以逗号分隔字符串,但忽略双引号内的逗号

现有数据格式如下，需要以逗号分隔字符串,但忽略双引号内的逗号，即 "Anemia, Hemolytic" 内的逗号不进行分割

a = '11-BETA-HSD3,100174880,"Anemia, Hemolytic",MESH:D000743,,"Water Pollutants, Chemical",4.49,,22425172'
需要的效果如下：
['11-BETA-HSD3', '100174880', '"Anemia, Hemolytic"', 'MESH:D000743', '', '"Water Pollutants, Chemical"', '4.49', '', '22425172']
目前想到两种方法，如下：
使用正则re.split进行字符串分割
使用csv模块的reader进行分割
使用正则re.split进行字符串分割
import re
a = '11-BETA-HSD3,100174880,"Anemia, Hemolytic",MESH:D000743,,"Water Pollutants, Chemical",4.49,,22425172'
b = re.split(r',\s*(?![^"]*\"\,)', a)
print(len(b))
print(b)
['11-BETA-HSD3', '100174880', '"Anemia, Hemolytic"', 'MESH:D000743', '', '"Water Pollutants, Chemical"', '4.49', '', '22425172']
r',\s*(?![^"]*\"\,)'
r表示原始字符串,,表示以,分割
\s表示匹配空白字符，*表示匹配前一个字符0次或者无数次,\s*即表示如果是,,会匹配为'',这样数据分割出来的列表长度一致，下标对应的字符串不会混乱
(?![^"]*\"\,)表示匹配后面有"asg",的逗号，但是不匹配后面是asg",的逗号，重点是这个，分为两部分解释
[^"]*\"\,在原始字符串模式下，表示匹配asdgdg",但是不匹配"asdgdg",
1(?!2)表示匹配13中的1,但是不匹配12中的1
Positive and Negative Lookahead

(pattern),(?:pattern),(?=pattern),(?!pattern)
import re
a = '11-BETA-HSD3,100174880,"Anemia, Hemolytic",MESH:D000743,,"Water Pollutants, Chemical",4.49,,"22425172,test"'
b = re.split(r',\s*(?![^"]*\"\,)', a)
print(len(b))
print(b)
['11-BETA-HSD3', '100174880', '"Anemia, Hemolytic"', 'MESH:D000743', '', '"Water Pollutants, Chemical"', '4.49', '', '"22425172', 'test"']
目前的解决办法是，在字符串后面加上,就可以让字符串尾部双引号内的逗号不被分割，但是这样分割出来的列表长度+1，列表最后一个元素为''
import re
a = '11-BETA-HSD3,100174880,"Anemia, Hemolytic",MESH:D000743,,"Water Pollutants, Chemical",4.49,,"22425172,test"'
a = a + ','
b = re.split(r',\s*(?![^"]*\"\,)', a)
print(len(b))
print(b)
['11-BETA-HSD3', '100174880', '"Anemia, Hemolytic"', 'MESH:D000743', '', '"Water Pollutants, Chemical"', '4.49', '', '"22425172,test"', '']
使用csv模块的reader进行分割
import csv
a = '11-BETA-HSD3,100174880,"Anemia, Hemolytic",MESH:D000743,,"Water Pollutants, Chemical",4.49,,22425172'
b = [a]
reader = csv.reader(b, delimiter=',')
for row in reader:
    print(len(row))
    print(row)