pandas透视表中文排序解决方案：映射法实现自定义顺序

 def get_special_sort_data(self, groupby, columns):
 # 获取需要特殊处理的字段的信息
 special_sort_cols = None
 cols_in_index_or_column = None # 判断特殊排序字段在index还是column中
 if self.datasource.has_special_sort_cols:
  # 获取表需要处理的特殊字段信息
  special_sort_cols = self.datasource.get_sort_columns() # {"grade_name": {}}

  if special_sort_cols:
  i_intersection = list(set(groupby) & set(special_sort_cols.keys()))
  c_intersection = list(set(columns) & set(special_sort_cols.keys()))
  if i_intersection:
   cols_in_index_or_column = ('index', i_intersection)
  elif c_intersection:
   cols_in_index_or_column = ('column', c_intersection)
 return cols_in_index_or_column, special_sort_cols
cols_in_index_or_column, special_sort_cols = self.get_special_sort_data(groupby, columns) # special_sort_cols:{"grade_name": {"一年级": 1, "二年级":2, "三年级": 3 ....}}

if cols_in_index_or_column:
 for col in cols_in_index_or_column[1]:
 df[col] = df[col].replace(special_sort_cols.get(col)) # 替换df

# 获取透视表
if cols_in_index_or_column:
 if cols_in_index_or_column[0] == 'index':
  if len(groupby) == 1:
   col_name = cols_in_index_or_column[1][0]
   sort_info = special_sort_cols.get(col_name)
   r_sort_info = {v:k for k, v in zip(sort_info.keys(), sort_info.values())}
   index_1 = df.index.tolist()
   index_1 = [r_sort_info.get(item) for item in index_1]
   df.index = Index(index_1, name=df.index.name)
  else:
   for item in cols_in_index_or_column[1]:
   ix = df.index.names.index(item)
   index_1 = df.index.levels[ix].tolist()
   sort_info = special_sort_cols.get(item)
   r_sort_info = {v: k for k, v in zip(sort_info.keys(), sort_info.values())}
   index_1 = [r_sort_info.get(item) for item in index_1]
   df.index = df.index.set_levels(index_1, level=ix)
 else:
  for item in cols_in_index_or_column[1]:
  ix = df.columns.names.index(item)
  col_1 = df.columns.levels[ix].tolist()
  sort_info = special_sort_cols.get(item)
  r_sort_info = {v: k for k, v in zip(sort_info.keys(), sort_info.values())}
  col_1 = [r_sort_info.get(item) for item in col_1]
  df.columns = df.columns.set_levels(col_1, level=ix)