在字符串中搜索逗号（，），如果存在逗号，则在python中打印逗号后的直接单词

Question

我是新的正则表达式匹配，我有如下字符串

"karthika has symptoms cold,cough her gender is female and his age is 45"

在第一个字符串匹配中，我将检查关键字“症状”，并选择关键字的下一个单词，如下所示：

regexp = re.compile("symptoms\s(\w+)")
symptoms = regexp.search(textoutput).group(1)

这将使症状值为“cold”，但我有多个症状在文本中出现，因此在第二步中，我需要在“cold”之后签入文本如果有逗号（，）出现，如果逗号出现意味着我需要使用正则表达式打印逗号i之后的值，即“咳嗽”。你知道吗

请帮我实现这个。。你知道吗

Answer 1

您可以使用正则表达式来查找'symptoms'之后的第一个单词，还可以选择使用以komma开头的更多匹配项、mabye空格和更多wordcharacters：

import re

pattern = r"symptoms\s+(\w+)(?:,\s*(\w+))*"
regex = re.compile(pattern)

t = "kathy has symptoms cold,cough her gender is female. john's symptoms  hunger, thirst."
symptoms = regex.findall(t)

print(symptoms)

输出：

[('cold', 'cough'), ('hunger', 'thirst')]

---

说明：

r"symptoms\s+(\w+)(?:,\s*(\w+))*"
# symptoms\s+                      literal symptoms followed by 1+ whitepsaces 
#            (\w+)                 followed by 1+ word-chars (first symptom) as group 1
#                 (?:,        )*   non grouping optional matches of comma+spaces
#                        (\w+)     1+ word-chars (2nd,..,n-th symptom) as group 2-n 

---

备用方式：

import re

pattern = r"symptoms\s+(\w+(?:,\s*\w+)*(?:\s+and\s+\w+)?)"

regex = re.compile(pattern)

t1 = "kathy has symptoms cold,cough,fever and noseitch her gender is female. "
t2 = "john's symptoms  hunger, thirst."
symptoms = regex.findall(t1+t2)

print(symptoms)

输出：

['cold,cough,fever and noseitch', 'hunger, thirst']

这只适用于“英式”英语-美国式的学习方式

"kathy has symptoms cold,cough,fever, and noseitch" 

只会导致cold,cough,fever, and匹配。你知道吗

您可以在','和" and "分割每个匹配项，以得到您的单一原因：

sym = [ inner.split(",") for inner in (x.replace(" and ",",") for x in symptoms)] 
print(sym)

输出：

[['cold', 'cough', 'fever', 'noseitch'], ['hunger', ' thirst']]