Question

我有两个DataFrame。 df1具有多索引，df2具有标准索引。如何在df2和df2.index的每个匹配项上重复来自df1.get的值来合并它们。

示例

import pandas as pd
import numpy as np
idx1 = pd.MultiIndex.from_product([['bar', 'baz', 'foo'],['one','two']])
idx2 = ['bar', 'baz']
df1 = pd.DataFrame(np.random.randn(6, 2), index=idx1, columns=['A', 'B'])
df2 = pd.DataFrame(np.random.randn(2, 1), index=idx2, columns=['C'])

如果df1是

                A         B
bar one  0.690827 -0.627957
    two -0.080936 -1.330712
baz one  1.395178 -0.099748
    two -0.116789  0.723990
foo one  0.313067  0.853808
    two  0.409727 -0.529002

和df2是

            C
bar -0.773924
baz  0.099662

如何进行合并？

                A         B         C
bar one  0.690827 -0.627957 -0.773924
    two -0.080936 -1.330712 -0.773924
baz one  1.395178 -0.099748  0.099662
    two -0.116789  0.723990  0.099662
foo one  0.313067  0.853808  NaN
    two  0.409727 -0.529002  NaN

Answer 1

我们可以分配

    <form [formGroup]="dataGroup">
  <div class="form-group">
    <label for="email">Email address:</label>
    <input type="email" formControlName="email" class="form-control" id="email">
  </div>
  <div class="form-group">
    <label for="pwd">Password:</label>
    <input type="password" formControlName="password" class="form-control" id="pwd">
  </div>
  <div class="form-group">
    <label for="pwd">List:</label>
    <select class="form-control" formControlName="list" id="list" multiple>
    <option value="one">one</option>
    <option value="two">two</option>
    <option value="three">three</option>
    </select>
  </div>
  <div class="form-group">
    <label for="pwd">Group:</label>
   <input type="checkbox" (change)="onCheckChange($event)" name="check1" value="group1">group1
   <input type="checkbox" (change)="onCheckChange($event)"  name="check2" value="group2">group2
  </div>

  <button type="submit" (click)="getFormData(dataGroup.value)" class="btn btn-default">Submit</button>
</form>

Answer 2

您可以为索引命名并在合并中使用它，而无需像下面那样重新索引或重置索引

df1.index.set_names(["id_1", "id_2"], inplace=True)
df1.merge(df2, left_on="id_1", right_index=True, how="left")

结果

                   A          B          C
id_1 id_2                                 
bar  one    0.690827  -0.627957  -0.773924
     two   -0.080936  -1.330712  -0.773924
baz  one    1.395178  -0.099748   0.099662
     two   -0.116789   0.723990   0.099662
foo  one    0.313067   0.853808        NaN
     two    0.409727  -0.529002        NaN

Answer 3

您可以这样做：

df1 = df1.reset_index().set_index('level_0')
result = df1.merge(df2, left_on='level_0', right_on=df2.index, how='left').set_index(['level_0', 'level_1'])
print(result)

输出

                        A         B         C
level_0 level_1                              
bar     one      0.692937  0.119553  0.941637
        two     -0.876270 -1.148878  0.941637
baz     one      1.413018  0.170197 -0.250836
        two      1.996977  1.184525 -0.250836
foo     one     -2.504001  0.591182       NaN
        two     -0.535933 -1.259659       NaN

Answer 4

我认为map的索引level_0上的df1也有用

df1['C'] = df1.index.get_level_values(0).map(df2.C)

Out[71]:
                A         B         C
bar one  0.690827 -0.627957 -0.773924
    two -0.080936 -1.330712 -0.773924
baz one  1.395178 -0.099748  0.099662
    two -0.116789  0.723990  0.099662
foo one  0.313067  0.853808       NaN
    two  0.409727 -0.529002       NaN

如何在熊猫中的部分多索引上合并

示例

4 个答案: