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Explore Stack Internal The issue is that the provided pattern also matches the outer list. Therefore ReplaceAll[] replaces the outer list and stops (as theirthere are no elements left).
To overcome this issue, Map[] must be used:
ReplaceAll[{x_, _} :> {x, "SOMETHING"}] /@ {{1, "2"}, {2, "3"}} The issue is that the provided pattern also matches the outer list. Therefore ReplaceAll[] replaces the outer list and stops (as their are no elements left).
To overcome this issue Map[] must be used:
ReplaceAll[{x_, _} :> {x, "SOMETHING"}] /@ {{1, "2"}, {2, "3"}} The issue is that the provided pattern also matches the outer list. Therefore ReplaceAll[] replaces the outer list and stops (as there are no elements left).
To overcome this issue, Map[] must be used:
ReplaceAll[{x_, _} :> {x, "SOMETHING"}] /@ {{1, "2"}, {2, "3"}} The issue is that the provided pattern also matches the outer list. Therefore ReplaceAll[] replaces the outer list and stops (as their are no elements left).
To overcome this issue Map[] must be used:
ReplaceAll[{x_, _} :> {x, "SOMETHING"}] /@ {{1, "2"}, {2, "3"}}