BUG: Modify rank calculation when for dense rank pandas-dev#20731

peterpanmj · May 22, 2018 · adfd619 · adfd619
1 parent 172ab7a
commit adfd619
Showing 1 changed file with 10 additions and 2 deletions.
diff --git a/pandas/_libs/groupby_helper.pxi.in b/pandas/_libs/groupby_helper.pxi.in
@@ -418,7 +418,7 @@ def group_rank_{{name}}(ndarray[float64_t, ndim=2] out,
                         bint is_datetimelike, object ties_method,
                         bint ascending, bint pct, object na_option):
     """
-	Provides the rank of values within each group.
+    Provides the rank of values within each group.
 
     Parameters
     ----------
@@ -453,6 +453,7 @@ def group_rank_{{name}}(ndarray[float64_t, ndim=2] out,
         TiebreakEnumType tiebreak
         Py_ssize_t i, j, N, K, val_start=0, grp_start=0, dups=0, sum_ranks=0
         Py_ssize_t grp_vals_seen=1, grp_na_count=0
+        Py_ssize_t total_tie_count=0
         ndarray[int64_t] _as
         ndarray[float64_t, ndim=2] grp_sizes
         ndarray[{{c_type}}] masked_vals
@@ -499,6 +500,9 @@ def group_rank_{{name}}(ndarray[float64_t, ndim=2] out,
 
     if not ascending:
         _as = _as[::-1]
+
+    if mask.any() and keep_na:
+        total_tie_count -= 1
 
     with nogil:
         # Loop over the length of the value array
@@ -560,6 +564,7 @@ def group_rank_{{name}}(ndarray[float64_t, ndim=2] out,
             if (i == N - 1 or
                     (masked_vals[_as[i]] != masked_vals[_as[i+1]]) or
                     (mask[_as[i]] ^ mask[_as[i+1]])):
+                total_tie_count += 1
                 dups = sum_ranks = 0
                 val_start = i
                 grp_vals_seen += 1
@@ -581,7 +586,10 @@ def group_rank_{{name}}(ndarray[float64_t, ndim=2] out,
 
         if pct:
             for i in range(N):
-                out[i, 0] = out[i, 0] / grp_sizes[i, 0]
+                if tiebreak == TIEBREAK_DENSE:
+                    out[i, 0] = out[i, 0] / total_tie_count
+                else:
+                    out[i, 0] = out[i, 0] / grp_sizes[i, 0]
 {{endif}}
 {{endfor}}