Update filter badges for multiple trees

The previous implementation only reflected counts in the LHS tree,
irregardless of any occurrences which may or may not be present in the
RHS tree. We now communicate both. This also fixes a bug introduced in
the previous commit where filtering to a traitName not present in the
LHS tree would cause an uncaught exception.

src/components/info/filtersSummary.js

@@ -33,6 +33,7 @@ const closeBracketSmall = <span style={{fontSize: "1.8rem", fontWeight: 300, pad
     metadata: state.metadata,
     nodes: state.tree.nodes,
     totalStateCounts: state.tree.totalStateCounts,
+    totalStateCountsSecondTree: state.treeToo?.totalStateCounts,
     visibility: state.tree.visibility,
     selectedClade: state.tree.selectedClade,
     dateMin: state.controls.dateMin,
@@ -86,7 +87,19 @@ class FiltersSummary extends React.Component {
       .sort((a, b) => a.value < b.value ? -1 : a.value > b.value ? 1 : 0)
       .map((item) => {
         let label = `${item.value}`;
-        if (filterName!==strainSymbol) label+= ` (${this.props.totalStateCounts[filterName].get(item.value)})`;
+        if (filterName!==strainSymbol) {
+          /* Add the _total_ occurrences in parentheses. We don't compute the intersections / visible values here -
+          e.g. we can have "England (5)" "North America (100)" even though such an intersection will deselect everything.
+          If we have two trees shown we show both values.
+          */
+          const tree1count = this.props.totalStateCounts[filterName]?.get(item.value) ?? 0;
+          if (this.props.totalStateCountsSecondTree) {
+            const tree2count = this.props.totalStateCountsSecondTree[filterName]?.get(item.value) ?? 0;
+            label+=` (L: ${tree1count}, R: ${tree2count})`;
+          } else {
+            label+=` (${tree1count})`;
+          }
+        }
         return this.createIndividualBadge({filterName, item, label, onHoverMessage});
       });
   }