我不敢确定如何处理你提供的投入数据,因为这种数据不正确格式。 然而,我认为,解决这一问题有两种办法。

Input data (as correct Python)

column = [
    [ 
 "Hello" 
 ],
    [ 
 "Hello" ,  
 "Hi"
 ],
    [ 
 "Hello" ,  
 "Hi"
 ,  
 ""
 ],
    [ 
 ""
 ,  
 "Hello" ,  
 "Hi"
 ],
    [ 
 "Hello" 
 ,  
 ""
 ],
    [ 
 ""
 ,  
 "Hello" 
 ]
]

Code: First map then List Comprehension

The map removes the whitespace including the newline characters. The list comprehension then removes the empty entries from each row ("").

def stripper(text):
    return text.strip().strip( " )

for row in column:
    output = list(map(stripper, row))
    print([i for i in output if i])

Output

[ Hello ]
[ Hello ,  Hi ]
[ Hello ,  Hi ]
[ Hello ,  Hi ]
[ Hello ]
[ Hello ]

请注意,最终结果有单一报价,而不是双重报价。 让我知道,这是否是你重新做的事。

For fun

就幸运而言,我从字面上完全拿到你的输入数据,并写了一套替换文件,确切地得出了你在该问题上的产出。

Input data

column = r"""[
 "Hello" 
]
[
 "Hello", 
 "Hi"
]
[
 "Hello", 
 "Hi"
, 
 ""
]
[
 ""
, 
 "Hello", 
 "Hi"
]
[
 "Hello" 
, 
 ""
]
[
 ""
, 
 "Hello" 
]""".splitlines()

Code

for row in column:
    print(row.replace( \n " ,  " ).replace( " \n ,  " ).replace( ""\n,  ,   ).replace( , ""\n ,   ).replace( "\n ,   ))

Output

["Hello"]
["Hello", "Hi]
["Hello", "Hi]
["Hello", "Hi]
["Hello"]
["Hello"]

举例来说,如果你提供了数据框架df,并附有栏目column,我们将使用以下代码。

def remove_empty_line(row):
    updated_list = list()
    for elem in row:
        updated_list.append(elem.replace("
", "").strip())
return updated_list

df["column"] = df["column"].apply(lambda row: remove_empty_line(row))

现在,你可以核对<代码>df.head(>。

如何做到这一点?

from ast import literal_eval

import pandas as pd


# Recreating data.
column = [
     [
 "Hello" 
] ,
     [
 "Hello", 
 "Hi"
] ,
     [
 "Hello", 
 "Hi"
, 
 ""
] ,
     [
 ""
, 
 "Hello", 
 "Hi"
] ,
     [
 "Hello" 
, 
 ""
] ,
     [
 ""
, 
 "Hello" 
] ,
]
df = pd.DataFrame({"column": column})

out = df.assign(
    # We map the following operations to "column".
    column=df.column
    .map(
        # Iterate over each value -- each will become a list of literals (in this case, strings).
        lambda x: [
            # We unpack the filtered values into the list.
            *filter(
                # Check the bool value of each item produced by `literal_eval`, keeping only those that are True.
                bool, literal_eval(x)
            )
        ]
    )
)
print(out)

        column
0      [Hello]
1  [Hello, Hi]
2  [Hello, Hi]
3  [Hello, Hi]
4      [Hello]
5      [Hello]