python - 索引不匹配时更新 Pandas 数据框的最有效方法

Question

我有两个 pandas DataFrame，我想用另一个更新一个...但我不能确定索引是否匹配。 (所以使用 DataFrame.update 是个问题!)

例子:

import pandas as pd
df1 = pd.DataFrame([('path1', 0, 0, 0),
                    ('path2', 0, 0, 0),
                    ('path3', 0, 0, 0),
                    ('path4', 0, 0, 0),],
                  columns=['path', 'class', 'manual', 'conf'],
                  index = [1,2,3,4])

df2 = pd.DataFrame([('path1', 1, 0, 0),
                    ('path2', 0, 1, 0),
                    ('path3', 0, 0, 1),
                    ('path5', 1, 1, 0),
                    ('path6', 1, 1, 0),],
                  columns=['path', 'class', 'manual', 'conf'],
                  index = [10,11,12,13,14])

期望的结果:

update_annotations(df1, df2)

    path  class  manual  conf
1  path1      1       0     0
2  path2      0       1     0
3  path3      0       0     1
4  path4      0       0     0

df1.update(df2) 可能存在风险，因为这些数据帧的索引可能不匹配。这样做最安全和最有效的方法是什么？

Answer 1

快速而肮脏

df1[['path']].merge(df2, 'left')

    path  class  manual  conf
0  path1    1.0     0.0   0.0
1  path2    0.0     1.0   0.0
2  path3    0.0     0.0   1.0
3  path4    NaN     NaN   NaN

---

较慢且较脏

df1[['path']].merge(df2, 'left').fillna(0).astype(df1.dtypes)

    path  class  manual  conf
0  path1      1       0     0
1  path2      0       1     0
2  path3      0       0     1
3  path4      0       0     0

---

迂腐

用df1填充NaN

df1[['path']].merge(df2, 'left').fillna({**df1}).astype(df1.dtypes)

    path  class  manual  conf
0  path1      1       0     0
1  path2      0       1     0
2  path3      0       0     1
3  path4      0       0     0

根据 Chris

df1.set_index('path').assign(**df2.set_index('path')).reset_index()

    path  class  manual  conf
0  path1    1.0     0.0   0.0
1  path2    0.0     1.0   0.0
2  path3    0.0     0.0   1.0
3  path4    NaN     NaN   NaN

---

保留索引

既然顺序保证是一样的，我们可以只使用set_index

df1[['path']].merge(df2, 'left').fillna({**df1}).astype(df1.dtypes).set_index(df1.index)

    path  class  manual  conf
1  path1      1       0     0
2  path2      0       1     0
3  path3      0       0     1
4  path4      0       0     0