Python Dataframe 将索引映射到列表的一列以提取元素

Question

我有一个由列表作为元素组成的数据框。另外，我有一个已知索引列表。现在我想提取每行中那些索引中存在的元素。我的代码：

df = pd.DataFrame({'A':[[7,8],[4,5,NaN],[NaN,1,9]],'match_idx':[1,0,NaN]})
df

      A             match_idx
0   [7, 8]          1
1   [4, 5, nan]     0
2   [nan, 1, 9]     NaN

# in each row, let's find the values located in the match_idx position

当前解决方案：

df['A_element'] = df.apply(lambda x: x['A'][x['match_idx']] if ~x['match_idx'].isnan() else np.nan,axis=1)

AttributeError: 'float' object has no attribute 'isnan'

预期解决方案：

df = 
      A             match_idx   A_element
0   [7, 8]          1           8
1   [4, 5, nan]     0           4
2   [nan, 1, 9]     NaN         NaN

Answer 1

对于 tet 非缺失值，使用 notna 并将索引转换为整数：

df['A_element'] = [a[int(i)] if pd.notna(i) else np.nan
                               for a, i in zip(df['A'], df['match_idx'])]

或者：

df['A_element'] = df.apply(lambda x: x['A'][int(x['match_idx'])] 
                           if pd.notna(x['match_idx']) else np.nan,axis=1)

print (df)
             A  match_idx  A_element
0       [7, 8]        1.0        8.0
1  [4, 5, nan]        0.0        4.0
2  [nan, 1, 9]        NaN        NaN

Python Dataframe 将索引映射到列表的一列以提取元素

Python Dataframe map indexes to a column of lists for extracting elements

python

lambda

list

dataframe

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