如何优化多DataFrame合并与函数参数？现有实现求改进方案

我已经成功生成了目标数据表，解决了之前的相关问题，但希望找到更简洁高效的实现方式。具体想请教：

• 能否将函数的多个列参数合并为单个参数传入？
• 是否需要先把CSV中的列转为列表，还是可以直接使用CSV列作为参数？

当前实现函数 #

def CHEGD_funct(countylist, Clist, Hlist, Elist, Glist, Dlist):
    Cdict = pd.DataFrame(list(zip(countylist, Clist)), columns = ['site', 'C'])
    Hdict = pd.DataFrame(list(zip(countylist, Hlist)), columns = ['site', 'H'])
    Edict = pd.DataFrame(list(zip(countylist, Elist)), columns = ['site', 'E'])
    Gdict = pd.DataFrame(list(zip(countylist, Glist)), columns = ['site', 'G'])
    Ddict = pd.DataFrame(list(zip(countylist, Dlist)), columns = ['site', 'D'])
    Cdf = Cdict.groupby('site').sum()
    Hdf = Hdict.groupby('site').sum()
    Edf = Edict.groupby('site').sum()
    Gdf = Gdict.groupby('site').sum()
    Ddf = Ddict.groupby('site').sum()
    dataframes = [Cdf, Hdf, Edf, Gdf, Ddf]
    mergedf = reduce(lambda  left,right: pd.merge(left,right,on=['site'],
                     how='outer'), dataframes).fillna('void')
    return mergedf

期望输出 #

type                                             C    H    E   G   D
site                                                            
Angus                                       20   92   25   0   0
Angus / East Perthshire                      1    8    3   0   0
Argyll                                     141  995   89  68   7
Ayrshire                                    71  336   68  17   9
Banffshire                                   7   86   19   0   0
Banffshire / Moray                           2   10    0   0   0
Banffshire / South Aberdeenshire             1    3    0   0   0
Berwickshire                                17   84   14   0   1
Caithness                                   43  374  202  36   5
Clyde Isles                                 29  142   38   9   5
Dumfriesshire                               34  336   69  21   8
Dunbartonshire                              38   85   19  10   2
East Inverness-shire & Nairn               162  879  318  45  10
East Inverness-shire & Nairn / Banffshire    1    8    4   0   0
East Inverness-shire / Moray                22   50   23   4   0
East Lothian                                22  123   36  16   4
East Perthshire                             31  149   79   7   5
East Ross                                   28  233   65  18   2
East Ross / East Inverness-shire             0    1    0   0   0
East Ross / East Sutherland                  1    3    1   0   0
East Sutherland                             34  208   71   8   4
Fifeshire                                   60  355   43  17   1
Kincardineshire                             30  160   27   9   1
Kintyre                                     15   75   11   3   5
Kirkcudbrightshire                          21  114   33   4   5
Lanarkshire                                 94  293   62  25   7
Lanarkshire / Peebleshire                    3    7    0   2   0
Mid Ebudes                                   9   97   30   1   1
Mid Perthshire                              46  378  176  26   8
Mid Perthshire / East Perthshire             3    9    3   1   0
Midlothain / Peebleshire                     5   23    1   0   0
Midlothian                                  52  194   47  12   2
Midlothian / Berwickshire                    8   20    3   2   1
Moray                                       60  311   77  11   2
Moray / East Inverness-shire & Nairn         5   14   11   0   1
North Aberdeenshire                         38  211   55  10   1
North Ebudes                               109  533  207  94  14
Orkney                                      50  230  126  23   1
Outer Hebrides                              26  265  110  29   0
Peebleshire                                 66  339  114  25  10
Peebleshire / Selkirkshire                   0    3    2   0   0
Renfrewshire                                76  312   40  39   4
Roxburghshire                               76  327   65  21   7
Selkirkshire                                28  168   68   3   3
Shetland                                    57  426  195  13   7
South Aberdeenshire                        159  791  320  64  21
South Aberdeenshire / East Perthshire        2   18   12   3   1
South Aberdeenshire / Kincardineshire        1    0    0   0   0
South Aberdeenshire / North Aberdeenshire    0    1    1   0   0
South Ebudes                                36  172   38   4   4
Stirlingshire                               33  121   26   5   4
West Inverness (Westerness)                  0    0    1   0   0
West Inverness-shire                        41  443  217  32   3
West Lothian                                15   57   11   1   1
West Lothian / Stirlingshire                 0    0    0   0   0
West Perthshire                             18  107   23   0   1
West Ross                                   19  277   84  22   2
West Ross / West Sutherland                  2    7    2   0   0
West Sutherland                             78  610  196  58   4
Wigtownshire                                 9   46   17   2   1 

优化方案 #

1. 直接传入DataFrame作为单个参数（推荐）

不需要拆分列成列表，直接读取CSV后把整个DataFrame传入函数，一步完成分组求和：

import pandas as pd

def CHEGD_funct(df):
    # 按site分组，对C、H、E、G、D列求和，空值填充为'void'
    mergedf = df.groupby('site')[['C', 'H', 'E', 'G', 'D']].sum().fillna('void')
    # 插入type列（若需要保留该列）
    mergedf.insert(0, 'type', '')
    return mergedf

# 使用示例：从CSV读取数据后直接调用
df = pd.read_csv('your_file.csv')
result = CHEGD_funct(df)

2. 若必须传入列表参数，合并为单个字典

如果场景限制只能用列表传入，可将所有列打包成字典作为单个参数：

from functools import reduce
import pandas as pd

def CHEGD_funct(data_dict):
    countylist = data_dict['site']
    cols = ['C', 'H', 'E', 'G', 'D']
    dataframes = []
    for col in cols:
        df = pd.DataFrame(list(zip(countylist, data_dict[col])), columns=['site', col])
        dataframes.append(df.groupby('site').sum())
    
    mergedf = reduce(lambda left, right: pd.merge(left, right, on=['site'], how='outer'), dataframes).fillna('void')
    mergedf.insert(0, 'type', '')
    return mergedf

# 使用示例
data_dict = {
    'site': countylist,
    'C': Clist,
    'H': Hlist,
    'E': Elist,
    'G': Glist,
    'D': Dlist
}
result = CHEGD_funct(data_dict)

关键优化点 #

• 避免重复创建单个列的DataFrame，直接批量处理目标列，代码更简洁、效率更高
• 无需多次merge操作，groupby.sum()会自动保留所有列的聚合结果
• 支持直接传入CSV读取的DataFrame，省去手动拆分列的繁琐步骤

内容的提问来源于stack exchange，提问作者Daisy