pandas 透视表中文字段排序方法
前几天有一个需求，透视表中的年级这一列要按照一年级，二年级这样的序列进行排序，但是用过透视表的人都知道，透视表对中文的排序不是太理想，放弃pandas自带的排序方法。测试了很久，想到一个办法。先把dataframe中需要特殊排序的列中的汉字转换成数字，然后生成透视表，生成透视表之后，再把透视表的index或者columns中的数字替换成相应的汉字，透视表的结果就会按照你想要的顺序进行排序。
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)
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