如何基于另一个数据框列为 True 创建函数？

Question

我有一个如下所示的数据框：

     Name   X    Y
0    A      False True
1    B      True  True
2    C      True  False

我想创建一个函数，例如：

example_function("A") = "A is in Y"
example_function("B") = "B is in X and Y"
example_function("C") = "C is in X"

这是我目前的代码（不正确，看起来效率不高）：

def example_function(name):
    for name in df['Name']:
        if df['X'][name] == True and df['Y'][name] == False:
            print(str(name) + "is in X")
        elif df['X'][name] == False and df['Y'][name] == True:
            print(str(name) + "is in Y")
        else:
            print(str(name) + "is in X and Y")

我最终想添加更多布尔列，因此它需要可扩展。我怎样才能做到这一点？创建字典而不是数据框会更好吗？

谢谢！

Answer 1

如果你真的想要一个功能，你可以这样做：

def example_function(label):
    s = df.set_index('Name').loc[label]
    l = s[s].index.to_list()
    return f'{label} is in {" and ".join(l)}'

example_function('A')
'A is in Y'

example_function('B')
'B is in X and Y'

您还可以将所有解计算为字典：

s = (df.set_index('Name').replace({False: pd.NA}).stack()
       .reset_index(level=0)['Name']
     )
out = s.index.groupby(s)

输出：

{'A': ['Y'], 'B': ['X', 'Y'], 'C': ['X']}

Answer 2

我认为你可以继续使用 DataFrame，使用这样的函数可以获得相同的输出：

def func (name, df):
    # some checks to verify that the name is actually in the df
    occurrences_name = np.sum(df['Name'] == name)
    if occurrences_name == 0: 
        raise ValueError('Name not found')
    elif occurrences_name > 1:
        raise ValueError('More than one name found')

    # get the index corresponding to the name you're looking for
    # and select the corresponding row
    index = df[df['Name'] == name].index[0]
    row = df.drop(['Name'], axis=1).iloc[index]
    outstring = '{} is in '.format(name)
    for i in range(len(row)):
        if row[i] == True:
            if i != 0: outstring += ', '
            outstring += '{}'.format(row.index[i])
    return outstring

当然你可以根据你的 df 的特定形状调整它，我假设包含名称的列实际上是 'Name'。

如何基于另一个数据框列为 True 创建函数？

How to create a function based on another dataframe column being True?

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