如何构建正则表达式来查找以\\ n和字母开头以及以数字或单词结尾的单词？

Question

Here's an example of string, spacing after digit could be different.

product_list = 'Buy:\n Milk \nYoughurt 4 \nBread  \nSausages 4     \nBanana '

I want to build a regexp with the following output:

import re

re.findall(r'some pattern', product_list)
['Milk', 'Youghurt 4', 'Bread', 'Sausages 4', 'Banana']

This is what I thought it should look like. However, it returns empty list:

re.findall(r'\n(\w+\w$))', product_list)

Answer 1

The approach of the below script is to first strip off the leading term:\\n in this case Buy:\\n . Then, we use re.findall with the following pattern to find all matches:

(.+?)\s*(?:\n|$)

This says to capture anything up until the first optional whitespace character, which is then followed by a newline, or the end of the string.

product_list = 'Buy:\n Milk \nYoughurt 4 \nBread  \nSausages 4     \nBanana '
product_list = re.sub(r'^[^\s]*\s+', '', product_list)

matches = re.findall(r'(.+?)\s*(?:\n|$)', product_list)
print(matches)

['Milk', 'Youghurt 4', 'Bread', 'Sausages 4', 'Banana']

Answer 2

I would suggest to use a non-regex (a regex seems expensive), if you can guarantee similar pattern of input:

list(map(lambda x: x.strip(), product_list.split('\n')))[1:]

Code :

product_list = 'Buy:\n Milk \nYoughurt 4 \nBread  \nSausages 4     \nBanana '

print(list(map(lambda x: x.strip(), product_list.split('\n')))[1:])
# ['Milk', 'Youghurt 4', 'Bread', 'Sausages 4', 'Banana']

Answer 3

This example can be done without a regex, split on : and then \\n

actual_list = 'Buy:\n Milk \nYoughurt 4 \nBread  \nSausages 4     \nBanana '
product_list = actual_list.split(':')[1]
processed_list = [product.strip() for product in product_list.split('\n') if product.strip() != '']
print(processed_list)
#['Milk', 'Youghurt 4', 'Bread', 'Sausages 4', 'Banana']