使用 ^ 匹配 Python 正则表达式中的行首

Question

I'm trying to extract publication years ISI-style data from the Thomson-Reuters Web of Science. The line for "Publication Year" looks like this (at the very beginning of a line):

PY 2015

For the script I'm writing I have defined the following regex function:

import re
f = open('savedrecs.txt')
wosrecords = f.read()

def findyears():
    result = re.findall(r'PY (\d\d\d\d)', wosrecords)
    print result

findyears()

This, however, gives false positive results because the pattern may appear elsewhere in the data.

So, I want to only match the pattern at the beginning of a line. Normally I would use ^ for this purpose, but r'^PY (\d\d\d\d)' fails at matching my results. On the other hand, using \n seems to do what I want, but that might lead to further complications for me.

Answer 1

re.findall(r'^PY (\d\d\d\d)', wosrecords, flags=re.MULTILINE)

应该管用

Answer 2

You can simply add (?m) inline modifier flag to the start of the pattern:

(?m)^PY\s+(\d{4})
^^^^

Do not confuse with (?s) ! (?s) is a DOTALL inline flag that makes . match any characters including line break characters.

Alternatively, you can use re.search with re.M or re.MULTILINE option :

import re
p = re.compile(r'^PY\s+(\d{4})', re.M)
test_str = "PY123\nPY 2015\nPY 2017"
print(re.findall(p, test_str)) 

See an IDEONE demo .

EXPLANATION :

• ^ - Start of a line (due to re.M )
• PY - Literal PY
• \s+ - 1 or more whitespace
• (\d{4}) - Capture group holding 4 digits

Answer 3

In this particular case there is no need to use regular expressions, because the searched string is always 'PY' and is expected to be at the beginning of the line, so one can use string.find for this job. The find function returns the position the substring is found in the given string or line, so if it is found at the start of the string the returned value is 0 (-1 if not found at all), ie.:

In [12]: 'PY 2015'.find('PY')
Out[12]: 0

In [13]: ' PY 2015'.find('PY')
Out[13]: 1

Perhaps it could be a good idea to strip the white spaces, ie.:

In [14]: '  PY 2015'.find('PY')
Out[14]: 2

In [15]: '  PY 2015'.strip().find('PY')
Out[15]: 0

And next if only the year is of interest it can be extracted with split, ie.:

In [16]: '  PY 2015'.strip().split()[1]
Out[16]: '2015'