你没有提供数据框的方式，我可以很容易地创建它，并与它弄乱。所以这只是一个猜测。
我会执行groupby + filter操作

result = (
  df.groupby("oldVoucherId")
    .filter(lambda g: g["valueRemaining"].eq(0).any())
)

可以比较0字符串：

ids = df.loc[df['valueRemaining'].eq('0'), 'oldVoucherId']

另一种方法是转换为数字并比较0数字：

ids = df.loc[pd.to_numeric(df['valueRemaining'], errors='coerce').eq(0), 'oldVoucherId']
print (ids)
5     2
8     5
10    3
Name: oldVoucherId, dtype: object

然后，如果ids中不存在oldVoucherId，则通过boolean indexing使用Series.isin过滤所有oldVoucherId，并通过~反转掩码：

out = df[~df['oldVoucherId'].isin(ids)]
print (out)
   oldVoucherId valuePurchased valueRemaining
0             1             60            NaN
1             1            NaN             50
2             1            NaN             40
11            8             50            NaN
12            9             70            NaN

对于预期的输出，删除重复项并保留最后一个值：

out = out.drop_duplicates('oldVoucherId', keep='last')
print (out)
   oldVoucherId valuePurchased valueRemaining
2             1            NaN             40
11            8             50            NaN
12            9             70            NaN

将valueRemaining中的缺失值替换为valuePurchased，并将NaN s设置为valuePurchased：

out['valueRemaining'] = out['valueRemaining'].fillna(out['valuePurchased'])
out['valuePurchased'] = np.nan
print (out)
   oldVoucherId  valuePurchased valueRemaining
2             1             NaN             40
11            8             NaN             50
12            9             NaN             70