如何为Pandas DataFrame的指定% Paid行添加后缀？

我想给Pandas DataFrame中的**% Paid**行所有值添加后缀（比如%），但目前只会给列名加后缀。有没有办法直接给特定行的所有值批量添加后缀？

示例代码 #

import pandas as pd

d={
    ("Payments","Jan","NOS"):[],
    ("Payments","Feb","NOS"):[],
    ("Payments","Mar","NOS"):[],
}
d = pd.DataFrame(d)

d.loc["Total",("Payments","Jan","NOS")] = 9991
d.loc["Total",("Payments","Feb","NOS")] = 3638
d.loc["Total",("Payments","Mar","NOS")] = 5433

d.loc["Paid",("Payments","Jan","NOS")] = 139
d.loc["Paid",("Payments","Feb","NOS")] = 123
d.loc["Paid",("Payments","Mar","NOS")] = 20

d.loc["% Paid",("Payments","Jan","NOS")] = round((d.loc["Paid",("Payments","Jan","NOS")] / d.loc["Total",("Payments","Jan","NOS")])*100)
d.loc["% Paid",("Payments","Feb","NOS")] = round((d.loc["Paid",("Payments","Feb","NOS")] / d.loc["Total",("Payments","Feb","NOS")])*100)
d.loc["% Paid",("Payments","Mar","NOS")] = round((d.loc["Paid",("Payments","Mar","NOS")] / d.loc["Total",("Payments","Mar","NOS")])*100)

我目前只能逐个单元格转换后加后缀，虽然可行但不够高效：

d.loc["% Paid",("Payments","Jan","NOS")] = str(round((d.loc["Paid",("Payments","Jan","NOS")] / d.loc["Total",("Payments","Jan","NOS")])*100)) + '%'
d.loc["% Paid",("Payments","Feb","NOS")] = str(round((d.loc["Paid",("Payments","Feb","NOS")] / d.loc["Total",("Payments","Feb","NOS")])*100)) + '%'
d.loc["% Paid",("Payments","Mar","NOS")] = str(round((d.loc["Paid",("Payments","Mar","NOS")] / d.loc["Total",("Payments","Mar","NOS")])*100)) + '%'

有两种常用方法可以批量给特定行添加后缀，根据需求选择：

方法1：直接修改数据类型为字符串并添加后缀 #

如果不需要后续对**% Paid**行的数值进行计算，可以直接将整行转为字符串后拼接后缀：

# 先批量计算% Paid行的数值
d.loc["% Paid"] = round((d.loc["Paid"] / d.loc["Total"]) * 100)
# 批量添加%后缀
d.loc["% Paid"] = d.loc["% Paid"].astype(str) + '%'

一行代码即可完成整行的后缀添加，无需逐个单元格处理。

方法2：用Styler格式化显示（不修改原始数据） #

如果需要保留**% Paid**行的数值类型以便后续计算，推荐用Pandas的Styler来格式化显示——原始数据仍为数值，仅展示时带后缀：

# 先批量计算% Paid行的数值
d.loc["% Paid"] = round((d.loc["Paid"] / d.loc["Total"]) * 100)

# 定义格式化函数，仅对% Paid行添加后缀
def format_percent(row):
    if row.name == "% Paid":
        return [f"{val}%" for val in row]
    return row

# 应用格式化并显示
styled_df = d.style.apply(format_percent, axis=1)
display(styled_df)

这种方式兼顾显示需求与数据可用性，灵活性更强。

内容的提问来源于stack exchange，提问作者Harish Osthe