[1]], res[[1, All, 2]]}] Cheers, MarcoPS: Do you really need ReplacePart?