Pandas dataframe groupby聚合后将多级索引列合并为带分隔符的字符串

实现方案 #

你可以通过遍历列多级索引的第一级（即cat1/cat2/cat3），分别取出对应分组的均值、标准差列，四舍五入后拼接即可得到目标格式，完整实现代码如下：

import pandas as pd

# 你原有生成原始DataFrame的代码
data = {
    "cat1": {
        (1, "class1", "metric1"): 0.9520103335380554,
        (1, "class1", "metric2"): 0.9596380591392517,
        (1, "class2", "metric1"): 0.9013115167617798,
        (1, "class2", "metric2"): 0.9917504191398621,
        (1, "class3", "metric1"): 0.9027230143547058,
        (1, "class3", "metric2"): 0.8536863327026367,
        (2, "class1", "metric1"): 0.8746241331100464,
        (2, "class1", "metric2"): 0.8844705820083618,
        (2, "class2", "metric1"): 0.7890198826789856,
        (2, "class2", "metric2"): 0.6964980363845825,
        (2, "class3", "metric1"): 0.9410034418106079,
        (2, "class3", "metric2"): 0.9601017236709595,
        (3, "class1", "metric1"): 0.9640659689903259,
        (3, "class1", "metric2"): 0.9766426682472229,
        (3, "class2", "metric1"): 0.893884003162384,
        (3, "class2", "metric2"): 0.9959416389465332,
        (3, "class3", "metric1"): 0.9533607363700867,
        (3, "class3", "metric2"): 0.9378591179847717,
    },
    "cat2": {
        (1, "class1", "metric1"): 0.9520103335380554,
        (1, "class1", "metric2"): 0.9596380591392517,
        (1, "class2", "metric1"): 0.9013115167617798,
        (1, "class2", "metric2"): 0.9917504191398621,
        (1, "class3", "metric1"): 0.9027230143547058,
        (1, "class3", "metric2"): 0.8536863327026367,
        (2, "class1", "metric1"): 0.8746241331100464,
        (2, "class1", "metric2"): 0.8844705820083618,
        (2, "class2", "metric1"): 0.7890198826789856,
        (2, "class2", "metric2"): 0.6964980363845825,
        (2, "class3", "metric1"): 0.9410034418106079,
        (2, "class3", "metric2"): 0.9601017236709595,
        (3, "class1", "metric1"): 0.9640659689903259,
        (3, "class1", "metric2"): 0.9766426682472229,
        (3, "class2", "metric1"): 0.893884003162384,
        (3, "class2", "metric2"): 0.9959416389465332,
        (3, "class3", "metric1"): 0.9533607363700867,
        (3, "class3", "metric2"): 0.9378591179847717,
    },
    "cat3": {
        (1, "class1", "metric1"): 0.8746241331100464,
        (1, "class1", "metric2"): 0.8844705820083618,
        (1, "class2", "metric1"): 0.7890198826789856,
        (1, "class2", "metric2"): 0.6964980363845825,
        (1, "class3", "metric1"): 0.9410034418106079,
        (1, "class3", "metric2"): 0.9601017236709595,
        (2, "class1", "metric1"): 0.9309893846511841,
        (2, "class1", "metric2"): 0.884644627571106,
        (2, "class2", "metric1"): 0.861851155757904,
        (2, "class2", "metric2"): 0.9180170893669128,
        (2, "class3", "metric1"): 0.8841384649276733,
        (2, "class3", "metric2"): 0.8577012419700623,
        (3, "class1", "metric1"): 0.8895564675331116,
        (3, "class1", "metric2"): 0.8351058959960938,
        (3, "class2", "metric1"): 0.832390308380127,
        (3, "class2", "metric2"): 0.8969333171844482,
        (3, "class3", "metric1"): 0.7883192300796509,
        (3, "class3", "metric2"): 0.8577012419700623,
    },
}
df = pd.DataFrame(data)
df = df.rename_axis(("experiment", "class", "metric"))
# 得到分组聚合后的结果
agg_df = df.groupby(["class", "metric"]).agg(["mean", "std"])

# 核心处理逻辑：拼接均值和标准差
decimal = 3  # 自定义保留的小数位数
sep = "±"    # 自定义拼接符号，可替换为括号、斜杠等
result = pd.DataFrame()
for cat in agg_df.columns.get_level_values(0).unique():
    mean_series = agg_df[(cat, "mean")].round(decimal).astype(str)
    std_series = agg_df[(cat, "std")].round(decimal).astype(str)
    result[cat] = mean_series + sep + std_series

处理后的result就是你需要的展示格式：行索引还是原来的class+metric，列是cat1/cat2/cat3，每个单元格都是均值和标准差拼接的字符串格式。

内容的提问来源于stack exchange，提问作者marco