有没有办法合并两个具有两个相似键列但不是相同键列的数据集？

您可以熔化、合并，然后进行透视：

out = (a
  .melt(ignore_index=False, value_name='variable_value')
  .reset_index()
  .merge(b, how='left')
  .pivot(index='index', columns='variable')
)

输出：

         variable_value                  value           
variable    have dinner have lunch have dinner have lunch
index                                                    
0                 set 1      set a         1.5        0.2
1                 set 2      set b         1.7        0.5

若要使格式更接近现有格式，请进一步重新处理列名：

out = (a.melt(ignore_index=False, value_name='variable_value')
  .reset_index()
  .merge(b, how='left')
  .pivot(index='index', columns='variable')
  .sort_index(level=1, axis=1)
  .rename_axis(index=None)
)

out.columns = out.columns.map(lambda x: f'{x[1]}{" value" if x[0] == "variable_value" else ""}')

输出：

   have dinner have dinner value  have lunch have lunch value
0          1.5             set 1         0.2            set a
1          1.7             set 2         0.5            set b

另类

或者，使用循环和 concat：

out = pd.concat([a[[col]].merge(b.loc[b['variable'].eq(col),
                                      ['variable_value', 'value']]
                                 .rename(columns={'variable_value': col,
                                                  'value': f'{col} value'
                                                 }),
                               how='left'
                               )
                 for col in a], axis=1)

输出：

  have lunch  have lunch value have dinner  have dinner value
0      set a               0.2       set 1                1.5
1      set b               0.5       set 2                1.7

尝试一下：

lunch = b[b['variable']=='have lunch'][['variable_value',   'value']]
dinner = b[b['variable']=='have dinner'][['variable_value', 'value']]

c = a.merge(lunch ,left_on='have lunch', right_on='variable_value', how='left').merge(
dinner,left_on='have dinner', right_on='variable_value', how='left')[
['have lunch', 'variable_value_x', 'have dinner', 'variable_value_y']].set_axis(
['have lunch', 'have lunch value', 'have dinner', 'have dinner value'], axis=1)
print(c)

输出：

   have lunch   have lunch value    have dinner have dinnervalue
0   set a          set a               set 1        set 1
1   set b          set b               set2         set2