name = 'User_id' # 折扣率排序 datas[name_prifix + 'discount_rateRank'] = datas.groupby(name)['discount_rate'].rank(ascending = False) datas[name_prifix + 'discount_rateRankAscend'] = datas.groupby(name)['discount_rate'].rank(ascending = True) # 距离排序 datas[name_prifix + 'distanceRank'] = datas.groupby(name)['Distance'].rank(ascending = False) datas[name_prifix + 'distanceRankAscend'] = datas.groupby(name)['Distance'].rank(ascending = True) # 领券日期排序 datas[name_prifix + 'date_receivedRank'] =datas.groupby(name)['Date_received'].rank(ascending = False) datas[name_prifix + 'date_receivedRankAscend'] = datas.groupby(name)['Date_received'].rank(ascending = True) return datas请帮我用groupby函数改写

可以改写如下：

def add_rank_features(df, id_col='User_id', rank_cols=['discount_rate', 'Distance', 'Date_received']):
    for col in rank_cols:
        df[f'{col}_rank_desc'] = df.groupby(id_col)[col].rank(method='dense', ascending=False)
        df[f'{col}_rank_asc'] = df.groupby(id_col)[col].rank(method='dense', ascending=True)
    return df

其中，`id_col` 为分组列名，`rank_cols` 为需要排序的列名列表。`method='dense'` 表示按照值的密度排序，相同值的排名相同，不会有空缺的排名。

相关问题

name = 'Merchant_id' # 折扣率排序 datas[name_prifix + 'discount_rateRank'] =datas.groupby(name)['discount_rate'].rank(ascending = False) datas[name_prifix + 'discount_rateRankAscend'] = datas.groupby(name)['discount_rate'].rank(ascending = True) # 距离排序 datas[name_prifix + 'distanceRank'] = datas.groupby(name)['Distance'].rank(ascending = False) datas[name_prifix + 'distanceRankAscend']=datas.groupby(name)['Distance'].rank(ascending = True) # 领券日期排序 datas[name_prifix + 'date_receivedRank'] =datas.groupby(name)['Date_received'].rank(ascending = False) datas[name_prifix + 'date_receivedRankAscend'] = datas.groupby(name)['Date_received'].rank(ascending = True)请帮我用groupby改写

好的，我可以帮你用groupby重写这段代码。你可以这样写：

cols = ['discount_rate', 'Distance', 'Date_received']
for col in cols:
    datas[name_prifix + col + 'Rank'] = datas.groupby(name)[col].rank(ascending=False)
    datas[name_prifix + col + 'RankAscend'] = datas.groupby(name)[col].rank(ascending=True)

这样，你就能用groupby对每一列进行操作，避免了多次重复写代码的问题。

name = 'Coupon_id' # 折扣率排序 datas[name_prifix + 'discount_rateRank'] = datas.groupby(name)['discount_rate'].rank(ascending = False) datas[name_prifix + 'discount_rateRankAscend'] = datas.groupby(name)['discount_rate'].rank(ascending = True) # 距离排序 datas[name_prifix + 'distanceRank'] = datas.groupby(name)['Distance'].rank(ascending = False) datas[name_prifix + 'distanceRankAscend'] = datas.groupby(name)['Distance'].rank(ascending = True) # 领券日期排序 datas[name_prifix + 'date_receivedRank'] =datas.groupby(name)['Date_received'].rank(ascending = False) datas[name_prifix + 'date_receivedRankAscend'] =datas.groupby(name)['Date_received'].rank(ascending = True) 请帮我用groupby函数改写

Sure! 这段代码可以使用 `groupby` 函数进行改写，如下所示：

cols_to_rank = ['discount_rate', 'Distance', 'Date_received']
grouped = datas.groupby(name)

for col in cols_to_rank:
    datas[name_prifix + col + 'Rank'] = grouped[col].rank(ascending=False)
    datas[name_prifix + col + 'RankAscend'] = grouped[col].rank(ascending=True)

这样，我们可以避免重复代码并且更加简洁明了。