Some optimizations for union formation (#521)

docs/changelog.md

@@ -1,5 +1,9 @@
 # Changelog
 
+## Unreleased
+
+- Optimize local variables with very complex inferred types (#521)
+
 ## Version 0.7.0 (April 13, 2022)
 
 Release highlights:

pyanalyze/stacked_scopes.py

@@ -1331,6 +1331,8 @@ def _get_value_from_nodes(
         ctx: _LookupContext,
         constraints: Iterable[Constraint] = (),
     ) -> Value:
+        # Deduplicate nodes to gain some performance.
+        nodes = OrderedDict.fromkeys(nodes)
         # If the variable is a nonlocal or composite, "uninitialized" doesn't make sense;
         # instead use an empty constraint to point to the parent scope.
         should_use_unconstrained = (

pyanalyze/value.py

@@ -2171,7 +2171,6 @@ def unite_values(*values: Value) -> Value:
     # sets have unpredictable iteration order.
     hashable_vals = OrderedDict()
     unhashable_vals = []
-    saw_unreachable = False
     for value in values:
         if isinstance(value, MultiValuedValue):
             subvals = value.vals
@@ -2184,21 +2183,22 @@ def unite_values(*values: Value) -> Value:
         else:
             subvals = [value]
         for subval in subvals:
-            if _is_unreachable(subval):
-                saw_unreachable = True
-                continue
             try:
                 # Don't readd it to preserve original ordering.
                 if subval not in hashable_vals:
                     hashable_vals[subval] = None
             except Exception:
                 unhashable_vals.append(subval)
     existing = list(hashable_vals) + unhashable_vals
-    num = len(existing)
+    reachabilities = [_is_unreachable(val) for val in existing]
+    num_unreachable = sum(reachabilities)
+    num = len(existing) - num_unreachable
     if num == 0:
-        if saw_unreachable:
+        if num_unreachable:
             return AnyValue(AnySource.unreachable)
         return NO_RETURN_VALUE
+    if num_unreachable:
+        existing = [val for i, val in enumerate(existing) if not reachabilities[i]]
     if num == 1:
         return existing[0]
     else: