Using ^ to match beginning of line in Python regex

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

should work

Use re.search with re.M:

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 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

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'