Python Pandas 基于hesapKodu1列条件计算两列差值并汇总的问题

你可以用numpy.where做条件判断，按规则计算每行差值后再汇总，代码逻辑清晰易维护：
首先导入依赖（如果尚未导入numpy）：

import numpy as np

完整实现代码：

# 保留你原有拆分列的逻辑
source_df['hesapKodu1_1']=source_df['hesapKodu1'].str[:1]
source_df['hesapKodu1_2']=source_df['hesapKodu1'].str[:2]
source_df['hesapKodu1_3']=source_df['hesapKodu1'].str[:3]

# 定义特殊编码集合
special_code_list = ['123', '125', '130']

# 按规则计算每行差值：特殊编码用BORC-ALACAK，其余用ALACAK-BORC
source_df['diff_val'] = np.where(
    source_df['hesapKodu1_3'].isin(special_code_list),
    source_df['BORÇ'] - source_df['ALACAK'],
    source_df['ALACAK'] - source_df['BORÇ']
)

# 汇总所有差值并保留两位小数
total_result = round(source_df['diff_val'].sum(), 2)

如果你的hesapKodu1列本身就是3位长度的字符串，不需要拆分列也可以直接判断原列，修改条件部分即可：

source_df['diff_val'] = np.where(
    source_df['hesapKodu1'].isin(special_code_list),
    source_df['BORÇ'] - source_df['ALACAK'],
    source_df['ALACAK'] - source_df['BORÇ']
)

如果不想新增临时列存储每行差值，可以分开计算两部分求和再合并：

special_code_list = ['123', '125', '130']
# 计算特殊编码的差值总和
special_sum = (source_df.loc[source_df['hesapKodu1_3'].isin(special_code_list), 'BORÇ'].sum() 
               - source_df.loc[source_df['hesapKodu1_3'].isin(special_code_list), 'ALACAK'].sum())
# 计算其余编码的差值总和
normal_sum = (source_df.loc[~source_df['hesapKodu1_3'].isin(special_code_list), 'ALACAK'].sum() 
              - source_df.loc[~source_df['hesapKodu1_3'].isin(special_code_list), 'BORÇ'].sum())
# 汇总结果
total_result = round(special_sum + normal_sum, 2)

内容的提问来源于stack exchange，提问作者Ali İlhami ÖZTAN