2

我有一个数据框（df）（最初来自一个 excel 文件），前 9 行是这样的：

      Control      Recd_Date/Due_Date                Action        Signature/Requester
0     2000-1703   2000-01-31 00:00:00           OC/OER/OPA/PMS/                 M WEBB
1           NaN   2000-02-29 00:00:00                       NaN              DATA CORP
2     2000-1776   2000-01-02 00:00:00            OC/ORA/OE/DCP/                  G KAN
3           NaN   2000-01-03 00:00:00           OC/ORA/ORO/PNC/              PALM POST
4           NaN                   NaN  FDA/OGROP/ORA/SE-FO/FLA-                    NaN
5           NaN                   NaN                DO/FLA-CB/                    NaN
6     2000-1983   2000-02-02 00:00:00  FDA/OGROP/ORA/CE-FO/CHI-                 M EGAN
7           NaN   2000-02-03 00:00:00                DO/CHI-CB/   BERNSTEIN LIEBHARD &
8           NaN                   NaN                       NaN             LONDON LLP
  • 类型(df['Control'][1])=float;
  • 类型(df['Recd_Date/Due_Date'][1])=datetime.datetime;
  • 类型（df['Action_Office'][1]）=浮动；
  • 类型（df['签名/请求者'][1]）=unicode

我想将此数据框（例如前 9 行）转换为：

      Control            Recd_Date/Due_Date                           Action                                                            Signature/Requester
0     2000-1703   2000-01-31 00:00:00,2000-02-29 00:00:00           OC/OER/OPA/PMS/                                                      M WEBB,DATA CORP
1     2000-1776   2000-01-02 00:00:00,2000-01-03 00:00:00           OC/ORA/OE/DCP/OC/ORA/ORO/PNC/FDA/OGROP/ORA/SE-FO/FLA-DO/FLA-CB/      G KAN,PALM POST
2     2000-1983   2000-02-02 00:00:00,2000-02-03 00:00:00           FDA/OGROP/ORA/CE-FO/CHI-DO/CHI-CB/                                   M EGAN,BERNSTEIN LIEBHARD & LONDON LLP

所以基本上：

  • 每次 pd.isnull(row['Control']) （这应该是唯一的 if 条件）为真时，然后将此行与前一行（其 'control' 值不为空）合并。
  • 对于'Recd_Date/Due_Date'和'Signature/Requester'，在每两个值之间添加','（或'/'）（来自两个合并的行）（例如'2000-01-31 00:00:00,2000- 02-29 00:00:00'和'G KAN，掌上邮报'）
  • 对于“Action”，只需合并它们而不添加任何标点符号（例如 FDA/OGROP/ORA/CE-FO/CHI-DO/CHI-CB/）

任何人都可以帮我吗？这是我试图让它工作的代码：

for i, row in df.iterrows():
    if pd.isnull(df.ix[i]['Control_#']):
       df.ix[i-1]['Recd_Date/Due_Date'] = str(df.ix[i-1]['Recd_Date/Due_Date'])+'/'+str(df.ix[i]['Recd_Date/Due_Date'])
       df.ix[i-1]['Subject'] = str(df.ix[i-1]['Subject'])+' '+str(df.ix[i]['Subject'])
       if str(df.ix[i-1]['Action_Office'])[-1] == '-':
           df.ix[i-1]['Action_Office'] = str(df.ix[i-1]['Action_Office'])+str(df.ix[i]['Action_Office'])
       else:
           df.ix[i-1]['Action_Office'] = str(df.ix[i-1]['Action_Office'])+','+str(df.ix[i]['Action_Office'])
       if pd.isnull(df.ix[i-1]['Signature/Requester']):
           df.ix[i-1]['Signature/Requester'] = str(df.ix[i-1]['Signature/Requester'])+str(df.ix[i]['Signature/Requester'])
       elif str(df.ix[i-1]['Signature/Requester'])[-1] == '&':
           df.ix[i-1]['Signature/Requester'] = str(df.ix[i-1]['Signature/Requester'])+' '+str(df.ix[i]['Signature/Requester'])
       else:
           df.ix[i-1]['Signature/Requester'] = str(df.ix[i-1]['Signature/Requester'])+','+str(df.ix[i]['Signature/Requester'])
       df.drop(df.index[i])

为什么 drop() 不起作用？我正在尝试删除当前行（如果其 ['Control_#'] 为空），因此可以将下一行（其 ['Control_#'] 为空）添加到上一行（其 ['Control_#'] 为NOT null) 迭代..

非常感激！！

4

1 回答 1

6

我认为您需要将行分组在一起，然后将列值连接起来。棘手的部分是找到一种方法以您想要的方式将行组合在一起。这是我的解决方案...

1) 将行分组在一起：静态变量

由于您的组取决于行中的序列，因此我在方法中使用静态变量将每一行标记为特定组

def rolling_group(val):
    if pd.notnull(val): rolling_group.group +=1 #pd.notnull is signal to switch group
    return rolling_group.group
rolling_group.group = 0 #static variable

此方法沿 Control 系列应用以将索引分类为组，然后用于拆分数据框以允许您合并行

#groups = df.groupby(df['Control'].apply(rolling_group),as_index=False)

这确实是唯一棘手的部分，您可以通过对每个组应用一个函数来合并行，从而为您提供所需的输出

完整的解决方案代码

def rolling_group(val):
    if pd.notnull(val): rolling_group.group +=1 #pd.notnull is signal to switch group
    return rolling_group.group
rolling_group.group = 0 #static variable

def joinFunc(g,column):
    col =g[column]
    joiner = "/" if column == "Action" else ","
    s = joiner.join([str(each) for each in col if pd.notnull(each)])
    s = re.sub("(?<=&)"+joiner," ",s) #joiner = " "
    s = re.sub("(?<=-)"+joiner,"",s) #joiner = ""
    s = re.sub(joiner*2,joiner,s)    #fixes double joiner condition
    return s

#edit above - str(each) - 转换为字符串...编辑上面的正则表达式以清除连接字符串连接

if __name__ == "__main__":
    df = """      Control      Recd_Date/Due_Date                Action        Signature/Requester
0     2000-1703   2000-01-31 00:00:00           OC/OER/OPA/PMS/                 M WEBB
1           NaN   2000-02-29 00:00:00                       NaN              DATA CORP
2     2000-1776   2000-01-02 00:00:00            OC/ORA/OE/DCP/                  G KAN
3           NaN   2000-01-03 00:00:00           OC/ORA/ORO/PNC/              PALM POST
4           NaN                   NaN  FDA/OGROP/ORA/SE-FO/FLA-                    NaN
5           NaN                   NaN                DO/FLA-CB/                    NaN
6     2000-1983   2000-02-02 00:00:00  FDA/OGROP/ORA/CE-FO/CHI-                 M EGAN
7           NaN   2000-02-03 00:00:00                DO/CHI-CB/   BERNSTEIN LIEBHARD &
8           NaN                   NaN                       NaN             LONDON LLP"""
    df =  pd.read_csv(StringIO.StringIO(df),sep = "\s\s+",engine='python')

    groups = df.groupby(df['Control'].apply(rolling_group),as_index=False)
    groupFunct = lambda g: pd.Series([joinFunc(g,col) for col in g.columns],index=g.columns)
    print groups.apply(groupFunct)

输出

     Control                       Recd_Date/Due_Date  \
0  2000-1703  2000-01-31 00:00:00,2000-02-29 00:00:00   
1  2000-1776  2000-01-02 00:00:00,2000-01-03 00:00:00   
2  2000-1983  2000-02-02 00:00:00,2000-02-03 00:00:00   

                                              Action  \
0                                    OC/OER/OPA/PMS/   
1  OC/ORA/OE/DCP/OC/ORA/ORO/PNC/FDA/OGROP/ORA/SE-...   
2                 FDA/OGROP/ORA/CE-FO/CHI-DO/CHI-CB/   

                      Signature/Requester  
0                        M WEBB,DATA CORP  
1                         G KAN,PALM POST  
2  M EGAN,BERNSTEIN LIEBHARD & LONDON LLP
于 2014-07-28T15:53:54.877 回答