Python 元组列表：如果第一个元组元素匹配，则合并两个数据帧列

你可以编写一个自定义的汇总函数来遍历一个组中的所有子帧并返回一个总和.

我并不完全喜欢这个解决方案，因为在将C转换为浮点型时，如果需要，您可以进一步使用它:

from itertools import groupby
import pandas as pd

ids = ['a','b','c','a','d']
dfs = [
    pd.DataFrame({"A": [0], "B": [0], "C": [0]}),
    pd.DataFrame({"A": [1], "B": [1], "C": [1]}),
    pd.DataFrame({"A": [2], "B": [2], "C": [2]}),
    pd.DataFrame({"A": [3], "B": [3], "C": [3]}),
    pd.DataFrame({"A": [4], "B": [4], "C": [4]})
]
df_groups = list(zip(ids, dfs))
df_groups = sorted(df_groups, key=lambda x: x[0])

def summarise_cols(group, cols_to_summarise=["A", "B"]):
    _, df = next(group)
    for _, next_df in group:
        df = df.add(next_df[cols_to_summarise], fill_value=0)
    return df

df_groups_b = ([summarise_cols(group)
                for _, group in groupby(df_groups, lambda x: x[0])])

for d in df_groups_b:
    print(d)

输出:

   A  B    C
0  3  3  0.0
   A  B  C
0  1  1  1
   A  B  C
0  2  2  2
   A  B  C
0  4  4  4

UPD:刚刚注意到你的更新

编辑:为澄清起见，原始数据框中的C列将保持不变.在第一个元组元素匹配的情况下，来自相应数据帧的列C将相同.

然后，您可以执行此操作，列类型将保持不变

def summarise_cols(group, cols_to_keep=["C"]):
    _, df = next(group)
    for _, next_df in group:
        df += next_df
        df[cols_to_keep] = next_df[cols_to_keep]
    return df

UPD2:将group_id和统一数据帧一起作为元组返回.

def summarise_cols(group, cols_to_keep=["C"]):
    group_id, df = next(group)
    for _, next_df in group:
        df += next_df
        df[cols_to_keep] = next_df[cols_to_keep]
    return group_id, df