append 如果 pandas 中没有重复，则从一个 df 到另一个的行值

Question

我有这两个 df

df1 = pd.DataFrame({'pupil': ["sarah", "john", "fred"],
                  'class': ["1a", "1a", "1a"]})


df2 = pd.DataFrame({'pupil_mixed': ["sarah", "john", "lex"],
                  'class': ["1a", "1c", "1a"]})

如果值不重复，我想将 append 从 df2 的“pupil_mixed”列到 df1 中的“pupil”列的行值

期望的结果：

df1 = pd.DataFrame({'pupil': ["sarah", "john", "fred", 'lex'],
                  'class': ["1a", "1a", "1a", NaN]})

我用append和loc

df1 = df1.append(df2.loc[df2['pupil_mixed'] != df1['pupil'] ])

它只是将另一列附加到具有匹配行值的 df，并将不匹配的行值更改为 NaN

    pupil   class   pupil_mixed
0   sarah   1a      NaN
1   john    1a      NaN
2   fred    1a      NaN
2   NaN     1a      lex

Answer 1

您可以使用concat + drop_duplicates ：

res = pd.concat((df1, df2['pupil_mixed'].to_frame('pupil'))).drop_duplicates('pupil')

print(res)

Output

   pupil class
0  sarah    1a
1   john    1a
2   fred    1a
2    lex   NaN

作为替代方案，您可以先过滤（使用isin ）然后连接：

# filter the rows in df2, rename the column pupil_mixed
filtered = df2.loc[~df2['pupil_mixed'].isin(df1['pupil'])]

# create a new single column DataFrame with the pupil column
res = pd.concat((df1, filtered['pupil_mixed'].to_frame('pupil')))

print(res)

两种解决方案都使用to_frame和 name 参数，有效地更改列名。

Answer 2

# distinct df1 & df2
df1['tag'] = 1
df2['tag'] = 2

# change the column name the same
df2.columns = df1.columns
df1 = df1.append(df2)
# drop_duplicates by keep df1
df1 = df1.drop_duplicates('pupil', keep='first')

# set tag == 2, class is null
cond = df1['tag'] == 2
df1.loc[cond, 'class'] = np.nan
del df1['tag']

print(df1)

output：

print(df1)

   pupil class
0  sarah    1a
1   john    1a
2   fred    1a
3    lex   NaN

Answer 3

在 df2 中重命名pupil_mixed后，您可以使用合并：

df1.merge(df2["pupil_mixed"].rename("pupil"), how="outer")

   pupil    class
0   sarah   1a
1   john    1a
2   fred    1a
3   lex    NaN