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, Marco

PS: Do you really need ReplacePart?

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

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