如何用Python re.sub()正则仅去除CSV指定行首尾空格/制表符

需求说明 #

需要编写正则表达式，通过Python的re.sub()方法，仅去除CSV文件中用户指定行内各字段的首尾空格/制表符，其余行保持完全不变。

示例数据 #

原始CSV文本：

"Column1 "," Column2"," Column3 ","Column4","Column5"
"Record11 "," Record12"," Record13 ","Record14","Record15"
"Record21 "," Record22"," Record23 ","Record24","Record25"
"Record31 "," Record32"," Record33 ","Record34","Record35"

预期输出 #

仅目标行字段的首尾空白被清理，其余行保持原样：

"Column1","Column2","Column3","Column4","Column5"
"Record11 "," Record12"," Record13 ","Record14","Record15"
"Record21 "," Record22"," Record23 ","Record24","Record25"
"Record31 "," Record32"," Record33 ","Record34","Record35"

解决方案 #

你之前的正则会全局匹配所有行的字段，问题出在没有锁定目标行范围。下面是精准匹配指定行的实现：

代码实现

import re

csv_text = '''"Column1 "," Column2"," Column3 ","Column4","Column5"
"Record11 "," Record12"," Record13 ","Record14","Record15"
"Record21 "," Record22"," Record23 ","Record24","Record25"
"Record31 "," Record32"," Record33 ","Record34","Record35"
'''

# 设定要处理的目标行号（从1开始计数）
target_line = 1

# 构建正则：精准锁定目标行，匹配该行内带首尾空白的字段
pattern = re.compile(
    rf'(?<=^(?:.*\n){{{target_line-1}}})"\s*(.*?)\s*"(?=,|$)',
    flags=re.DOTALL | re.MULTILINE
)

# 执行替换
cleaned_csv = pattern.sub(r'"\1"', csv_text)

print(cleaned_csv)

正则逻辑拆解

• (?<=^(?:.*\n){{{target_line-1}}})：正向后顾断言，用来定位目标行的起始位置。比如要处理第1行时，这个断言就是(?<=^)（匹配行首）；处理第3行时，就是(?<=^.*\n.*\n)（跳过前两行）。
• "\s*(.*?)\s*"：匹配字段中被引号包裹、且首尾带有空格/制表符的内容，.*?是非贪婪匹配，避免跨字段匹配。
• (?=,|$)：正向前瞻断言，确保匹配的引号后面是字段分隔符（逗号）或行尾，防止误匹配。
• re.DOTALL + re.MULTILINE：让.能匹配换行符，同时让^匹配每行的开头，保证行号定位准确。

灵活适配 #

如果需要处理其他行，比如第3行，只需要修改target_line = 3即可，正则会自动定位到目标行并清理字段空白，其他行不受影响。

内容的提问来源于stack exchange，提问作者Suraj raghu