Question

我需要在数据框中的列之间映射值。我的数据框中的值类型为list。这是我的数据框的具体示例：

    date      tablenameFrom  tablenameJoin                   attribute
 1 01-03-2019 [item]         []                              [itemID, itemName]
 2 02-03-2019 [order as o]   [customer as c, payment as p]   [o.oderID, p.paymentID,c.firstName]
 3 03-03-2019 [customer c]   [order o]                      [c.customerID, o.orderID]

这是我尝试过的：How to match the value of the column between columns in a dataframe, if there is an empty column。但是解决方案无法处理数据框的第一行，即该行具有多个属性名称。

这是我尝试过的：

import numpy as np
def unnesting(df, explode):
    idx = df.index.repeat(df[explode[0]].str.len())
    df1 = pd.concat([pd.DataFrame({x: np.concatenate(df[x].values)}) for x 
    in explode], axis=1)
    df1.index = idx
    return df1.join(df.drop(explode, 1), how='left')

df['tablename']=df['tablenameFrom']+df['tablenameJoin']
yourdf=unnesting(df[['date','tablename','attribute']], ['tablename','attribute'])
yourdf['tablename']=yourdf['tablename'].str.split().str[0]# fix the output
yourdf['attribute']=yourdf['attribute'].str.split(r'[''|.]').str[-1]
yourdf

我收到错误消息：ValueError: Length mismatch: Expected axis has 561 elements, new values have 412 elements

我需要将属性映射到表名。我的预期结果：

   date         tablename   attributename
 1 01-03-2019   item        itemID
 2 01-03-2019   item        itemName
 3 02-03-2019   order       orderID
 4 30-03-2019   customer    firstName
 5 30-03-2019   payment     paymentID
 6 31-03-2019   customer    customerID
 7 31-03-2019   order       orderID

我想知道是否有人可以给我一些解决方案。非常感谢。

Answer 1

将zip_longest与None一起用于缺失值，最后每组分别按ffill和bfill替换它们：

d = {'date': ['29-03-2019', '30-03-2019', '31-03-2019'],
     'tablenameFrom': [['customer'], ['order as o'], ['customer']], 
     'tablenameJoin': [[], ['customer as c', 'payment as p'], ['order']], 
     'attribute': [['customerID', 'another'], ['o.oderID', 'p.paymentID', 'c.firstName'],
                   ['customerID', 'orderID']]}
df = pd.DataFrame(d, index=[1,2,3])
print (df)
         date tablenameFrom                  tablenameJoin  \
1  29-03-2019    [customer]                             []   
2  30-03-2019  [order as o]  [customer as c, payment as p]   
3  31-03-2019    [customer]                        [order]   

                              attribute  
1                 [customerID, another]  
2  [o.oderID, p.paymentID, c.firstName]  
3                 [customerID, orderID]

from  itertools import zip_longest

x = df['attribute']
y = df['tablenameFrom'] + df['tablenameJoin']
a = [(m, l, k) for m, (i, j) 
               in enumerate(zip_longest(x, y, fillvalue=[None])) 
               for k, l 
               in zip_longest(i, j, fillvalue=None)]
#print (a)

df1 = pd.DataFrame(a, columns=['date','tablename','attributename'])
df1['date'] = df['date'].values[df1['date'].values]
df1['tablename'] = df1.groupby('date')['tablename'].apply(lambda x: x.ffill().bfill())

使用map可获得正确的匹配值：

df2 = df1['tablename'].str.split(' as ', expand=True)
s = df2.dropna().drop_duplicates(1).set_index(1)[0]
print (s)
1
o       order
c    customer
p     payment
Name: 0, dtype: object

df1['attributename'] = df2[1].map(s).fillna(df1['attributename'])
df1['tablename'] = df2[0]
print (df1)
         date tablename attributename
0  29-03-2019  customer    customerID
1  29-03-2019  customer       another
2  30-03-2019     order         order
3  30-03-2019  customer      customer
4  30-03-2019   payment       payment
5  31-03-2019  customer    customerID
6  31-03-2019     order       orderID

Answer 2

d = {'date': ['29-03-2019', '30-03-2019', '31-03-2019'],
     'tablenameFrom': [['item'], ['order as o'], ['customer']], 
     'tablenameJoin': [[], ['customer as c', 'payment as p'], ['order']], 
     'attribute': [['customerID', 'another'], ['o.oderID', 'p.paymentID', 'c.firstName'],
                   ['customerID', 'orderID']]}

d = pd.DataFrame(d, index=[1,2,3])

>>> d
date    tablenameFrom   tablenameJoin   attribute
1   29-03-2019  [item]  []  [customerID, another]
2   30-03-2019  [order as o]    [customer as c, payment as p]   [o.oderID, p.paymentID, c.firstName]
3   31-03-2019  [customer]  [order] [customerID, orderID]

df_list = []
cols = d.columns
for col in d.columns:
    df_ = d[col].apply(pd.Series).stack().reset_index().set_index('level_0')
    df_ = df_.drop(columns='level_1')
    df_list.append(df_)

nw_df = df_list[0]
for df_ in df_list[1:]:
    nw_df = pd.merge(nw_df,df_,on='level_0',how='outer')

nw_df.columns = cols

>>> nw_df

如何将数据框中的列之间的值与空列匹配

2 个答案: