ValueError：无法将字符串转换为浮点数：（pd.Series）

Question

我未能在下面的代码 sn-p 上执行“lambda 函数”。

我想要的目标是分别拆分列（ btts_x 和 btts_y ）以进行进一步的数学计算。

lambda 函数在第一个位置列 btts_x 上成功（参见 btts_x_1 和 btts_x_2 ）；但在追溯重新 ValueError 中显示的 btts_y 列上失败。我想我需要在 lambda 函数中传递一个 re.sub() ，但是我坚持下去，希望能得到帮助！

注意：Team_x 中的特殊字符 \n\n 和 Team_y 中的 \n；因此 re.sub() 问题

def results(frame):
    frame[['btts_x_1', 'btts_x_2']] = frame['btts_x'].apply(lambda x: x.split('\n\n')).apply(pd.Series).astype(float)
    frame[['btts_y_1', 'btts_y_2']] = frame['btts_y'].apply(lambda x: x.split('\n\n')).apply(pd.Series).astype(float)



                             Teams_x          btts_x    Teams_y                             btts_y  btts_x_1    btts_x_2
0   Leicester City vs Manchester United    1.55\n\n2.40 Leicester City vs Manchester United 1.50\n2.40  1.55    2.40
1   Aston Villa vs Crystal Palace          1.68\n\n2.14 Aston Villa vs Crystal Palace       1.60\n2.20  1.68    2.14
2   Fulham vs Southampton                  1.72\n\n2.08 Fulham FC vs Southampton FC         1.70\n2.00  1.72    2.08
3   Arsenal vs Chelsea                     1.79\n\n1.98 Arsenal FC vs Chelsea FC            1.70\n2.00  1.79    1.98
...

TraceBack....
4     frame[['btts_x_1', 'btts_x_2']] = frame['btts_x'].apply(lambda x: x.split('\n\n')).apply(pd.Series).astype(float)
----> 5     frame[['btts_y_1', 'btts_y_2']] = frame['btts_y'].apply(lambda x: x.split('\n\n')).apply(pd.Series).astype(float)
      6 
      7 #     frame[['btts_x_1', 'btts_x_2']] = frame['btts_x'].apply(lambda x: x.split('\n')).apply(pd.Series).astype(float)

D:\Anaconda\envs\web_scraping\lib\site-packages\pandas\core\generic.py in astype(self, dtype, copy, errors)
   5546         else:
   5547             # else, only a single dtype is given
-> 5548             new_data = self._mgr.astype(dtype=dtype, copy=copy, errors=errors,)
   5549             return self._constructor(new_data).__finalize__(self, method="astype")
   5550 

D:\Anaconda\envs\web_scraping\lib\site-packages\pandas\core\internals\managers.py in astype(self, dtype, copy, errors)
    602         self, dtype, copy: bool = False, errors: str = "raise"
    603     ) -> "BlockManager":
--> 604         return self.apply("astype", dtype=dtype, copy=copy, errors=errors)
    605 
    606     def convert(

D:\Anaconda\envs\web_scraping\lib\site-packages\pandas\core\internals\managers.py in apply(self, f, align_keys, **kwargs)
    407                 applied = b.apply(f, **kwargs)
    408             else:
--> 409                 applied = getattr(b, f)(**kwargs)
    410             result_blocks = _extend_blocks(applied, result_blocks)
    411 

D:\Anaconda\envs\web_scraping\lib\site-packages\pandas\core\internals\blocks.py in astype(self, dtype, copy, errors)
    593             vals1d = values.ravel()
    594             try:
--> 595                 values = astype_nansafe(vals1d, dtype, copy=True)
    596             except (ValueError, TypeError):
    597                 # e.g. astype_nansafe can fail on object-dtype of strings

D:\Anaconda\envs\web_scraping\lib\site-packages\pandas\core\dtypes\cast.py in astype_nansafe(arr, dtype, copy, skipna)
    995     if copy or is_object_dtype(arr) or is_object_dtype(dtype):
    996         # Explicit copy, or required since NumPy can't view from / to object.
--> 997         return arr.astype(dtype, copy=True)
    998 
    999     return arr.view(dtype)

ValueError: could not convert string to float: '1.50\n2.40'

抱歉拖了这么久。

Answer 1

让我们尝试按原始字符串 '\n\n' 拆分，展开并重命名列

 df['x']=df.btts_x+ " "+df.btts_y
df1=df.join(df['x'].str.split(r'\n\n|\n|\s|\n', expand=True)).rename(columns={0:'btts_x_1',1:'btts_x_2',2:'btts_y_1',3:'btts_y_2'}).drop(columns=['x'])
print(df1)

                            Teams_x        btts_x  \
0  Leicester City vs Manchester United  1.55\n\n2.40   
1        Aston Villa vs Crystal Palace  1.68\n\n2.14   
2                Fulham vs Southampton  1.72\n\n2.08   
3                   Arsenal vs Chelsea  1.79\n\n1.98   

                               Teams_y      btts_y btts_x_1 btts_x_2 btts_y_1  \
0  Leicester City vs Manchester United  1.50\n2.40     1.55     2.40     1.50   
1        Aston Villa vs Crystal Palace  1.60\n2.20     1.68     2.14     1.60   
2          Fulham FC vs Southampton FC  1.70\n2.00     1.72     2.08     1.70   
3             Arsenal FC vs Chelsea FC  1.70\n2.00     1.79     1.98     1.70   

  btts_y_2  
0     2.40  
1     2.20  
2     2.00  
3     2.00