compare pandas NA ambiguous objects correct

[KRRT7]

User description

when trying to compare pandas NA values, which have ambiguous boolean behavior, codeflash crashed, fix

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PR Type

Bug fix, Tests

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Description

• Fix NA comparison with pandas.isna check

• Prevent crashes on comparing pd.NA values

• Add extensive tests for pd.NA edge cases

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Changes walkthrough 📝

	Relevant files
Bug fix	comparator.py NA handling using pandas.isna check                                            codeflash/verification/comparator.py Added condition to check both values with pandas.isna Returns True when both NA values are detected +2/-0
Tests	test_comparator.py Add tests for pd.NA and None comparisons                                  tests/test_comparator.py Introduced tests for comparing pd.NA with pd.NA and None Validated NA behavior in Series, DataFrames, and dicts Extended edge case coverage for NA sparse arrays +30/-0

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[github-actions]

PR Reviewer Guide 🔍

Here are some key observations to aid the review process:

⏱️ Estimated effort to review: 2 🔵🔵⚪⚪⚪

🧪 PR contains tests

🔒 No security concerns identified

⚡ Recommended focus areas for review Condition Placement The NA check using pandas.isna is a good addition. However, ensure that its position in the branching logic does not inadvertently override other type-specific comparisons. It may be beneficial to explicitly document the intended priority of this condition. if HAS_PANDAS and pandas.isna(orig) and pandas.isna(new): return True Test Robustness Extensive tests for pd.NA edge cases are added. Consider including tests where pd.NA is mixed with other representations of missing data (e.g. numpy.nan) to fully validate comparator behavior. assert comparator(pd.NA, pd.NA) assert not comparator(pd.NA, None) assert not comparator(None, pd.NA) s1 = pd.Series([1, 2, pd.NA, 4]) s2 = pd.Series([1, 2, pd.NA, 4]) s3 = pd.Series([1, 2, None, 4]) assert comparator(s1, s2) assert not comparator(s1, s3) df1 = pd.DataFrame({'a': [1, 2, pd.NA], 'b': [4, pd.NA, 6]}) df2 = pd.DataFrame({'a': [1, 2, pd.NA], 'b': [4, pd.NA, 6]}) df3 = pd.DataFrame({'a': [1, 2, None], 'b': [4, None, 6]}) assert comparator(df1, df2) assert not comparator(df1, df3) d1 = {'a': pd.NA, 'b': [1, pd.NA, 3]} d2 = {'a': pd.NA, 'b': [1, pd.NA, 3]} d3 = {'a': None, 'b': [1, None, 3]} assert comparator(d1, d2) assert not comparator(d1, d3) s1 = pd.Series([1, 2, pd.NA, 4]) s2 = pd.Series([1, 2, pd.NA, 4]) filtered1 = s1[s1 > 1] filtered2 = s2[s2 > 1] assert comparator(filtered1, filtered2)

[github-actions]

PR Code Suggestions ✨

Explore these optional code suggestions:

Category	Suggestion	Impact
Possible issue	Ensure matching types Add a type equality check to ensure that both values are of the same missing-value type to prevent pd.NA and None from being treated as equivalent. codeflash/verification/comparator.py [158-159] -if HAS_PANDAS and pandas.isna(orig) and pandas.isna(new): +if HAS_PANDAS and type(orig) is type(new) and pandas.isna(orig) and pandas.isna(new): return True Suggestion importance[1-10]: 8 __ Why: The suggestion addresses a potential bug by ensuring that only missing values of the same type are considered equal, which prevents pd.NA and None from being erroneously evaluated as equivalent. This added type check increases correctness while the score is capped due to it being a type checking improvement.	Medium