当pandas中的值为null时，不使用pandas的to_列表

Question

df = pd.DataFrame({'a':[None,1, 2], 'b':[None, (1,2), (3,4)]}) 


    a   b
0   NaN None
1   1.0 (1, 2)
2   2.0 (3, 4)

我想将列中的元组设置为每个都有自己的列。然而，我对我的方法有一个问题

df[['b1', 'b2']] = pd.DataFrame(df['b'].tolist(), index=df.index)

ValueError: Columns must be same length as key

我试图填充一个空的元组，但它不需要元组。我怎样才能做到这一点

Answer 1

如果元组具有不同数量的元素，则更通用的解决方案是创建如下所示的自定义函数

def create_columns_from_tuple(df, tuple_col):
    
    # get max length of tuples
    max_len = df[tuple_col].apply(lambda x: 0 if x is None else len(x)).max()
    
    # select rows with non-empty tuples
    df_full = df.loc[df[tuple_col].notna()]
    
    # create dataframe with exploded tuples
    df_full_exploded = pd.DataFrame(df_full[tuple_col].tolist(),
                                    index=df_full.index, 
                                    columns=[tuple_col + str(n) for n in range(1, max_len+1)])
    
    # merge the two dataframes by index
    result = df.merge(df_full_exploded, left_index=True, right_index=True, how='left')
    
    return result

在此函数中，传递数据帧和元组列的名称。该函数将自动创建与元组最大长度相同的列

create_columns_from_tuple(df, tuple_col='b')
#      a       b   b1   b2
# 0  NaN    None  NaN  NaN
# 1  1.0  (1, 2)  1.0  2.0
# 2  2.0  (3, 4)  3.0  4.0

---

如果元组的元素数不同：

df = pd.DataFrame({'a':[None,1, 2], 'b':[None, (1,2,42), (3,4)]}) 
create_columns_from_tuple(df, tuple_col='b')
#      a           b   b1   b2    b3
# 0  NaN        None  NaN  NaN   NaN
# 1  1.0  (1, 2, 42)  1.0  2.0  42.0
# 2  2.0      (3, 4)  3.0  4.0   NaN