Implement arithmetic.test_numeric and arithmetic.test_object (#22254)

Author: jbrockmendel

pandas/tests/arithmetic/conftest.py

@@ -4,13 +4,49 @@
 import numpy as np
 import pandas as pd
 
+from pandas.compat import long
+
 
 @pytest.fixture(params=[1, np.array(1, dtype=np.int64)])
 def one(request):
     # zero-dim integer array behaves like an integer
     return request.param
 
 
+zeros = [box_cls([0] * 5, dtype=dtype)
+         for box_cls in [pd.Index, np.array]
+         for dtype in [np.int64, np.uint64, np.float64]]
+zeros.extend([np.array(0, dtype=dtype)
+              for dtype in [np.int64, np.uint64, np.float64]])
+zeros.extend([0, 0.0, long(0)])
+
+
+@pytest.fixture(params=zeros)
+def zero(request):
+    # For testing division by (or of) zero for Index with length 5, this
+    # gives several scalar-zeros and length-5 vector-zeros
+    return request.param
+
+
+@pytest.fixture(params=[pd.Float64Index(np.arange(5, dtype='float64')),
+                        pd.Int64Index(np.arange(5, dtype='int64')),
+                        pd.UInt64Index(np.arange(5, dtype='uint64'))],
+                ids=lambda x: type(x).__name__)
+def idx(request):
+    return request.param
+
+
+@pytest.fixture(params=[pd.Timedelta('5m4s').to_pytimedelta(),
+                        pd.Timedelta('5m4s'),
+                        pd.Timedelta('5m4s').to_timedelta64()],
+                ids=lambda x: type(x).__name__)
+def scalar_td(request):
+    """
+    Several variants of Timedelta scalars representing 5 minutes and 4 seconds
+    """
+    return request.param
+
+
 # ------------------------------------------------------------------
 
 @pytest.fixture(params=[pd.Index, pd.Series, pd.DataFrame],