如果目标是一个 Dataframe ，为什么不只做这个而不是两个列表呢。如果您将字符串转换为序列，您可以使用pandas.Series.str.extract()将其拆分为所需的列：

import pandas as pd

s = '''0                   cheese    100
1                   cheddar cheese    1100
2                   gorgonzola    1300
3                   smoked cheese    200'''

pd.Series(s.split('\n')).str.extract(r'.*?\s+(?P<type>.*?)\s+(?P<value>\d+)')

这给出了一个 Dataframe ：

type             value
0   cheese           100
1   cheddar cheese   1100
2   gorgonzola       1300
3   smoked cheese    200

IIUC字符串是列表的元素。您可以使用re.split在找到两个或更多空格的位置进行拆分：

import re
import pandas as pd

your_list = [
  "0                   cheese    100",
  "1                   cheddar cheese    1100",
  "2                   gorgonzola    1300",
  "3                   smoked cheese    200",
]

df = pd.DataFrame([re.split(r'\s{2,}', s)[1:] for s in your_list], columns=["type", "value"])

输出：

type value
0          cheese   100
1  cheddar cheese  1100
2      gorgonzola  1300
3   smoked cheese   200

我认为以下内容可能有用：

import pandas as pd
import re
mylist=['0 cheese 100','1 cheddar cheese 200']

numbers = '[0-9]'

list1=[i.split()[-1] for i in mylist]
list2=[re.sub(numbers, '', i).strip() for i in mylist]

your_df=pd.DataFrame({'name1':list1,'name2':list2})
your_df

我可以建议这个简单的解决方案吗

lines = [
         "1                   cheddar cheese    1100 ",
         "2                   gorgonzola    1300 ",
         "3                   smoked cheese    200",
        ]

for line in lines:
  words = line.strip().split()
  print( ' '.join( words[1:-1]), words[-1])

结果：

cheddar cheese 1100
gorgonzola 1300
smoked cheese 200

您可以通过使用切片来实现这一点：

from curses.ascii import isdigit

inList = ['0                   cheese    100', '1                   cheddar cheese    1100', '2                   gorgonzola    1300', '3                   smoked cheese    200']

cheese = []
prices = []

for i in inList:
    temp = i[:19:-1] #Cuts out first number and all empty spaces until first character and reverses the string
    counter = 0
    counter2 = 0
    for char in temp: #Temp is reversed, meaning the number e.g. '100' for 'cheese' is in front but reversed
        if char.isdigit(): 
            counter += 1
        else:   #If the character is an empty space, we know the number is over
            prices.append((temp[:counter])[::-1]) #We know where the number begins (at position 0) and ends (at position counter), we flip it and store it in prices

            cheeseWithSpace = (temp[counter:]) #Since we cut out the number, the rest has to be the cheese name with some more spaces in front
            for char in cheeseWithSpace:
                if char == ' ': #We count how many spaces are in front
                    counter2 += 1
                else:   #If we reach something other than an empty space, we know the cheese name begins.
                    cheese.append(cheeseWithSpace[counter2:][::-1]) #We know where the cheese name begins (at position counter2) cut everything else out, flip it and store it
                    break
            break

print(prices)
print(cheese)

查看代码注解以了解方法。基本上，您可以使用[：：-1]来翻转字符串，使其更容易处理。然后逐个移除每个零件。

如果您有：

text = '''0                   cheese    100
1                   cheddar cheese    1100
2                   gorgonzola    1300
3                   smoked cheese    200'''

# OR

your_list = [
 '0                   cheese    100',
 '1                   cheddar cheese    1100',
 '2                   gorgonzola    1300',
 '3                   smoked cheese    200'
]

text = '\n'.join(your_list)

正在执行：

from io import StringIO

df = pd.read_csv(StringIO(text), sep='\s\s+', names=['col1', 'col2'], engine='python')
print(df)

输出：

col1  col2
0          cheese   100
1  cheddar cheese  1100
2      gorgonzola  1300
3   smoked cheese   200

• 这是将第一个数字作为索引，但如果需要，可以使用df=df.reset_index(drop=True)重置它。