如何使用字典更改另一列

Question

这里有df

df = pd.DataFrame({'Menu':['Salad', 'Buger', 'Buger', 'Buger'], 
                    'Combo':['Salad A', 'Buger A', 'Buger B', 'Buger C'], 
                    'Status':['Regular', 'Regular', 'Regular', 'Promotion']}) 

    Menu    Combo       Status
0   Salad   Salad A     Regular
1   Buger   Buger A     Regular
2   Buger   Buger B     Regular
3   Buger   Buger C     Promotion

我想使用dic来基于Combo值更改Status。为此：

dic = {'Buger B': 'Promotion', 'Buger C': 'Unavailable'}

    Menu    Combo       Status
0   Salad   Salad A     Regular
1   Buger   Buger A     Regular
2   Buger   Buger B     Promotion
3   Buger   Buger C     Unavailable

Answer 1

for key in dic.keys():
    df.loc[df['Combo'] == key, 'Status'] = dic[key]

输出：

    Menu    Combo       Status
0  Salad  Salad A      Regular
1  Buger  Buger A      Regular
2  Buger  Buger B    Promotion
3  Buger  Buger C  Unavailable

Answer 2

您可以使用^{}和^{}：

df["Status"] = df["Combo"].map(dic).fillna(df["Status"])

输出：

    Menu    Combo   Status
0   Salad   Salad A Regular
1   Buger   Buger A Regular
2   Buger   Buger B Promotion
3   Buger   Buger C Unavailable

Answer 3

我会这样做的。总是建议使用itertuples()，因为在pandas.DataFrame上迭代要快得多

for val in df.itertuples():
    combo_val = getattr(val, 'Combo')
    if combo_val in dic.keys():
        df.loc[val[0], 'Status'] = dic[combo_val]

print(df)

输出：

    Menu    Combo       Status
0  Salad  Salad A      Regular
1  Buger  Buger A      Regular
2  Buger  Buger B    Promotion
3  Buger  Buger C  Unavailable