Python :从字符串中提取卷(mL)

Question

我有以下字符串来提取卷(只匹配毫升，而不是毫克/毫升)

test = [
"10ML", # 10
"10 ML", # 10
"10.5ML", # 10.5
"1MG/1ML", # [] not match
"1MG/10ML", # [] not match
"10MG/0.5ML", # [] not match
"   10ML and 15ML  ", # 10, 15
"LODEXA (DEXAMETHASONE) 5 MG/ML INJECTION", # []
"NSS.0.9% 1000 ML (PLASTIC BAG)", # 1000
"110 MLM HIDRASEC (RACECADOTIL)10 ML POWDER FOR 1-30 ML", # 10
]

这是我目前的模式和结果。

pattern = re.compile("(?<!\/)([0-9]*[.]*[0-9]+)\s*ML(?![\/A-z])")

for i, s in enumerate(test):
    print(test[i], '>>' , pattern.findall(s))

10ML >> ['10']
10 ML >> ['10']
10.5ML >> ['10.5']
1MG/1ML >> []
1MG/10ML >> ['0'] # Wrong []
10MG/0.5ML >> ['.5'] # Wrong []
   10ML and 15ML   >> ['10', '15']
LODEXA (DEXAMETHASONE) 5 MG/ML INJECTION >> []
NSS.0.9% 1000 ML (PLASTIC BAG) >> ['1000']
110 MLM HIDRASEC (RACECADOTIL)10 ML POWDER FOR 1-30 ML >> ['10', '30'] # Wrong ['10']

正如您所看到的，我从["1MG/10ML", "10MG/0.5ML", "110 MLM HIDRASEC (RACECADOTIL)10 ML POWDER FOR 1-30 ML"]中得到了错误的结果。应该是[[], [], ['10']。

我已经试过修正我的模式，但还是找不出答案。请帮我改正一下我的图案。谢谢!

Answer 1

您可以使用

(?<![/\d])(?<!\d[.-])(\d+(?:\.\d+)?)\s*ML\b(?!/)

见Python regex演示。

详细信息

• (?<![/\d]) -当前位置左侧不允许使用/或数字
• (?<!\d[.-]) -当前位置左侧不允许数字+ .或-
• (\d+(?:\.\d+)?) -第1组:一个或多个数字，一个.和一个或多个数字的可选序列
• \s* -零或更多空格字符
• ML\b - ML作为一个整体
• (?!/) -不允许在当前位置的右侧立即使用/。

见Python演示

import re
pattern = re.compile(r'(?<![/\d])(?<!\d[.-])(\d+(?:\.\d+)?)\s*ML\b(?!/)', re.A)
test = ["10ML", "10 ML", "10.5ML", "1MG/1ML", "1MG/10ML", "10MG/0.5ML", "   10ML and 15ML  ",
"LODEXA (DEXAMETHASONE) 5 MG/ML INJECTION", "NSS.0.9% 1000 ML (PLASTIC BAG)", 
"110 MLM HIDRASEC (RACECADOTIL)10 ML POWDER FOR 1-30 ML"]
for i, s in enumerate(test):
    print(test[i], '>>' , pattern.findall(s))

输出：

10ML >> ['10']
10 ML >> ['10']
10.5ML >> ['10.5']
1MG/1ML >> []
1MG/10ML >> []
10MG/0.5ML >> []
   10ML and 15ML   >> ['10', '15']
LODEXA (DEXAMETHASONE) 5 MG/ML INJECTION >> []
NSS.0.9% 1000 ML (PLASTIC BAG) >> ['1000']
110 MLM HIDRASEC (RACECADOTIL)10 ML POWDER FOR 1-30 ML >> ['10']

Answer 2

有关以下正则表达式组件的详细信息，请参阅这个RegExr链接。

import re

test = [
    "10ML", # 10
    "10 ML", # 10
    "10.5ML", # 10.5
    "1MG/1ML", # [] not match
    "1MG/10ML", # [] not match
    "10MG/0.5ML", # [] not match
    "   10ML and 15ML  ", # 10, 15
    "LODEXA (DEXAMETHASONE) 5 MG/ML INJECTION", # []
    "NSS.0.9% 1000 ML (PLASTIC BAG)", # 1000
    "110 MLM HIDRASEC (RACECADOTIL)10 ML POWDER FOR 1-30 ML", # 10
]

for s in test:
    re.findall(r'(?<![\-\/])(\d+(?:\.?\d+)) *ML\b', s)

输出

['10']
['10']
['10.5']
[]
[]
[]
['10', '15']
[]
['1000']
['10']

Answer 3

另一个也许更容易读懂：

(?<![/\d-])(\d+\.*\d+)\s*ML\b