如何使用比较其他两列的值创建一个新列？

Question

我正在开展一个项目，该项目将对拥有计算机帐户的员工进行审计。我想打印一个包含两个新列的数据框。这与 Comparing Columns in Dataframes 问题不同，因为我正在使用字符串。我还需要做一些模糊逻辑，但那是进一步的。

我收到的数据在 Excel 表格中。它来自两个我无法控制的来源，因此我将它们格式化为 [First Name, Last Name] 并将它们打印到控制台以确保我正在使用的数据是正确的。我将 .xls 转换为 .csv 文件，格式化信息，并能够在具有两列的单个数据框中输出两个名称列表，但无法将我想要的值放在最后两列中。我使用了查询（返回真/假，而不是名称）、差异和正则表达式。我认为我只是错误地使用了这些工具。

    import pandas as pd

    nd = {'col1': ["Abraham Hansen","Demetrius McMahon","Hilary 
          Emerson","Amelia H. Hayden","Abraham Oliver"],
          'col2': ["Abraham Hansen","Abe Oliver","Hillary Emerson","DJ 
          McMahon","Amelia H. Hayden"]}
    info = pd.DataFrame(data=nd)

    for row in info:
    if info.col1.value not in info.col2:
        info["Need Account"] = info.col1.value

    if info.col2.value not in info.col1:
        info["Delete Account"] = info.col2.value

    print(info)

我想要一个包含 2 列的新数据框：需要帐户和删除帐户，并根据数据框中的其他列填写适当的值。在这种情况下，我收到“系列”没有属性“值”的错误。 这是我的预期输出示例：

    df_out: 
    Need Account       Delete Account
    Demetrius McMahon  Abe Oliver
    Abraham Oliver     Hillary Emerson
    Hilary Emerson     DJ McMahon

从这个列表中，我可以查看出现了谁的昵称，然后从那里减少列表。

Answer 1

您想使用isin 和np.where 有条件地分配新值：

info['Need Account'] = np.where(~info['col1'].isin(info['col2']), info['col1'], np.NaN)
info['Delete Account'] = np.where(~info['col2'].isin(info['col1']), info['col2'], np.NaN)

                col1              col2       Need Account   Delete Account
0     Abraham Hansen    Abraham Hansen                NaN              NaN
1  Demetrius McMahon        Abe Oliver  Demetrius McMahon       Abe Oliver
2     Hilary Emerson   Hillary Emerson     Hilary Emerson  Hillary Emerson
3   Amelia H. Hayden        DJ McMahon                NaN       DJ McMahon
4     Abraham Oliver  Amelia H. Hayden     Abraham Oliver              NaN

---

或者如果你想要一个新的数据框，就像你在问题中所说的那样：

need = np.where(~info['col1'].isin(info['col2']), info['col1'], np.NaN)
delete = np.where(~info['col2'].isin(info['col1']), info['col2'], np.NaN)

newdf = pd.DataFrame({'Need Account':need,
                      'Delete Account':delete})

        Need Account   Delete Account
0                NaN              NaN
1  Demetrius McMahon       Abe Oliver
2     Hilary Emerson  Hillary Emerson
3                NaN       DJ McMahon
4     Abraham Oliver              NaN

Answer 2

我没有看到您的预期输出，而是在阅读您在代码中尝试的内容。让我知道这是否是您要找的？

nd = {'col1': ["Abraham Hansen","Demetrius McMahon","Hilary Emerson","Amelia H. Hayden","Abraham Oliver"],
      'col2': ["Abraham Hansen","Abe Oliver","Hillary Emerson","DJ McMahon","Amelia H. Hayden"], 
      'Need Account':"", 
      'Delete Account':""
     }
info = pd.DataFrame(data=nd)

print(info)

               col1              col2 Need Account Delete Account
0     Abraham Hansen    Abraham Hansen                            
1  Demetrius McMahon        Abe Oliver                            
2     Hilary Emerson   Hillary Emerson                            
3   Amelia H. Hayden        DJ McMahon                            
4     Abraham Oliver  Amelia H. Hayden    

不要使用循环，使用向量...

info.loc[info['col1'] != info['col2'], 'Need Account'] = info['col1']
info.loc[info['col2'] != info['col1'], 'Delete Account'] = info['col2']

print(info)

               col1              col2       Need Account    Delete Account
0     Abraham Hansen    Abraham Hansen                                     
1  Demetrius McMahon        Abe Oliver  Demetrius McMahon        Abe Oliver
2     Hilary Emerson   Hillary Emerson     Hilary Emerson   Hillary Emerson
3   Amelia H. Hayden        DJ McMahon   Amelia H. Hayden        DJ McMahon
4     Abraham Oliver  Amelia H. Hayden     Abraham Oliver  Amelia H. Hayden

Answer 3

IIUC，您的输入数据框中似乎没有太多“结构”需要维护，因此您可以使用集合直接比较组中的成员资格。

nd = {'col1': ["Abraham Hansen","Demetrius McMahon","Hilary Emerson","Amelia H. Hayden","Abraham Oliver"],
      'col2': ["Abraham Hansen","Abe Oliver","Hillary Emerson","DJ McMahon","Amelia H. Hayden"]}
df = pd.DataFrame(data=nd)

col1 = set(df['col1'])
col2 = set(df['col2'])

need = col1 - col2
delete = col2 - col1

print('need = ', need)
print('delete =  ', delete)

产量

need =  {'Hilary Emerson', 'Demetrius McMahon', 'Abraham Oliver'}
delete =   {'Hillary Emerson', 'DJ McMahon', 'Abe Oliver'}

然后你可以放入一个新的数据框：

data = {'need':list(need), 'delete':list(delete)}
new_df = pd.DataFrame.from_dict(data, orient='index').transpose()

（编辑以考虑need 和delete 长度不等的可能性。）