Removing all sublists containing any arbitrary string as one or more of its list component

Question

so, basically I have a list of sublists of varying length. For example,

list={{#,a,b,c,d},{#,e,f,g},{#,h,1},{#,2},{#},{#,k,l,m,n,o},{#,a,b,1,2},{1,2},{2,3},{3,4},{4,5}}

And I would like to end up with

listFiltered={{1,2},{2,3},{3,4},{4,5}}

So, what I would like to filter out is all the sublist containing any string. Now, instead of filtering for any sublist containing any string, I also know that each sublist that I want to filter out has the "#" character as its first sublist element.

This is then the repeating pattern, a "#" character as the first element of every sublist of arbitrary length, that contains a string somewhere, and therefore should be removed from the main list.

How should I go about achieving this? I have tried a number of things with DeleteCases and StringMatchQ but couldn't really get the syntax right. Thank you very much in advance!

Answer 1

Cases[{__Integer}] @ list

{{1, 2}, {2, 3}, {3, 4}, {4, 5}}

DeleteCases[{"#",___}] @ list

{{1, 2}, {2, 3}, {3, 4}, {4, 5}}

DeleteCases[{"#", ___}|{___,_Symbol,___}] @ list

{{1, 2}, {2, 3}, {3, 4}, {4, 5}}

Cases[Except[{"#" ,___}|{___,_Symbol,___}]] @ list

{{1, 2}, {2, 3}, {3, 4}, {4, 5}}

Answer 2

list = {{"#", a, b, c, d}, {"#", e, f, g}, {"#", h, 1}, {"#", 
    2}, {"#"}, {"#", k, l, m, n, o}, {"#", a, b, 1, 2}, {1, 2}, {2, 
    3}, {3, 4}, {4, 5}};

Pick[list, MemberQ[Except[__Integer]] /@ list, False]

Extract[list, Position[list, {__Integer}]]

SequenceReplace[list, k : {{___, Except[_Integer], ___}} :> Nothing]

Pick[list, NoneTrue[Head@# =!= Integer &] /@ list]

Pick[list, AllTrue[Head@# === Integer &] /@ list]

---

Result

{{1, 2}, {2, 3}, {3, 4}, {4, 5}}