如果 pandas 中没有重复项，则将行值从一个 df 附加到另一个

Question

我有这两个 dfs

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


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

我想从 df2 中附加“pupil_mixed”列中的行值 如果值不重复，则到 df1 中的“瞳孔”列

期望的结果：

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

我使用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

你可以使用concat + drop_duplicates:

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

print(res)

输出

   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，带有name参数，有效地改变了列的名称。

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)

输出：

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