Question

我有一个pandas数据框，我更喜欢使用lambda函数而不是循环来解决我的问题。

问题是这样的;

df = pd.DataFrame({'my_fruits':['fruit', 'fruit', 'fruit', 'fruit', 'fruit'],
         'fruit_a': ['apple', 'banana', 'vegetable', 'vegetable', 'cherry'],
         'fruit_b': ['vegetable', 'apple', 'vegeatble', 'pineapple', 'pear']})

如果我应用以下循环;

for i in np.arange(0,len(df)):
    if df['fruit_a'][i] == 'vegetable' or df['fruit_b'][i] == 'vegetable':
        df['my_fruits'][i] = 'not_fruit'

我能够得到我想要的结果。这是因为如果fruit_a或fruit_b列包含值vegetable，我希望my_fruits列等于not_fruit。

我怎样才能在lamda函数中设置它。无法理解两列输入如何用于更改不同的列值。谢谢！

Answer 1

您可以.live使用Series.mask：

boolean mask

mask = (df['fruit_a'] == 'vegetable') | (df['fruit_b'] == 'vegetable') print (mask) 0 True 1 False 2 True 3 True 4 False dtype: bool df.my_fruits = df.my_fruits.mask(mask, 'not_fruits') print (df) fruit_a fruit_b my_fruits 0 apple vegetable not_fruits 1 banana apple fruit 2 vegetable vegetable not_fruits 3 vegetable pineapple not_fruits 4 cherry pear fruit的另一个解决方案是按mask比较所有选定的列，然后any至少在一列中获取所有vegetable：

True

Answer 2

使用pd.Series.where并在一步中检查'vegetable'是否与any相结合 where与mask相反，这就是我使用cond的否定的原因否则，这与jezrael的回答非常相似

cond = df[['fruit_a', 'fruit_b']].eq('vegetable').any(1)
df.my_fruits = df.my_fruits.where(~cond, 'not_fruit')

从我的手机回答。请原谅错别字。

Answer 3

您可以使用apply方法执行此操作：

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或者你可以这样做：

>>> df.my_fruits = df.apply(lambda x: 'not_fruit' if x['fruit_a'] == 'vegetable' or x['fruit_b'] == 'vegetable' else x['my_fruits'], axis=1)
0    not_fruit
1        fruit
2    not_fruit
3    not_fruit
4        fruit

Python - 多列上的Lambda函数

3 个答案: