re.findall not returning full match?

The problem you have is that if the regex that re.findall tries to match captures groups (i.e. the portions of the regex that are enclosed in parentheses), then it is the groups that are returned, rather than the matched string.

One way to solve this issue is to use non-capturing groups (prefixed with ?:).

>>> import re
>>> s = 'size=50;size=51;'
>>> re.findall('size=(?:50|51);', s)
['size=50;', 'size=51;']

If the regex that re.findall tries to match does not capture anything, it returns the whole of the matched string.

Although using character classes might be the simplest option in this particular case, non-capturing groups provide a more general solution.

When a regular expression contains parentheses, they capture their contents to groups, changing the behaviour of findall() to only return those groups. Here's the relevant section from the docs:

(...)

Matches whatever regular expression is inside the parentheses, and indicates the start and end of a group; the contents of a group can be retrieved after a match has been performed, and can be matched later in the string with the \number special sequence, described below. To match the literals '(' or ')', use \( or \), or enclose them inside a character class: [(] [)].

To avoid this behaviour, you can use a non-capturing group:

>>> print re.findall(r'size=(?:50|51);',myfile)
['size=51;', 'size=51;', 'size=51;', 'size=50;', 'size=50;', 'size=50;', 'size=50;']

Again, from the docs:

(?:...)

A non-capturing version of regular parentheses. Matches whatever regular expression is inside the parentheses, but the substring matched by the group cannot be retrieved after performing a match or referenced later in the pattern.

In some cases, the non-capturing group is not appropriate, for example with regex which detects repeated words (example from python docs)

r'(\b\w+)\s+\1'

In this situation to get whole match one can use

[groups[0] for groups in re.findall(r'((\b\w+)\s+\2)', text)]

Note that \1 has changed to \2.