如何用 Pandas 计算逗号分隔列的平均值?

让我们考虑以下 CSV 文件test.csv:


"x","y","A","B"

8000000000,"0,1","0.113948,0.113689",0.114042

8000000000,"0,1","0.114063,0.113823",0.114175

8000000000,"0,1","0.114405,0.114366",0.114524

8000000000,"0,1,2,3","0.167543,0.172369,0.419197,0.427285",0.427576

8000000000,"0,1,2,3","0.167784,0.172145,0.418624,0.426492",0.428736

8000000000,"0,1,2,3","0.168121,0.172729,0.419768,0.427467",0.428578


我的目标是按列"x"和来对行进行分组,并计算列和"y"的算术平均值。"A""B"


我的第一个方法是在 Pandas 中使用groupby()和 的组合:mean()


import pandas


if __name__ == "__main__":

    data = pandas.read_csv("test.csv", header=0)

    data = data.groupby(["x", "y"], as_index=False).mean()

    print(data)

运行此脚本会产生以下输出:


            x        y         B

0  8000000000      0,1  0.114247

1  8000000000  0,1,2,3  0.428297

正如我们所看到的,实现单值列的目标"B"很简单。然而,该列"A"被省略。相反,我希望该列带有"A"一个字符串,其中包含每个逗号分隔值的算术平均值。所需的输出应如下所示:


            x        y                                    A         B

0  8000000000      0,1                    0.114139,0.113959  0.114247

1  8000000000  0,1,2,3  0.167816,0.172414,0.419196,0.427081  0.428297

有人知道怎么做这个吗?


回首忆惘然
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1回答

哆啦的时光机

您可以创建一个自定义聚合函数,将这些字符串解析为列表,查找每列的平均值,并将它们格式化回字符串:def string_mean(rows):    data_list = []    for row in rows:        data_list.append([float(item) for item in row.split(",")])    data = np.array(data_list)    return ",".join([f"{item:.6f}" for item in data.mean(axis=0)])    df.groupby(["x", "y"], as_index=False).agg({"A": string_mean, "B": "mean"})返回            x        y                                    A         B0  8000000000      0,1                    0.114139,0.113959  0.1142471  8000000000  0,1,2,3  0.167816,0.172414,0.419196,0.427081  0.428297请注意,如果 A 中的字符串在单个组中具有不同数量的列,则会出错。顺便说一句,你可能可以大大清理我上面的函数
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