Question

嗨，我有2个非常大的CSV文件

df1

x   y  z      keywords
a   b  c  [apple,iphone,watch,newdevice]
e   w  q   NaN
w   r  t  [pixel,google]
s   t  q  [india,computer]
d   j  o  [google,apple]

df2

name       stockcode   
apple.inc      appl   
lg.inc          weew   
htc.inc         rrr    
google.com     ggle

现在我需要检查df1中的m个值是否与df2中的新值匹配，我需要将新值的详细信息组合到df1中，否则我们需要填充空值

我需要使用python，请帮助我

样本输出

x   y  z      keywords                        stockcode    
a   b  c  [apple,iphone,watch,newdevice]       aapl    
e   w  q   NaN                                 null    
w   r  t  [pixel,google,]                      ggle    
s   t  q  [india,computer]                     null    
d   j  o  [google,apple]                      aapl,ggle

我已经编写了这段代码，但是它只是比较一个关键字并给出一个股票代码，如果我们有2个在df2中匹配的关键字，我就需要2个股票代码

df1['stockcode'] = np.nan
#mapping data 
for indexKW,valueKW in df1.keyword.iteritems():
    for innerVal in valueKW.split():
        for indexName, valueName in df2['Name'].iteritems():
            for outerVal in valueName.split():
                if outerVal.lower() == innerVal.lower():
                    df1['stockcode'].loc[indexKW] = df2.Identifier.loc[indexName]

上述程序的输出

x   y  z      keywords                        stockcode    
a   b  c  [apple,iphone,watch,newdevice]       aapl    
e   w  q   NaN                                 null    
w   r  t  [pixel,google,]                      ggle    
s   t  q  [india,computer]                     null    
d   j  o  [google,apple]                       ggle

对于最后一行，我有2个在df2中匹配的关键字，但是我只得到一个与关键字google匹配的股票代码，我也需要获取苹果的股票代码，如示例输出所示。

示例输出：-

x   y  z      keywords                        stockcode    
a   b  c  [apple,iphone,watch,newdevice]       aapl    
e   w  q   NaN                                 null    
w   r  t  [pixel,google,]                      ggle    
s   t  q  [india,computer]                     null    
d   j  o  [google,apple]                      aapl,ggle

请帮助我

Answer 1

您可以将df2转换为查找字典，然后将其映射到df1;）

import numpy as np
import pandas as pd


data1 = {'x':'a,e,w'.split(','),
         'keywords':['apple,iphone,watch,newdevice'.split(','),
                    np.nan,
                    'pixel,google'.split(',')]}
data2 = {'name':'apple lg htc google'.split(),
        'stockcode':'appl weew rrr ggle'.split()}

df1 = pd.DataFrame(data1)
df2 = pd.DataFrame(data2)

mapper = df2.set_index('name').to_dict()['stockcode']

df1['stockcode'] = df1['keywords'].replace(np.nan,'').apply(lambda x : [mapper[i] for i in x if (i and i in mapper.keys())])
df1['stockcode'] = df1['stockcode'].apply(lambda x: x[0] if x else np.nan)

Answer 2

您可以将apply和map和join配合使用：

df2.set_index('name',inplace=True)
df1.apply(lambda x: pd.Series(x['keywords']).map(df2['stockcode']).dropna().values,1)

0          [appl]
1              []
2          [ggle]
3              []
4    [ggle, appl]
dtype: object

或者：

df1.apply(lambda x: ','.join(pd.Series(x['keywords']).map(df2['stockcode']).dropna()),1)

0         appl
1             
2         ggle
3             
4    ggle,appl
dtype: object

或者：

df1.apply(lambda x: ','.join(pd.Series(x['keywords']).map(df2['stockcode']).dropna()),1)\
                       .replace('','null')
0         appl
1         null
2         ggle
3         null
4    ggle,appl
dtype: object

df1['stockcode'] = df1.apply(lambda x: ','.join(pd.Series(x['keywords'])\
                                          .map(df2['stockcode']).dropna()),1)\
                             .replace('','null')
print(df1)
   x  y  z                           keywords  stockcode
0  a  b  c  [apple, iphone, watch, newdevice]       appl
1  e  w  q                                NaN       null
2  w  r  t                    [pixel, google]       ggle
3  s  t  q                  [india, computer]       null
4  d  j  o                    [google, apple]  ggle,appl

如果需要匹配，则需要匹配2个不同的熊猫数据框的2列，我们需要附加新数据

2 个答案: