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nan min/max to handle possible nan ghosts #923

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merged 2 commits into from
Nov 12, 2024
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nan min/max to handle possible nan ghosts #923

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nan min/max to handle possible nan ghosts
nicolasaunai Nov 11, 2024
58570ba
flatten
nicolasaunai Nov 11, 2024
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10 changes: 6 additions & 4 deletions pyphare/pyphare/pharesee/hierarchy/hierarchy.py
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Original file line number Diff line number Diff line change
Expand Up @@ -275,10 +275,11 @@ def global_min(self, qty, **kwargs):
for patch in lvl.patches:
pd = patch.patch_datas[qty]
if first:
m = pd.dataset[:].min()
m = np.nanmin(pd.dataset[:])
first = False
else:
m = min(m, pd.dataset[:].min())
data_and_min = np.concatenate(([m], pd.dataset[:].flatten()))
m = np.nanmin(data_and_min)
Comment on lines +278 to +282
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🛠️ Refactor suggestion

Consider optimizing the min calculation and handling edge cases.

While using np.nanmin correctly handles NaN values, the current implementation has some performance concerns:

  1. Concatenating arrays in a loop is inefficient (O(n²) complexity).
  2. No explicit handling for the case where all values are NaN.

Consider this more efficient implementation:

-                    m = np.nanmin(pd.dataset[:])
-                    first = False
-                else:
-                    data_and_min = np.concatenate(([m], pd.dataset[:].flatten()))
-                    m = np.nanmin(data_and_min)
+                    values = pd.dataset[:]
+                    first = False
+                else:
+                    values = np.append(values, pd.dataset[:])
+        if np.all(np.isnan(values)):
+            raise ValueError("All values are NaN")
+        return np.nanmin(values)

Committable suggestion skipped: line range outside the PR's diff.


return m

Expand All @@ -289,10 +290,11 @@ def global_max(self, qty, **kwargs):
for patch in lvl.patches:
pd = patch.patch_datas[qty]
if first:
m = pd.dataset[:].max()
m = np.nanmax(pd.dataset[:])
first = False
else:
m = max(m, pd.dataset[:].max())
data_and_max = np.concatenate(([m], pd.dataset[:].flatten()))
m = np.nanmax(data_and_max)

return m

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