如何将数字从列隔离并创建3列？

Question

我正在尝试访问列，过滤其数字，然后在3列中拆分。但是我只是遇到错误。我正在尝试这个：

dsc = df["Descricao"].str.findall("\d+")
dsc

The Output:
0                   []
1       [475, 2000, 3]
2        [65, 2000, 2]
3        [51, 2000, 3]
4       [320, 2000, 3]
             ...      
2344               NaN
2345    [480, 2000, 1]
2346     [32, 2000, 6]
2347    [250, 2000, 1]
2348               NaN
Name: Descricao, Length: 2349, dtype: object

然后，我正在尝试拆分，每次得到这种错误时：

df[['Larg','comp', 'qtd']] = dsc.str.split(',',expand=True)
df.head(5)

The Error:
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
~\AppData\Local\Temp/ipykernel_15388/2481153233.py in <module>
----> 1 df[['Larg','comp', 'qtd']] = dsc.str.split(',',expand=True)
      2 df.head(5)

~\anaconda3\lib\site-packages\pandas\core\frame.py in __setitem__(self, key, value)
   3598             self._setitem_frame(key, value)
   3599         elif isinstance(key, (Series, np.ndarray, list, Index)):
-> 3600             self._setitem_array(key, value)
   3601         elif isinstance(value, DataFrame):
   3602             self._set_item_frame_value(key, value)

~\anaconda3\lib\site-packages\pandas\core\frame.py in _setitem_array(self, key, value)
   3637         else:
   3638             if isinstance(value, DataFrame):
-> 3639                 check_key_length(self.columns, key, value)
   3640                 for k1, k2 in zip(key, value.columns):
   3641                     self[k1] = value[k2]

~\anaconda3\lib\site-packages\pandas\core\indexers.py in check_key_length(columns, key, value)
    426     if columns.is_unique:
    427         if len(value.columns) != len(key):
--> 428             raise ValueError("Columns must be same length as key")
    429     else:
    430         # Missing keys in columns are represented as -1

ValueError: Columns must be same length as key

我认为与str.findall生成列表有关。 有人知道如何解决这个问题吗？ 有关信息，我的所有列都是对象。

Answer 1

您可以尝试以下操作：

dsc = pd.DataFrame(df["Descricao"].str.findall("\d+").tolist(), columns=['Larg','comp', 'qtd'])

df = pd.concat([df, dsc], axis=1)

请注意，如果任何时候都有三个以上的列，这可能不起作用（考虑到您的尝试，情况并非如此）。

该方法来自在这里。

Answer 2

一般情况下，某些输入可能没有解析为 3 个数值的字符串。

这里有一种方法可以完成问题所要求的操作，同时用 NaN 填充任何异常行的新列。如果非标准行的所需行为不同，则可以根据需要调整逻辑。

import pandas as pd
import numpy as np
df = pd.DataFrame({
    'Descricao' : ['', '475,2000,3', '65,2000,2', np.nan, 'abc,def,ghi', '1,2', '1']
})
print(f"\nInput dataframe:\n{df}")

df2 = df["Descricao"].str.findall("\d+").to_frame()
print(f"\nDataframe with lists of 3 where possible:\n{df2}")

df2["Descricao"] = df2.apply(lambda x: 
    x["Descricao"] 
        if (len(x["Descricao"]) if isinstance(x["Descricao"], list) else 0) == 3 else 
    [np.NaN]*3, 
    axis=1)
print(f"\nDataframe with lists include NaNs for incomplete data:\n{df2}")

df2[['Larg','comp', 'qtd']] = pd.DataFrame(df2["Descricao"].tolist(), columns=['Larg','comp', 'qtd'])
df2 = df2.drop(['Descricao'], axis=1)
print(f"\nResult dataframe with NaNs for incomplete inputs:\n{df2}")

示例输出：

Input dataframe:
     Descricao
0
1   475,2000,3
2    65,2000,2
3          NaN
4  abc,def,ghi
5          1,2
6            1

Dataframe with lists of 3 where possible:
        Descricao
0              []
1  [475, 2000, 3]
2   [65, 2000, 2]
3             NaN
4              []
5          [1, 2]
6             [1]

Dataframe with lists include NaNs for incomplete data:
         Descricao
0  [nan, nan, nan]
1   [475, 2000, 3]
2    [65, 2000, 2]
3  [nan, nan, nan]
4  [nan, nan, nan]
5  [nan, nan, nan]
6  [nan, nan, nan]

Result dataframe with NaNs for incomplete inputs:
  Larg  comp  qtd
0  NaN   NaN  NaN
1  475  2000    3
2   65  2000    2
3  NaN   NaN  NaN
4  NaN   NaN  NaN
5  NaN   NaN  NaN
6  NaN   NaN  NaN

Answer 3

谢谢大家！遵循@ConstantStranger解决方案，该解决方案的零件并开发了新版本。但这很容易开始。最后，我的解决方案是：

dsc = ndf['descricao'].str.findall('\d+')        #Separated only the numerical elements
# Created 3 lists for the elements
larg = []
comp = []
qtd = []
for lines in dsc:                    
    for item in enumerate(lines):
        if len(lines) != 3:       #If the length of the elements is not 3, does nothing.
            continue
        if item[0] == 0:
            larg.append(item[1])
        if item[0] == 1:
            comp.append(item[1])
        if item[0] == 2:
            qtd.append(item[1])
#Then i checked for the length of all
print(len(larg), len(comp), len(qtd))

lis = [larg, comp, qtd]
df1 = pd.DataFrame(lis).transpose()
df1.columns = ['larg', 'comp', 'qtd']
df1

输出：

    larg    comp    qtd
0   32  2000    6
1   46  1000    1
2   320 100 20
3   220 100 50
4   220 50  30
... ... ... ...
1404    50  2000    1
1405    52  200 2
1406    48  2000    1
1407    325 3000    1
1408    33  2000    2
1409 rows × 3 columns

我想，这不是大数据的理想解决方案，但目前正在起作用。我尝试使用to_frame（）尝试了.findall表达式，但是由于某种原因，每个长度都归功于零。
因此，现在我将寻找一种优化的方法。